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Outside energy had to be introduced for the twin towers to collapse the way they did

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posted on Sep, 19 2011 @ 04:34 AM
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Originally posted by Darkwing01

The acceleration of the body was 9.8m/s^2 in 1970, it was 9.8m/s^2 when it the planes hit. It was 9.8m/s^2 when the supports failed.


At these points in time it was stationary. It had the force of 1 G.


Originally posted by Darkwing01
I was 9.8m/s^2 when it fell one whole story. It was 9.8m/s^2 when it impacted the lower story. It was 9.8m/s^2 when the lower story failed.



At these points in time it was dynamic. It had a force greater than 1 G.



Originally posted by Darkwing01


Do you see velocity anywhere in there? I sure don't.



Did you know that acceleration is defined as any change in velocity.

Force is MASS times ACCELERATION.

F=ma



Funny Truthers.




posted on Sep, 19 2011 @ 04:42 AM
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reply to post by -PLB-
 



You confuse deceleration with a lower acceleration.

Darkwing01 was correct in saying the plate would decelerate the brick as the brick broke the plate.


There are 3 options. The acting force is larger than average resistance: acceleration. The acting force is equal to average resistance: constant speed. The acting force is smaller than average resistance: deceleration.

There is one option: the plate provides an equal and opposite force to the brick (the magnitude being the force required to break the plate).


So what you need to show, using physics and math, is that the average resistance is equal or greater than the acting force. If it is not, there will be a net acceleration. So you need to know the resistance. This is very complex to determine. I have yet to see anyone make a realistic approximation. Obviously no truther has done so either. All we ever hear is baseless assertions based on general ignorance of physics.

I would agree that the design of the towers and (official) mode of collapse makes approximating the available resistance somewhat more complex, however in WTC7 it is a lot more straight forward.

Regardless of if you believe NIST's version of the internal structural collapse eight seconds prior, the building collapsed through its own columns (which we know should have been able to provide at least as great of a resistance force as they did when the structure was stationary). Instead they provided the resistance of wet cardboard.


edit on 19-9-2011 by DrinkYourDrug because: (no reason given)



posted on Sep, 19 2011 @ 05:13 AM
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Originally posted by DrinkYourDrug
Darkwing01 was correct in saying the plate would decelerate the brick as the brick broke the plate.


No he is not, it is one of the three possibilities. It depends on several factors such as how heavy the brick is, from which height it falls, how much resistance the plate is offering, etc.


There is one option: the plate provides an equal and opposite force to the brick (the magnitude being the force required to break the plate).


Equal and opposite force doesn't say a thing about whether there is acceleration, deceleration or constant speed. Air molecules hitting your windshield also cause an equal opposite force on your car. Still the result can be acceleration, constant speed, or deceleration.


I would agree that the design of the towers and (official) mode of collapse makes approximating the available resistance somewhat more complex, however in WTC7 it is a lot more straight forward.

Regardless of if you believe NIST's version of the internal structural collapse eight seconds prior, the building collapsed through its own columns (which we know should have been able to provide at least as great of a resistance force as they did when the structure was stationary). Instead they provided the resistance of wet cardboard.


Is good that you ""know" that. I am awaiting your publication.



posted on Sep, 19 2011 @ 05:20 AM
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Originally posted by Nonchalant
www.drjudywood.com...


I assume you have read her book, well I just finished it.

It is without doubt a literary engineering masterpiece.

I just wish everybody on this planet could read it.

Fellas, if you want to find out the truth, BUY THE BOOK AND THE TRUTH WILL SET YOU FREE.



posted on Sep, 19 2011 @ 05:39 AM
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Originally posted by Darkwing01
Force is MASS times ACCELERATION.

F=ma

Do you see kinetic energy or velocity in that calculation PLB, because I sure don't. Perhaps you should contact Newton and inform him of his error because WTC sure proved that velocity should be in there..... Or not....

The acceleration of the body was 9.8m/s^2 in 1970, it was 9.8m/s^2 when it the planes hit. It was 9.8m/s^2 when the supports failed. I was 9.8m/s^2 when it fell one whole story. It was 9.8m/s^2 when it impacted the lower story. It was 9.8m/s^2 when the lower story failed.

Do you see velocity anywhere in there? I sure don't.

The lower structure was resisting that mass at that acceleration, in fact was was capable of resisting twice to three times that. The only way that can overcome the resistance is by slowing the original mass down (decelerating it).

You can observe this in countless impacts and Verinage demolitions, to deny it is to deny reality.

____________________________________________________________________________________________

Take those concrete slabs you were talking about earlier, now take the broken slab and drop it the same distance on another concrete slab. Does it still break?

Can you see why it doesn't?


Ouch, you have the physics completely wrong again.

The force is indeed calculated by f=ma. But 'a' is not the acceleration as result of gravity. It is the 'a' as result of the support trying to stop the falling mass. So you have a mass with an initial velocity, a resistance trying to stop this mass, and a resulting 'a' determined by the distance this mass is tried to be stopped in. In case of the WTC towers there was a local momentary decelleration as result of this force. The average motion of all the mass together is still acceleration. Since this force is localized and momentarily, you do not see any reaction of it in the top section.

Try answering this: A completely rigid body with a mass of 10kg is falling with 10m/s on a completely rigid ground. What is the impact force? What 'a' did you use to calculate this? I bet you do not know the answer.

I have seen a java snippit somewhere that explains this in more detail. I will see if I can find it.

Edit: here is one: www.livephysics.com...

It also contains the formula to calculate the impact force. Notice how the velocity is in that formula. Play with it, and try to answer my question. Good luck.
edit on 19-9-2011 by -PLB- because: (no reason given)



posted on Sep, 19 2011 @ 05:49 AM
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reply to post by -PLB-
 



No he is not, it is one of the three possibilities. It depends on several factors such as how heavy the brick is, from which height it falls, how much resistance the plate is offering, etc.

The scenario is that the plate could support the brick at rest but not when dropped from sufficient height. I stand by my post above.


Equal and opposite force doesn't say a thing about whether there is acceleration, deceleration or constant speed. Air molecules hitting your windshield also cause an equal opposite force on your car. Still the result can be acceleration, constant speed, or deceleration.

The resistance provided by a fluid is not the same as inelastic collisions. This statement proves nothing.

If a plate can support a brick at rest it means the plate has a breaking strength which is greater than the weight of the brick. Therefore the brick will decelerate upon providing this equal and opposite force at the moment of impact because it has a net upwards force (its self weight downwards due to gravity countered by a greater than self weight force upwards provided by the plate).



Is good that you ""know" that. I am awaiting your publication.

Articles outlining very common and basic engineering knowledge are not usually accepted for publication.

edit on 19-9-2011 by DrinkYourDrug because: (no reason given)



posted on Sep, 19 2011 @ 06:19 AM
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reply to post by -PLB-
 




But 'a' is not the acceleration as result of gravity.


No, indeed it is not. But when both objects are at rest its value is exactly the same as that of gravity.

Because every action has an equal and opposite reaction.

The resistance offered by the lower structure is in the form of a normal force
en.wikipedia.org... and the value of that normal force is the same as the force being exerted on the stationary object (but in the opposite direction). The force exerted on the downward object is the force required to accelerate that object at 9.8m/s^2, and since the mass doesn't change the force doesn't change.

The velocity is irrelevant.



posted on Sep, 19 2011 @ 06:31 AM
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Originally posted by DrinkYourDrug
reply to post by -PLB-
 



No he is not, it is one of the three possibilities. It depends on several factors such as how heavy the brick is, from which height it falls, how much resistance the plate is offering, etc.

The scenario is that the plate could support the brick at rest but not when dropped from sufficient height. I stand by my post above.

The resistance provided by a fluid is not the same as inelastic collisions. This statement proves nothing.

If a plate can support a brick at rest it means the plate has a breaking strength which is greater than the weight of the brick. Therefore the brick will decelerate upon providing this equal and opposite force at the moment of impact because it has a net upwards force (its self weight downwards due to gravity countered by a greater than self weight force upwards provided by the plate).



You are correct, it is indeed a possible error in my explanation. There are a couple more factors involved though. The material the plate is made of and the angle of the brick hitting the plate is of relevance here. If you replace the plate with a sheet of paper and you drop the brick under an angle in my example, then it all works out again. The paper will fail locally, and not as a whole at once. You can argue that the paper would not be able to hold the brick when it is placed on it in that specific angle though. But that is not required for my example. (we know for example that the top section in the WTC tilted).

If you assume that the plate completely fails at once, you are correct. Though the deceleration is a very short instance, the exact time between the brick hitting the plate, and the plate being broken. You probably wont even notice it when you look at it.

But well done in spotting this, you seem to actually understand the physics
. Just a question, what do you think of all the claims by Anok, Bob and Darkwing?


Articles outlining very common and basic engineering knowledge are not usually accepted for publication.


I think an article that disproves NIST findings will be quite uncommon and non-basic.
edit on 19-9-2011 by -PLB- because: (no reason given)



posted on Sep, 19 2011 @ 06:45 AM
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reply to post by -PLB-
 




If you assume that the plate completely fails at once, you are correct. Though the deceleration is a very short instance, the exact time between the brick hitting the plate, and the plate being broken. You probably wont even notice it when you look at it.


You might not notice the deceleration, but you certainly would the change in velocity


Seriously, look at any number of collisions or Verinages...

Also I would like an answer as whether 1 pound of unconfined concrete dust would do the same amount of damage as one pound of solid concrete. Answers on the back of a postcard please.



posted on Sep, 19 2011 @ 06:53 AM
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Originally posted by Darkwing01
The force exerted on the downward object is the force required to accelerate that object at 9.8m/s^2, and since the mass doesn't change the force doesn't change.


Lets try to decipher this. By "downward object" you probably mean "object falling down". So the force exerted on the object falling down is the same as the force required to accelerate the object.

This is wrong. An object falling 1 meter will have a much greater impact force than an object falling 10 meter, while the gravitational force does not change. You should be able to understand this intuitively. The difference is the impact velocity.

If you had visited the link I gave you and had played with the snippet, you should understand this. You should also have been able to answer my question.



posted on Sep, 19 2011 @ 06:56 AM
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Originally posted by Darkwing01
You might not notice the deceleration, but you certainly would the change in velocity


Seriously, look at any number of collisions or Verinages...

Also I would like an answer as whether 1 pound of unconfined concrete dust would do the same amount of damage as one pound of solid concrete. Answers on the back of a postcard please.


And guess what, the resistance did change the velocity of the falling top section in the WTC collapse
. And your question is pretty much irrelevant. It was not unconfined concrete dust that was causing the supports to fail.



posted on Sep, 19 2011 @ 07:28 AM
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reply to post by -PLB-
 





This is wrong. An object falling 1 meter will have a much greater impact force than an object falling 10 meter


The equation, PLB, is F=ma.

This mass of this object has not changed? Gravity has not appreciably been altered has it?

Then the force exerted by an object travelling at one bazillion miles an hour is the same as the that same when it is stationary.

The kinetic energy is vastly different, but that is not a term in this equation...

In practice the object travel faster does exert more force because there is less time for the object to deform, but that is a different matter than the one we are discussing. There is no intrinsic difference in force exerted due to speed ceteris paribus.
edit on 19-9-2011 by Darkwing01 because: (no reason given)



posted on Sep, 19 2011 @ 07:47 AM
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reply to post by Darkwing01
 


We have been talking about impact forces. You completely fail to grasp the physics. You refuse to look at an educational link I gave you. You embrace ignorance.



posted on Sep, 19 2011 @ 08:22 AM
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reply to post by -PLB-
 


The impact force is the same PLB.

The MAXIMUM force that the top of the building can possibly exert on the lower structure is the force required to accelerate the body 9.8m/s^2.

Any deformation can only ever REDUCE that number by effectively reducing the amount of mass participating in the collision any given instant, but it will never increase it, because gravity stays the same and the mass is not increasing.

The force can never be anything other than mass times acceleration, velocity is not a term in the equation.



posted on Sep, 19 2011 @ 08:26 AM
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Originally posted by Darkwing01
reply to post by -PLB-
 


The impact force is the same PLB.

The MAXIMUM force that the top of the building can possibly exert on the lower structure is the force required to accelerate the body 9.8m/s^2.

Any deformation can only ever REDUCE that number by effectively reducing the amount of mass participating in the collision any given instant, but it will never increase it, because gravity stays the same and the mass is not increasing.

The force can never be anything other than mass times acceleration, velocity is not a term in the equation.


Actually, the mass of the collapsing floors are increasing, as each collapsed floor adds to the equation, minus whatever percentage is ejected.

The floors' masses don't just disappear.



posted on Sep, 19 2011 @ 08:29 AM
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reply to post by Varemia
 





Actually, the mass of the collapsing floors are increasing, as each collapsed floor adds to the equation, minus whatever percentage is ejected.

The floors' masses don't just disappear.
.

One thing at a time, we are only talking about how much force is being exerted at the moment of collision and whether THAT collision would decelerate the falling mass.

There is no additional mass yet. But see my earlier question about whether one pound of concrete dust does the same damage as a one pound block of concrete. Put another way, would you rather have a pound of lead or a pound of feathers dropped on your head.

Mass is not the only variable here.


edit on 19-9-2011 by Darkwing01 because: (no reason given)



posted on Sep, 19 2011 @ 08:33 AM
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reply to post by Darkwing01
 


Yes. That would be factoring in density and air resistance. Basically mass. We know it had approximately 12 feet to accelerate after the collapse initialized, so there is velocity which begets kinetic energy which applies an increased force. If you drop a bowling ball from different heights, it will have a different effect, because it has different energy built up which applies a different level of force.

Gravity as acceleration is very important in determining the applicable force.



posted on Sep, 19 2011 @ 08:36 AM
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Originally posted by Darkwing01
The impact force is the same PLB.

The MAXIMUM force that the top of the building can possibly exert on the lower structure is the force required to accelerate the body 9.8m/s^2.

Any deformation can only ever REDUCE that number by effectively reducing the amount of mass participating in the collision any given instant, but it will never increase it, because gravity stays the same and the mass is not increasing.

The force can never be anything other than mass times acceleration, velocity is not a term in the equation.


If you do not accept it from me, accept it from another source. You are directly contradicted by the link I gave you. I don't feel like explaining it any further to you, you are in complete denial, since it does not fit in your fantasized nonsense physics you require for your conspiracy theory.



posted on Sep, 19 2011 @ 08:39 AM
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reply to post by -PLB-
 




You refuse to look at an educational link I gave you.


hyperphysics.phy-astr.gsu.edu...

What is the problem here PLB, what term in the equation is the one we are trying to establish?


Even though the application of conservation of energy to a falling object allows us to predict its impact velocity and kinetic energy, we cannot predict its impact force without knowing how far it travels after impact.


In other words, this equation cannot be solved on the basis you are trying to solve it.

At the MOMENT OF IMPACT the objects have not moved at all, we are trying to establish what the value of the resistance is.

What I am saying is that the resistance must be at least 9.8m/s^2, so the object will decelerate. You impact force calculator in no way contradicts that.

The additional force can only come from the change in kinetic energy, which is exactly what we have been trying to get through to you.


Note that the above calculation of impact force is accurate only if the height h includes the stopping distance, since the process of penetration is further decreasing its gravitational potential energy.


We WANT TO KNOW what the stopping distance is.

Impact is not a "real force" like gravity is, it is just an extension of the normal force.
edit on 19-9-2011 by Darkwing01 because: (no reason given)



posted on Sep, 19 2011 @ 08:44 AM
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Originally posted by Darkwing01

Then the force exerted by an object travelling at one bazillion miles an hour is the same as the that same when it is stationary.



en.wikipedia.org...

"Statics is the branch of mechanics concerned with the analysis of loads on physical systems in static equilibrium, that is, in a state where the relative positions of subsystems do not vary over time, or where components and structures are at a constant velocity. When in static equilibrium, the system is either at rest, or its center of mass moves at constant velocity.

By Newton's first law, this situation implies that the net force and net torque on every body in the system is zero "



Do you understand that?

Net force is zero.

f=ma

0=ma

there is mass, therefore ...

a=0

Which explains your woeful belief that a body at rest is accelerating at 9.8m/s/s.

It's NOT accelerating !! It's stationary !!!



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