If it doesn't agree with experiment, it's wrong. Part II

[Crazy Chainsaw]

It is sad, but the explosion in India of the black powder fireworks, shows just what a carbon fueled fuel air blast could have done in the Towers collapses. If ignited and contained by the collapse
Front.
Fortunately for the towers the rain of debris would have muffled the explosion.

 

[PhantomWolf]

Everyone keeps forgetting that whatever was at the bottom has mass, even the tissue.

Therefore conservation of linear momentum applies, and unless the materials are so flexible as for the maximum deceleration during their compression to be < g, which happens very seldom (think crash tests, for example - typical accelerations of the occupants are >g), the net result is that the top object must decelerate. And that's independent of the connections. Even if the tissue is thrown upwards in a vacuum in such a way that when the brick reaches it, the tissue is on the top of its ballistic trajectory (vertical speed zero), you'll get deceleration of the brick. Probably not measurable, because of the differences in mass, but it will be there. In the case of a floor slab, the mass will be far bigger, and not negligible at all, and the materials are obviously not compressible enough to dampen the fall. The deceleration must be present.

No, because in a falling object meets a resistant object scenario, you have a three body system not a two body system. You are forgetting the Earth. The falling mass is accelerated by the Earth at g, so unless the resistance applies an acceleration greater than g, the object continues to accelerate. The amount of resistance is based on the object's ability to absorb energy and deform as well as it's over all strength. If the tissue cannot hold up the brick when it is placed on it, then it can't place a force required to nullify gravity on the falling object, let along one that is greater.

The reason that crash test dummies get put under extreme g's is because the object the car they are in hits has more resistance then the car has energy to pass through it. This means that the car's velocity is arrested in a matter of millimeters and fraction of a second.

since a = v/t the faster the car is stopped, the greater the acceleration

A car that going from 100 km/h and being stopped in 0.1s will undergo -28g.

This doesn't happen to the falling object though. The resistant item can only return as much force to the falling object as it can withstand. Since we know that is less than mg, the negative acceleration must be less than g.

The only time that we need to consider mass is after the resistance with momentum transfer where the falling object continues to assert a force to accelerate the now attached object to the same velocity as the falling one, but here gravity also helps as it's accelerating both items.

The conservation of momentum in this case involves three objects, the falling object the resistant object and the Earth. The momentum is conserved between the three of them.

 

[Crazy Chainsaw]

No, because in a falling object meets a resistant object scenario, you have a three body system not a two body system. You are forgetting the Earth. The falling mass is accelerated by the Earth at g, so unless the resistance applies an acceleration greater than g, the object continues to accelerate. The amount of resistance is based on the object's ability to absorb energy and deform as well as it's over all strength. If the tissue cannot hold up the brick when it is placed on it, then it can't place a force required to nullify gravity on the falling object, let along one that is greater.

The reason that crash test dummies get put under extreme g's is because the object the car they are in hits has more resistance then the car has energy to pass through it. This means that the car's velocity is arrested in a matter of millimeters and fraction of a second.

since a = v/t the faster the car is stopped, the greater the acceleration

A car that going from 100 km/h and being stopped in 0.1s will undergo -28g.

This doesn't happen to the falling object though. The resistant item can only return as much force to the falling object as it can withstand. Since we know that is less than mg, the negative acceleration must be less than g.

The only time that we need to consider mass is after the resistance with momentum transfer where the falling object continues to assert a force to accelerate the now attached object to the same velocity as the falling one, but here gravity also helps as it's accelerating both items.

The conservation of momentum in this case involves three objects, the falling object the resistant object and the Earth. The momentum is conserved between the three of them.

Exactly why I proposed the two hammers experiment, that FF, admitted would have different results based on the energy values, I simply changed mass, and left resistance the same.

He admitted results would vary so he admitted both he and Cole were wrong, he just will never admit I proposed an experiment that debunked him.

Actually I can't take all the credit, he is very good at self debunking.

 

[Myriad]

Everyone keeps forgetting that whatever was at the bottom has mass, even the tissue.

Therefore conservation of linear momentum applies, and unless the materials are so flexible as for the maximum deceleration during their compression to be < g, which happens very seldom (think crash tests, for example - typical accelerations of the occupants are >g), the net result is that the top object must decelerate. And that's independent of the connections. Even if the tissue is thrown upwards in a vacuum in such a way that when the brick reaches it, the tissue is on the top of its ballistic trajectory (vertical speed zero), you'll get deceleration of the brick. Probably not measurable, because of the differences in mass, but it will be there. In the case of a floor slab, the mass will be far bigger, and not negligible at all, and the materials are obviously not compressible enough to dampen the fall. The deceleration must be present.

A deceleration due to momentum transfer must be present, but only in the simplifying (and therefore unrealistic) case where it is assumed that the momentum transfer is instantaneous.

If the momentum transfer takes place over a finite interval of time, as must be the case in all real collisions, then there may or may not be a deceleration. The more rapidly the momentum transfer occurs, the greater the collision force (which increases in proportion to 1/delta-t) and the more likely that force will exceed the gravitational force (mg) on the falling mass, in which case there will be an actual deceleration during the collision.

To paraphrase a maritime adage: if the collision goes "clang!" or "wham!" then there is probably a deceleration. If it goes "crunch!" or "thunk!" there might not be.

 

[FalseFlag]

The point is that once a collapse begins, nothing in the world will stop it from continuing.

Are you sure about that?

https://www.youtube.com/watch?v=BwPD0HF3GPM

 

[PhantomWolf]

Are you sure about that?

https://www.youtube.com/watch?v=BwPD0HF3GPM

And yet again you prove your inability to understand that different structures act differently under different loads. Colour me surprised.

 

[Myriad]

Are you sure about that?

https://www.youtube.com/watch?v=BwPD0HF3GPM

Cole's firecracker experiment does not duplicate the motions of the controlled demolition in that video, proving that the 9/11 collapses were not CD.

 

[Myriad]

Thread re-opened following thread closure that appears to have been accidental. Reported for mod staff attention to determine the cause.

Posted By: Myriad

 

[Crazy Chainsaw]

Thanks - I've read the paper from the link Dave gave. And recalled how I read it many years back. And realised why it did not stick in my memory.
That is the first part of the paper and if that is what the spreadsheet was based on it will be in the right ballpark - i.e. right answers for right or near enough right reasons. I haven't seen discussion of the spread sheet.

?? So? The joists stripped off. Whether by Maxwell's theories or Santa's Custard.

So that is all in the early part of paper. BUT I also read the later part of the paper - did you?

"experimentally" - why? Reasoned analysis by many persons has explained the mechanism - and with better clarity of the staged mechanisms and without the modelling of mechanism confusions Greening reveals in the second part of his paper. It was good work for the day. Whether it stands the test of time is a different issue.

Bottom line AFAICS he was correct on the first part of the paper - as have been several others. And that is the aspect relevant to the reference a few posts back in this thread.

FalseFlag asked it the Claims were experimentally proven Ozeco41.

 

[ozeco41]

FalseFlag asked it the Claims were experimentally proven Ozeco41.

But why give any credence to FF? Who cares what nonsense he claims/asks/asserts.

I don't know about you but my focus is on legitimately explaining WTC collapses - plus any applied physics which is necessary for that purpose. Don't need to chase trolls down rabbit burrows to do that - in fact it is counter productive.

 

[Crazy Chainsaw]

But why give any credence to FF? Who cares what nonsense he claims/asks/asserts.

I don't know about you but my focus is on legitimately explaining WTC collapses - plus any applied physics which is necessary for that purpose. Don't need to chase trolls down rabbit burrows to do that - in fact it is counter productive.

Please don't mention Rabbits Just getting over Rabbit fever, and I agree, just was being respectful of a member of the forum giving him the benefit of the doubt, hoping he was capable of being honest.

 

[pgimeno]

No, because in a falling object meets a resistant object scenario, you have a three body system not a two body system. You are forgetting the Earth.

No, I'm not. I'm analysing the acceleration experienced by the falling floor slab, and I count gravity as a force that is external to it. I don't need to consider another object; I need to consider an external force. A force equal to m_floor·g.

The falling mass is accelerated by the Earth at g, so unless the resistance applies an acceleration greater than g, the object continues to accelerate. The amount of resistance is based on the object's ability to absorb energy and deform as well as it's over all strength.

Agreed.

If the tissue cannot hold up the brick when it is placed on it, then it can't place a force required to nullify gravity on the falling object, let along one that is greater.

Disagreed. We can discuss that later. For now let me make my point.

The reason that crash test dummies get put under extreme g's is because the object the car they are in hits has more resistance then the car has energy to pass through it. This means that the car's velocity is arrested in a matter of millimeters and fraction of a second.

since a = v/t the faster the car is stopped, the greater the acceleration

A car that going from 100 km/h and being stopped in 0.1s will undergo -28g.

Agreed.

This doesn't happen to the falling object though. The resistant item can only return as much force to the falling object as it can withstand. Since we know that is less than mg, the negative acceleration must be less than g.

Disagreed.

To simplify the problem, we can take gravity out of the equation, by looking at it as a problem of an object impacting another horizontally at some speed, and determining if the deceleration experimented by the impacting object is >g or not. If the deceleration is >g, then in the case with gravity we will have net deceleration. If the deceleration is <g, we won't.

We are now in the exact same realm as that of a car impacting another object, no matter if that object is free to move (think parked bike, for example).

Imagine how long it will take to decelerate on impact a floor slab that has fallen in free fall for 0.9 s on another (0.9 seconds is about the time it takes for an object to fall 4m, which is approx. the height of each floor). There's no cushion to dampen the fall; in the WTC case it would typically be steel against concrete, both hardly compressible at all, or steel against furniture (it's very unlikely that the furniture included very elastic cushions).

It should be obvious that for the instant deceleration at any time to be < g, which is required for the object not to experiment a net deceleration at any point when gravity is considered, the total deceleration time, which determines the average deceleration, should be at least 0.9 seconds.

For that deceleration to be at least 0.9 seconds on impact (which matches the time of the fall), we need really deformable and elastic materials. Concrete and steel are not.

Now, I admit I might have been wrong on the tissue case. To try to imagine if there would be deceleration, I tried to imagine how much a stack of them would be compressed. This is the wrong approach, because force increases as the object compresses, and tissue is quite compressible, so in the first impact the deceleration might indeed be <g. But let's take a different example: a very fine and brittle glass pane, say 1mm thick. I'm sure a brick will necessarily decelerate when going falling through one. [ETA: And here's one point of contention: that will happen even if the pane can't hold the brick statically.]

To test my understanding of the subject, I made a program using a 2D physics engine. The results agree with me. I published the details and the results here: http://www.internationalskeptics.com/forums/showthread.php?p=11220654#post11220654

The only time that we need to consider mass is after the resistance with momentum transfer where the falling object continues to assert a force to accelerate the now attached object to the same velocity as the falling one, but here gravity also helps as it's accelerating both items.

No. We need to consider mass and material compressibility. A glass pane is barely compressible. Steel and concrete are even less compressible. Maybe the rule of thumb that Myriad has mentioned makes some sense.

The conservation of momentum in this case involves three objects, the falling object the resistant object and the Earth. The momentum is conserved between the three of them.

Not relevant. The momentum is conserved between the falling object and the impacted object to enough degree as to make taking Earth into consideration unnecessary.

 

[pgimeno]

A deceleration due to momentum transfer must be present, but only in the simplifying (and therefore unrealistic) case where it is assumed that the momentum transfer is instantaneous.

Um, no.

If the momentum transfer takes place over a finite interval of time, as must be the case in all real collisions, then there may or may not be a deceleration.

OK, yes, that's more like it.

And very big accelerations induced by impacts are common.

You even pointed that out here:

http://www.internationalskeptics.com/forums/showthread.php?p=9014181#post9014181

 

[pgimeno]

A deceleration due to momentum transfer must be present, but only in the simplifying (and therefore unrealistic) case where it is assumed that the momentum transfer is instantaneous.

I see the problem now.

I interpreted your sentence as meaning that it's only possible to have deceleration if the momentum transfer is instantaneous.

Now I think you actually meant that the momentum transfer only causes deceleration for sure if the momentum transfer is instantaneous.

Sorry.

 

[rwguinn]

The problem isn't calculus. The problem is in looking at the reference frame of a falling body as though it were an inertial reference frame. It isn't.

Step back and observe a falling body from an inertial reference frame. The falling body is constantly under the force mg. For a body falling through air, or hitting an insubstantial object that can't support even its static load, like a brick on a piece of tissue paper, the NET force is mg - air resistance, or resistance of the tissue. The NET force remains positive, and there is no deceleration.

If, OTOH, our brick encounters an object that could support its static load of mg (say a pane of glass), the glass will apply a force equal to its ability to carry a static load. If the brick, by dint of its dynamic load, exerts a greater force, the glass breaks but the brick experiences a net force temporarily greater than g and the brick actually decelerates, then accelerates after the glass is broken.

This just brings me back to my main point: Newton's Laws of Motion are useless for studying the WHY of a progressive collapse!

And if you read the other thread (In Science, medicine, etc) You will note that this is the same conclusion I reached after working thru the math. I was looking at the problem posed here from my experience as a structures guy, not from the physics standpoint, and was having difficulty assimilating the stuff being thrown around.

So true so often.

Good advice in many situations - not just this one.

Maybe not entirely useless BUT only relevant AFTER you have decided HOW the collapse progressed - otherwise you don't know WTF you are discussing or explaining. THEN - a step or two later - the "WHY" comes into play - and Newton's MAY be relevant for some of the details.

BUT NOT for the whole big picture.....

Safest default position - if a truther mentions "Newton" you can be near certain that what follows will be wrong.

On that we can agree.

 

[rwguinn]

I just built a tower of dominoes, using a playing card as a "floor" between levels. Dropped a single domino on the last floor. Dominoes all over the damn room.
Next?

 

[Myriad]

I see the problem now.

I interpreted your sentence as meaning that it's only possible to have deceleration if the momentum transfer is instantaneous.

Now I think you actually meant that the momentum transfer only causes deceleration for sure if the momentum transfer is instantaneous.

Sorry.

That is what I meant, but the wording was tricky. Thanks for puzzling it out!

And yes, on short time scales large accelerations can occur in impacts. Vibrations are an example, and can entail large accelerations even for modest-seeming movements. Ring a 100 Hz bell and then drop it. Arrange to drop it horizontally in an orientation that aligns its fundamental vibrational mode with the vertical. If the deflection of the mouth of the bell in that mode has an amplitude of half a millimeter, the accelerations of the top and bottom edges of the mouth of the bell will be about 4g upward and downward alternately relative to the bell's center of gravity, and so even taking the bell's fall into account, those points would be accelerating upward almost half the time.

In that case, the center of gravity of the bell would still be falling smoothly and accelerating at g (minus air resistance, which we can't completely neglect if we want to be able to hear the bell!) So it would be overstating to say that the entire bell was accelerating and decelerating at 100 Hz on the way down. However, if the bell hit an upward-tossed ball bearing in mid-air, hard enough to make the bell ring with the same amplitude as in the previous case, well… that's not quite enough information to prove that the bell's CG must have decelerated by the impact without examining the actual masses and velocities involved, but it becomes a quite reasonable possibility. Hence, my half-joke about things going "clang."

In such short time scales, also have to consider what part or portion of the falling mass we're talking about when we speak of its acceleration or deceleration. Especially when the mass isn't rigid or even necessarily fully connected to begin with. Making some part vibrate or otherwise momentarily decelerate isn't the same thing (and doesn't have the same significance to the overall dynamics) as making its overall center of gravity do so. If the part of the falling mass that makes the initial contact with the obstacle immediately breaks off, that part probably decelerates but it doesn't follow necessarily that the CG of the whole does at that moment. The same if the collision induces a vibration but not in a primary mode of the whole mass (so some parts vibrate upward and some downward at any given moment). If a series of parts breaks off as the collision continues… well, obviously it gets complicated. That's the "crunch" case.

 

[pgimeno]

Thanks for your elaboration on the subject. Fully agreed.

In such short time scales, also have to consider what part or portion of the falling mass we're talking about when we speak of its acceleration or deceleration.

Fair point, I didn't state explicitly that I was considering the CoM of the falling floor.

As for the short time scales, as I've explained, for the case in point the momentum transfer time should be at least 0.9 seconds (rough approximation) for having deceleration. There are several other considerations that can change it, but not reduce it by orders of magnitude. Yet the momentum transfer time expected in most realistic scenarios between the steel trusses and the concrete slabs is << 0.9 seconds (as in, orders of magnitude less).

 

[fuelair]

Hyperbole.

and, if one had he would have barely but moreso unlikely at all have survived and he would have lost several functions!!!!! Did he???

If not, nothing "threw him " up two stories.

 

[Myriad]

Fair point, I didn't state explicitly that I was considering the CoM of the falling floor.

As for the short time scales, as I've explained, for the case in point the momentum transfer time should be at least 0.9 seconds (rough approximation) for having deceleration. There are several other considerations that can change it, but not reduce it by orders of magnitude. Yet the momentum transfer time expected in most realistic scenarios between the steel trusses and the concrete slabs is << 0.9 seconds (as in, orders of magnitude less).

I agree, for the case presented of a single slab falling onto another one (with or without office contents/furniture). If I recall correctly, measurements taken from vérinage demolition videos also show clear decelerations for at least the first few floor impacts.

For the case of the twin towers, as far as I can tell neither the video evidence nor calculations using best-guess estimates of the parameters seem to be able to decisively answer the question of whether or not brief decelerations of the CoM of the falling mass took place early in the collapses.

By the way, your calculations for the two slab case remind me of the very first modeling and calculations I did to understand the tower collapses, long before I arrived at this sub forum. Instead of posing the question in terms of energy absorption as Bazant did, or on the load capacity of the floor-column connections as is currently the standard line of argument, I set it up as whether or not the lower floors could displace downward over enough distance to decelerate the falling mass to a stop, without breaking. (Answer: not.)