Hardfire: Physics of 9/11

[Dave Rogers]

I don't need to think about it in the slightest.

There's your problem, right there.

That's not as flippant and sarcastic as it sounds. That is, in fact, your problem. You're starting from the assumption that your understanding is perfect and doesn't need re-examination.

It is perfectly clear and simple.
The kinetic energy is consumed by deceleration of the initial mass, and acceleration of the impacted mass.
The sum of these two KE sinks is exactly the amount of KE consumed during the inelastic collision, as is clearly shown in my calcs.

Let's look at the mechanism by which that acceleration and deceleration occurs. An object moving at constant speed strikes a stationary object, and in the collision one of those objects is accelerated and the other decelerated. However, this is not an instantaneous process, because there cannot be any such thing as a perfectly rigid body.

Let's suppose we have two blocks, each of length L in the direction of motion. Block 1 is moving, and block 2 is stationary. At the instant of impact, the centres of the two blocks are separated by a distance L. Now, from that instant to the instant that the two blocks are moving at the same speed, block 1 decelerates and block 2 accelerates. Throughout this time interval, block 1 is moving faster than block 2. The distance between their centres is decreasing. At the time when the blocks can be considered to be moving at the same speed, the distance between their centres is less than it was at the moment of collision, when both blocks were undeformed. Therefore, the loss of energy in an inelastic collision is equal to the deformation energy, because the deformation and the inelastic collision are the same process.

You can go through the calculations the same way assuming that the two blocks deform is a perfectly plastic fashion, with a constant energy for a given crushing distance, and you'll get the same energy loss in the collision as you got from your conservation of momentum calculation. I'll leave that for you to work out for yourself, but it's relatively simple Newtonian dynamics. The force deforming the two blocks is the same force, by Newton's Third Law, that is accelerating one of them and decelerating the other.

Let's assume, though, that you can't be bothered to work that out, and you're still convinced you're right and everybody else in the world is wrong. Simply answer me one question.

Energy can't be created or destroyed. The kinetic energy of the two objects moving together after the collision, as you've correctly stated, is less than the kinetic energy of one object before the collision. If the lost kinetic energy didn't go into deformation, where did it go? You can't say it went into decelerating one object; that's an energy loss, not an energy gain. Deceleration doesn't consume energy, it releases it. Where did it go?

Dave

 

[femr2]

There's your problem, right there.

That's not as flippant and sarcastic as it sounds. That is, in fact, your problem. You're starting from the assumption that your understanding is perfect and doesn't need re-examination.

Let's look at the mechanism by which that acceleration and deceleration occurs. An object moving at constant speed strikes a stationary object, and in the collision one of those objects is accelerated and the other decelerated.

However, this is not an instantaneous process, because there cannot be any such thing as a perfectly rigid body.

That, I think, is where the conflict arises. I am looking at a virtual model, in which the pile-driving mass is a perfectly rigid body, and it is an instantaneous process.

Let's suppose we have two blocks, each of length L in the direction of motion. Block 1 is moving, and block 2 is stationary. At the instant of impact, the centres of the two blocks are separated by a distance L. Now, from that instant to the instant that the two blocks are moving at the same speed, block 1 decelerates and block 2 accelerates. Throughout this time interval, block 1 is moving faster than block 2. The distance between their centres is decreasing. At the time when the blocks can be considered to be moving at the same speed, the distance between their centres is less than it was at the moment of collision, when both blocks were undeformed. Therefore, the loss of energy in an inelastic collision is equal to the deformation energy, because the deformation and the inelastic collision are the same process.

The inclusion of time implies quite a change from the virtual mechanism I've been 'rigid' about, does it not ?

You can go through the calculations the same way assuming that the two blocks deform is a perfectly plastic fashion, with a constant energy for a given crushing distance, and you'll get the same energy loss in the collision as you got from your conservation of momentum calculation. I'll leave that for you to work out for yourself, but it's relatively simple Newtonian dynamics. The force deforming the two blocks is the same force, by Newton's Third Law, that is accelerating one of them and decelerating the other.

Taking 'impossible' perfectly rigid bodies and the inclusion of time into account, why is it that the energy loss is exactly the amount of energy required to decelerate+accelerate ? In the circumstance (within a crush-down model) where the support structure requires less than the c-o-m calc 'loss' to fail, what is supposed to happen to the excess energy ? (Add: In the model I've assumed it to be used to enable finer concrete crush/possible additional mass loss/faster debris ejecta, otherwise it results in further acceleration which makes no sense at all.)

Let's assume, though, that you can't be bothered to work that out, and you're still convinced you're right and everybody else in the world is wrong.

Again, I think the issue is more about the application to 'real' world against 'virtual' world, but I am listening.

Simply answer me one question.

Energy can't be created or destroyed. The kinetic energy of the two objects moving together after the collision, as you've correctly stated, is less than the kinetic energy of one object before the collision. If the lost kinetic energy didn't go into deformation, where did it go? You can't say it went into decelerating one object; that's an energy loss, not an energy gain. Deceleration doesn't consume energy, it releases it. Where did it go?

If we are including time, and removing perfectly rigid bodies, it is a different picture, is it not ?

In the virtual model of perfectly rigid 'floating' slabs (no support) impacting with perfect inelastic collisions, is my position not correct ?

I'll look at the implications of what you are saying of course.

In addition, as stated a few posts ago, I've included a switch on the model to allow inclusion of the c-o-m energy loss in subsequent deformation calcs, so the model now handles both scenarios, which I hope will suffice.

(Though my preference will be to specifically include the time and non-rigid body mechanics instead, as it seems to be a big point of contention)

 

[Dave Rogers]

That, I think, is where the conflict arises. I am looking at a virtual model, in which the pile-driving mass is a perfectly rigid body, and it is an instantaneous process.

Then your collisions cannot be inelastic. Your model is self-contradictory from the start, so it's unlikely to yield any significant results. I can't stress this too strongly; irreversible deformation is the mechanism by which inelastic collisions occur, so by excluding irreversible deformation you are excluding the possibility of inelastic collision. A model that tries to include both elements is a physical impossibility.

Taking 'impossible' perfectly rigid bodies and the inclusion of time into account, why is it that the energy loss is exactly the amount of energy required to decelerate+accelerate ?

Because the physical process is the same. The force acting on each body does work upon it, and the reaction force accelerates the body. Both must translate through the same distance as they act on the same surface. The result (that the energy loss is exactly the amount of energy required to accelerate less the energy released by deceleration) follows inevitably from Newton's Third Law.

In the circumstance (within a crush-down model) where the support structure requires less than the c-o-m calc 'loss' to fail, what is supposed to happen to the excess energy ?

That question doesn't mean a lot. There are energy losses from conservation of momentum in an inelastic collision, which are equal to the energy sunk into deformation (including breakage). There will also be energy losses from breakage of the support structure; this can be seen as a separate inelastic collision, but one in which the stationary mass is effectively infinite (because the support structure is anchored to the ground, so the acceleration of the Earth may be neglected). That component of the excess energy is sunk into fracture energy and deformation energy of the lower structure. Again, by Newton's Third Law, this is the same as the kinetic energy loss of the falling block.

In the virtual model of perfectly rigid 'floating' slabs (no support) impacting with perfect inelastic collisions, is my position not correct ?

As I've stated above, the model of perfectly rigid floating slabs impacting with perfect inelastic collisions is a fundamental physical contradiction, and cannot therefore produce any instructive results.

Dave

 

[Newtons Bit]

Femr2, you never responded to my question.

 

[Horatius]

why is it that the energy loss is exactly the amount of energy required to decelerate+accelerate ?

Putting it a simpler way: If the amount of energy involved in deforming was different from the energy involved in the changing motion of the bodies, we would have either created additional energy from somewhere, if the difference was positive, or destroyed that same energy, if the difference was negative, both of which we know to be impossible.

 

[femr2]

As I've stated above, the model of perfectly rigid floating slabs impacting with perfect inelastic collisions is a fundamental physical contradiction, and cannot therefore produce any instructive results.

Dave

Fine, though that is the fundamental basis of many-a crush-down model.

The only real solution I think, is to move to elastic collisions soon. The implications of a non-rigid pile-driving mass and the time factors required are going to be interesting to implement within a spreadsheet.

In the meantime, I assume the inclusion of the model switch to make the c-o-m energy loss available for deformation addresses concerns. It will also allow comparison of the difference it makes.

 

[Dave Rogers]

In the meantime, I assume the inclusion of the model switch to make the c-o-m energy loss available for deformation addresses concerns. It will also allow comparison of the difference it makes.

It ought to, since that's the actual basis of every correct crush-down model. In effect, it's a way of compensating for the contradiction between rigidity and inelasticity; or, put another way, it's a way of circumventing the detailed processes taking place in an inelastic collision, and simply accounting for their effects on the global energy balance.

In effect, though, that'll simply make your model the same as Greening's, Urich's, Newtons Bit's, and everyone else's that predicts a collapse time within the bounds of observation.

Dave

 

[femr2]

In effect, though, that'll simply make your model the same as Greening's, Urich's, Newtons Bit's, and everyone else's that predicts a collapse time within the bounds of observation.

Dave

From experimentation, and the inclusion of the fairly detailed source data (specific masses, specific floor heights etc), the most significant factor is still the mass of the pile-driver, and the ensuing effect on collapse time.

Specific floor-by-floor steel masses will greatly improve the accuracy, especially the exterior columns.

As I've said from the outset however, my own personal intention is to create a model which enables users to experiment and see the effect of various parameters upon collapse time, rather than produce something which 'proves' one side of the argument or the other. The more accurate the better, but at the end of the day a full blown FEA is required to even approach the word 'proof'. My personal view is that with mass loss data 'I' see as appropriate, the actual 'collapse' time was far too short. I'm sure you will disagree. If I end up with a model and data-set which can enable useful experimentation, great.

 

[Dave Rogers]

In the virtual model of perfectly rigid 'floating' slabs (no support) impacting with perfect inelastic collisions, is my position not correct ?

I've put a little more thought into this, and the answer is no.

The energy lost as kinetic energy in an inelastic collision is converted to deformation energy. The deformation involves a force moving the material of the deformed body through a distance, so energy = force x distance. In your non-physical model of perfectly rigid bodies, you're suggesting that, because there can be no deformation, then the distance is zero, and hence the work done is zero. However, if there can be no deformation, then the velocities of the two bodies must both change instantly from V1 and 0 to V2 and V2 respectively; since dt=0, then the acceleration dV/dt must be infinite, and hence the force between the two bodies must be infinite. The work done is therefore the multiple of an infinite force by a zero distance, which is not zero; it is indeterminate.

Therefore, using basic principles of calculus, we must conclude that the deformation energy in an inelastic collision between perfectly rigid bodies must be the limit as the deformation tends to zero of the multiple of the deformation distance and the deformation force. Since this is not a variable, but is a constant, equal to the change in total kinetic energy, then we have the paradoxical result that, in an inelastic collision between two perfectly rigid bodies, there is an energy loss to deformation equal to the difference between initial and final kinetic energies. The result is paradoxical because the situation is physically unreasonable, but it's the only result that has any kind of validity.

Therefore, in any scenario, even that of rigid, undeformable blocks, the kinetic energy lost due to inelastic collision is converted to deformation energy in the colliding bodies. There are no exceptions.

Dave

 

[Newtons Bit]

I've put a little more thought into this, and the answer is no.

The energy lost as kinetic energy in an inelastic collision is converted to deformation energy. The deformation involves a force moving the material of the deformed body through a distance, so energy = force x distance. In your non-physical model of perfectly rigid bodies, you're suggesting that, because there can be no deformation, then the distance is zero, and hence the work done is zero. However, if there can be no deformation, then the velocities of the two bodies must both change instantly from V1 and 0 to V2 and V2 respectively; since dt=0, then the acceleration dV/dt must be infinite, and hence the force between the two bodies must be infinite. The work done is therefore the multiple of an infinite force by a zero distance, which is not zero; it is indeterminate.

Therefore, using basic principles of calculus, we must conclude that the deformation energy in an inelastic collision between perfectly rigid bodies must be the limit as the deformation tends to zero of the multiple of the deformation distance and the deformation force. Since this is not a variable, but is a constant, equal to the change in total kinetic energy, then we have the paradoxical result that, in an inelastic collision between two perfectly rigid bodies, there is an energy loss to deformation equal to the difference between initial and final kinetic energies. The result is paradoxical because the situation is physically unreasonable, but it's the only result that has any kind of validity.

Therefore, in any scenario, even that of rigid, undeformable blocks, the kinetic energy lost due to inelastic collision is converted to deformation energy in the colliding bodies. There are no exceptions.

Dave

By definition, a rigid body cannot undergo either a inelastic collision or an elastic collision as it can neither deform elastically or inelastically. It is quite the paradox, isn't it?

 

[Dave Rogers]

By definition, a rigid body cannot undergo either a inelastic collision or an elastic collision as it can neither deform elastically or inelastically. It is quite the paradox, isn't it?

Physics gets a bit tricky in the presence of infinities, for sure.

Dave

 

[alienentity]

Femr2, I'm heartened that you are proceeding to improve your model. Hopefully it will eventually be complete enough to help dispel the dogma which seems to have you in a mind-numbing grip.

I'm curious to know,since you've had your maths corrected (but have rejected the corrections) can at least accept, in theory, that you may be in error. Not only that, realize such an error is excusable and perfectly fine, and you would be better off without it.
Honestly, if you consider yourself reasonably intelligent, then act intelligently. This is an opportunity. Perhaps that's even part of the reason you came to this forum, and began consulting Ryan.
It's your call.

 

[Furcifer]

Therefore, using basic principles of calculus, we must conclude that the deformation energy in an inelastic collision between perfectly rigid bodies must be the limit as the deformation tends to zero of the multiple of the deformation distance and the deformation force.

What happens if two mass less, frictionless, perfectly rigid pulleys collide on a frictionelss surface?

 

[Furcifer]

In effect, though, that'll simply make your model the same as Greening's, Urich's, Newtons Bit's, and everyone else's that predicts a collapse time within the bounds of observation.

Dave

Quit giving away the answers Dave. It's about the process

 

[psikeyhackr]

Quit giving away the answers Dave. It's about the process

.
I thought it was about figuring out why a 150 ton airliner could not destroy a 400,000 ton building in less than 2 hours.

Apparently a lot of people think it is about playing pseudo-intellectual games FOREVER.

This crap should have been resolved in less than a year.

psik

 

[Furcifer]

.
This crap should have been resolved in less than a year.
psik

It has been for almost 8. It's just that several people have been a little slow on the up-take. We're trying to work that out. Please bare with us.

 

[psikeyhackr]

It has been for almost 8. It's just that several people have been a little slow on the up-take. We're trying to work that out. Please bare with us.

.
It's difficult.

It took them less than SEVEN YEARS to figure out how much steel and concrete to put on every level of the towers. But some people aren't even asking the question after that much time. What's are wheat chex.

psik

 

[aggle-rithm]

.
I thought it was about figuring out why a 150 ton airliner could not destroy a 400,000 ton building in less than 2 hours.

Elbow grease, plus the airliner doesn't stop for bathroom breaks.

 

[Toke]

Bazant showed collapse within a factor of 5-10, under the most optimistic conditions.
I don´t see how you can keep up the insistance on excat weights, what diffrence would it make.

I have pointed out where your washer model fail to relate to the towers, without responce.

What is your point?

 

[Toke]

.
I thought it was about figuring out why a 150 ton airliner could not destroy a 400,000 ton building in less than 2 hours.
psik

There were a stundie candidate where somebody calculated the impact to something like a shotgun slug and a human body to show how harmless it was.

And then divided the original shotgun speed by 1000 to get it in ridicules range.