Hardfire: Physics of 9/11

[femr2]

There is no embarrassment, and certainly no need for any further education.
That you so readily resort to character assassination is not unexpected.

The point of contention is one we will not agree upon, however, as the model is indeed accurate enough for valid analysis of the events, I'll paint a scenario for you.

Given correctly specified floor by floor mass loss and associated effect on the mass of the 'pile driver' the model clearly shows that the 'collapse' time witnessed was far too low. (Even when the pile driving mass is still increasing, though not by a full floor mass on each floor impact)

I'll be releasing an updated version of the model fairly shortly, and will quite happily switch off the structural support entirely in one scenario to prove the point.

What I think you misunderstand is that I am not someone who does not understand the significant effect of gravity upon the descent, and that ultimately, once at appropriate velocity, the structural support posed a 'relatively' small obstacle.

The real issues result from the KE available through the initiation process and the rate of descent given the witnessed volumes of mass loss.

I'll leave you to it until I release the updated model.

http://femr2.ucoz.com

 

[NutCracker]

There is no embarrassment, and certainly no need for any further education.
That you so readily resort to character assassination is not unexpected.

No CA. It follows as a logical conclusion from the observation of you failing to understand elementary physics.

 

[femr2]

Well here, then, is the question...

In Ryans outline model inelastic collisions are used, with resistance posed by the structural supports.
Ryan is suggesting that the supports must be present for valid usage of inelastic collisions.

So what happens to Ryans model if the resistance is low ?

Scenario 1: The supports fail with 1000 GJ KE impact over their static load bearing capacity.

Inelastic collisions okay ?

Scenario 2: The supports fail with 1 J KE impact over their static load bearing capacity.

Still okay to use inelastic collisions ?

If we move to elastic collisions the picture becomes very interesting indeed, and major issues with the behaviour of the actual events become very pronounced.

Anyway. Mass loss data to compile.

 

[Heiwa]

If you google my name, it's the second link - WTC Collapse Simulator.

All the calculations are housed in a relatively simple, open, xl spreadsheet, which is focussed upon the generation of a collapse time for a gravity only driven collapse. The spreadsheet is available for download.

The visualisation module simply imports data directly from the spreadsheet, so until I 'finalise' the model, it's not currently available for download, though there are a few YT video scenarios available.

The forum is linked on the home page, and includes fairly detailed answers to Ryans' initial queries.

Let's get started with what happens between first impact and first failure ... and second impact.

You are kindly requested to explain why the moving red Upper Part bottom floor and supports above are not damaged at the contact with the static first/top floor of the Lower Part and its support below!

The Upper Part is much weaker than the Lower Part at the first impact interface; the Upper Part gets weaker higher up, and the Lower Part gets stronger lower down.

Before impact Lower Part is decompressed (it does not carry the Upper Part) so stress in Lower Part top columns is very small - it carries only the Lower Part top floor and not the Upper Part. Likewise Upper Part is completely decompressed - it is under free fall!

What is the energy applied at first contact and how much energy is added due to elastic displacements before failures (this is total energy applied before first failure).

How much energy is required to elastically compress the Upper Part and the Lower Part before first failure.

How much are the Upper and Lower parts compressed or shortened before first failures occur?

What is the velocity of Upper Part roof and Upper Part bottom floor*, when first failure occur. Are they same or different?

(*Same as top floor of Lower part)

What is the time between impact and first failure, i.e. time for elastic compression of Upper and Lower Parts only?

What is time between first failure and second impact? Have the Parts time to decompress?

Are you really certain that it is the Lower Part that gets damaged at first failure? In my opinion based on long experience it would be the Upper Part!

 

[psikeyhackr]

YES exactly!!! Which is why we have been asking you to pick an upper and lower bound!!

.
Why am I supposed to pick anything when we should just be given the data about the actual building? We aren't trying to design something new here we are trying to scale something that used to exist to try to figure out why it no longer exists.

The people who want to BELIEVE the Impossible throw up excuses for not having the actual data. Not having the quantities and weights of the perimeter wall panels is one of the most obviously ridiculous incompetencies of the NIST.

You change the point of the discussion from "let's solve this problem" to "prove to us you are smart" by doing this irrelevant trivia. The real objective being to NOT SOLVE the problem by having everyone TALK FOREVER.

How does Mackey expect any scaled model without accurate data on the building or was he just throwing up a challenge he knew could not be met without the data? But then he makes himself look silly for having complained about my "broken record" of TONS of STEEL and TONS of CONCRETE on every level because we can't scale a model without that information.

psik

 

[Heiwa]

Look, you're trying to get me to buy that (1) when the descending mass contacts a floor, the energy loss at collision does zero damage to materials, yet (2) there's an additional energy sink that corresponds to damaging the materials.

You're double-counting the energy sink, and you're double-talking me. Don't think you can fool me with contrived examples of infinitely rigid plates, or collapses in which there are no structures at all. Learn the above, if you can, and fix your model accordingly. If you do, you'll see why the collapses are completely ordinary. But, of course, we couldn't have that, could we?

Energy analysis is very easy to understand. My kids' audience likes it a lot!

Say we have two objects C and A of same structure (structural sub-elements joined together to make up a structure), where C=1/10A.

From total strength point of view you could say that object A can absorb 10x more strain energy than object C. Let's assume the structure doesn't deform elastically. Elements/joints break at once when sufficent energy is applied to break/shred them.

So let's assume C can absorb X Joule and A 10X Joule strain energy and then something happens. What does it mean?

Say we apply X joule energy on C alone and that the result is that all joints are broken and that all the elements are shredded into rubble. C has collapsed 100% due to the energy X applied on it. It happens!

Now, what happens, if we apply X Joule energy to object A? Wouldn't only 10% of object A be damaged?

Evidently we can drop object C 'as a descending mass' on object A, a gravity driven collision, to test the energy calculations with various energy inputs Y.

Say Y is 0.2X! At contact C on A both objects fail identically, i.e. 0.1X is used to produce failures in C and 0.1X to do same thing in A. The energy losses at collision are split 50/50! Result? Right! Local failures in both C and A.

Say Y=X! Result? Same as before - just more local failures.

Say Y=2X! Result? Well, gentlemen/women reading this and others who are not so fortunate, I am sorry to tell you that C is completely broken ... and A is not!

OK, I know many ignorant persons will now object and suggest that it is the rubble of C and some 10% failed parts of A that then, magically, destroys 90% of what remains of A, but ... my kids' audience doesn't accept that. They know rubble cannot damage intact joints and elements.

Mackey has proposed a Challenge where C (with mass M) shall destroy A (with mass >>M).

Is above sufficient to show that C cannot collapse A?

 

[alienentity]

Femr2, to be fair, if you read the Hardfire Modeling Challenge, you haven't met the criteria yet. Is that so difficult to understand?

And is is unreasonable? If so I'd like to know why. You seem to be getting sidetracked with the idea of a special debate forum for yourself and Mr. Mackey, without having met the basic challenge itself. (although if Ryan starts getting more responses, where are they going to be hosted?)

 

[alienentity]

Energy analysis is very easy to understand. My kids' audience likes it a lot!

Say we have two objects C and A of same structure (structural sub-elements joined together to make up a structure), where C=1/10A. blah blah blah etc...

Is above sufficient to show that C cannot collapse A?

Heiwa, consider the moderator warning about derails from your crush-down thread.

If you have a model to propose in keeping with the challenge - which I posted above for your convenience - why don't you do so? Here you're just reposting the same material you post on every discussion....which has already been rejected, btw. It's borderline spam IMHO.

 

[Myriad]

The fact is that the laws of physics do not change.

A simple inelastic collision calculation as described, including no deformation of materials and no resistance results in the specified energy 'usage'.

In purest physics terms, where are you suggesting that energy is consumed, given that it CANNOT be deformation of materials, as there is none.

A collision that includes no deformation of materials is physically impossible.

There is always deformation of materials. The only question is how much of the deformation is elastic and how much is inelastic.

An "inelastic collision including no deformation of materials" is a direct contradiction. It makes no more sense than "a blackbody that absorbs no light" or "a tornado with zero wind speeds."

That is why your titanium plates example does not help your argument. It starts by assuming an impossibility.

Respectfully,
Myriad

 

[Heiwa]

50/50 or 0/100?

Heiwa, consider the moderator warning about derails from your crush-down thread.

If you have a model to propose in keeping with the challenge - which I posted above for your convenience - why don't you do so? Here you're just reposting the same material you post on every discussion....which has already been rejected, btw. It's borderline spam IMHO.

Don't worry. The Mackey post was about energy loss at collision upper part C/lower part A and damages associated with it, and I just repeated/clarified again that damages are often split 50/50 between objects C/A involved. As Mackey in his Hardfire show suggests damages C/A are split 0/100, it seems we have different opinions.
If Mackey can repost his ideas in sub-posts, why can't I? This is a friendly and lively discussion!

 

[aggle-rithm]

.
We aren't trying to design something new here we are trying to scale something that used to exist to try to figure out why it no longer exists.

I don't know about you, but if an airliner crashed into ME, I sure as hell wouldn't exist.

 

[femr2]

GlennB

Is that a fair representation of your 'wooden blocks' analogy?

Yes.

Heiwa

Are you really certain that it is the Lower Part that gets damaged at first failure? In my opinion based on long experience it would be the Upper Part!

In reality, both would be damaged in my opinion, however, it is a virtual model, with stated limitations. Divergence from 'reality' will always be present, even in the most complex FEA simulation, which my particular model is not. It is a relatively simple extension of the basic crush-down model, in a form that can be used to give users the ability to see what effect additional factors have upon the collapse time.

psikeyhackr

TONS of STEEL and TONS of CONCRETE on every level because we can't scale a model without that information

The concrete mass can be calculated to reasonable approximation. The floor-by-floor steel masses would be extremely useful, in order to calculate more accurate mass loss and resistance values.

alienentity

although if Ryan starts getting more responses, where are they going to be hosted?

He wants to have private email conversation. In that instance they would not be hosted anywhere at all. aka...pointless.

Myriad

A collision that includes no deformation of materials is physically impossible.

There is always deformation of materials. The only question is how much of the deformation is elastic and how much is inelastic.

I think we are all aware of that pure inelastic collision is improbable, however, Ryans base model uses inelastic collisions only, so I think the approach can, in a virtual model, be appropriately used, as long as the limitations are known. Also, collision between two titanium blocks at low velocity effectively results in no deformation at all...

---

So, again here, then, is the question...

In Ryans outline model inelastic collisions are used, with resistance posed by the structural supports.
Ryan is suggesting that the supports must be present for valid usage of inelastic collisions.

So what happens to Ryans model if the resistance is low ?

Scenario 1: The supports fail with 1000 GJ KE impact over their static load bearing capacity.

Inelastic collisions okay ?

Scenario 2: The supports fail with 1 J KE impact over their static load bearing capacity.

Still okay to use inelastic collisions ?

Also, what if the energy required to fail the supports is still quite high, but significantly less than the energy 'consumed' through the conservation of momentum calculation ? What is supposed to have happened to the energy 'left over' ?

The energy consumed during a conservation of momentum calculation is used in velocity change of the masses involved. eos.

 

[psikeyhackr]

I don't know about you, but if an airliner crashed into ME, I sure as hell wouldn't exist.

.
If you weight 400,000 TONS you had better do some SERIOUS DIETING.

psik

 

[Heiwa]

Heiwa

In reality, both would be damaged in my opinion, however, it is a virtual model, with stated limitations. Divergence from 'reality' will always be present, even in the most complex FEA simulation, which my particular model is not. It is a relatively simple extension of the basic crush-down model, in a form that can be used to give users the ability to see what effect additional factors have upon the collapse time.

Good! So in your model the red upper part will be damaged at first impact. I could not agree more.

 

[psikeyhackr]

In reality, both would be damaged in my opinion, however, it is a virtual model, with stated limitations. Divergence from 'reality' will always be present, even in the most complex FEA simulation, which my particular model is not.

.
That is the nice thing about PHYSICAL MODELS.

They cannot DIVERGE FROM REALITY. It is just a question of what the physical model demonstrates.

If only computer models can be MADE TO DO what happened to the WTC on 9/11 is that supposed to PROVE SOMETHING? So in order for a model to be correct it must conform to what some people have chosen to BELIEVE. Where is the data on the real building? We can't have decent computer or physical models without that.

We have to SCALE things for Mackey.

psik

 

[Jonnyclueless]

Yes and stacking pizza boxes or making a oscillating toy in your living room IS supposed to prove something. And you guys wonder why people laugh at you...

 

[GlennB]

Sorry to butt in, but could you clarify this, so that I can follow thie discussion better? You seem to be suggesting that a snooker ball in perfect axial collision with a second ball will 'join' it and the two will move along together? Is that a fair representation of your 'wooden blocks' analogy?

femr2 - You answered 'yes' . But snooker balls do not behave like this. When the cue ball hits the object ball square-on, the cue ball stops dead and the object ball starts moving with slight energy loss through deformation and attendant heating of the material (assuming no topspin or backspin on the cue ball, which is a reasonable assumption in this analogy).

So. What are you trying to demonstrate with the wooden blocks that supposedly 'weld' ? They won't weld any more than snooker balls, unless the wood is rotten and wet. Blocks of putty or dough might well 'weld', however, but then we're talking about adhesion afaics.

 

[femr2]

You answered 'yes' . But snooker balls do not behave like this. When the cue ball hits the object ball square-on, the cue ball stops dead and the object ball starts moving (assuming no topspin or backspin on the cue ball, which is a reasonable assumption in this analogy).

So. What are you trying to demonstrate with the wooden blocks that supposedly 'weld' ? They won't weld any more than snooker balls. Blocks of putty or dough might well 'weld', however.

I am explaining the energy exchange within a conservation of momentum calculation, as I think I have made perfectly clear.

You could ask the same question of Ryans' outline model, which also uses inelastic collisions.

Both are 'virtual' models, with known limitations, though of course my model is rather more advanced in terms of the factors and accuracy of source data used.

The discussion stems from Ryan attempting to imply that the energy used in an inelastic collision is used for deformation of materials, which is simply incorrect. I'll go back to the calcs, with the final 'nail in the coffin' included to prove the point.

38.67e6 kg drops 3.6576m impacting a floor of mass 2.47e6 kg at 8.47m/s
Energy 'usage' due to conservation of momentum = 83.33e6 J
Resultant velocity of whole mass = 7.96m/s

What we have is 2 bodies of mass changing velocity, both actions consuming energy.

1) 38.67e6 kg decelerating from 8.47m/s to 7.96m/s (5.01e6 J)
2) 2.47e6kg accelerating from rest to 7.96m/s (78.32e6 J)

Add them up, we get 83.33e6 J

What a surprise. No energy used in deformation of materials.

 

[Furcifer]

(snip)
Add them up, we get 83.33e6 J

What a surprise. No energy used in deformation of materials.

That 83.33e6 J is the energy used in deformation.

You must realize energy can't be destroyed, only converted into another form. If not deformation then where?

You really need to think about this before you answer.

 

[femr2]

That 83.33e6 J is the energy used in deformation.

You must realize energy can't be destroyed, only converted into another form. If not deformation then where?

You really need to think about this before you answer.

I don't need to think about it in the slightest. I have presented the absolute facts in the post you are responding to:

38.67e6 kg drops 3.6576m impacting a floor of mass 2.47e6 kg at 8.47m/s
Energy 'usage' due to conservation of momentum = 83.33e6 J
Resultant velocity of whole mass = 7.96m/s

What we have is 2 bodies of mass changing velocity, both actions consuming energy.

1) 38.67e6 kg decelerating from 8.47m/s to 7.96m/s (consumes 5.01e6 J)
2) 2.47e6kg accelerating from rest to 7.96m/s (consumes 78.32e6 J)

Add them up, we get 5.01e6 + 78.32e6 = 83.33e6 J

ALL the energy is used in change of velocity of the 2 bodies of mass. Simple physics fact. Zero used in the deformation of materials. End of story.