Hardfire: Physics of 9/11

[femr2]

6. Your mass numbers for the structure look pretty good. Have you considered varying the live load applied to the structure? NIST and some researchers, such as Purdue, use a relatively low figure as you do here, but there are others such as Dr. Quintiere who suppose the live load was up to three times as high. There is some support for either argument, and it is also suggested that NIST's live load refers only to accessible flammable materials rather than actual mass. I accept your assumption here as reasonable, but it's a parameter that remains largely unknown and has the potential to affect your results, so you may want to make it controllable as well.

The live loads in the spreadsheet are set to the maximum rated loads for all floors, and are also specified separately for the inner core and outside core areas.

(MAXIMUM Live Load as per ASCE 7-02 as applied to original design specification for WTC. NISTNCSTAR1-2A P102)

I feel that without further definite data, that the usage of the maximum design rated load is more than reasonable, however, as the data is housed in a spreadsheet, the loads can easily be changed by the user on a per-floor basis. (Sheet Tower Mass, Columns P thru V)

 

[femr2]

In summary, it would help if you explained your model in narrative rather than burying calculations in the spreadsheet. Nonetheless, it looks like you've done a good job on it, nothing a little documentation can't fix. You are one of the very few who has done so, and thus far yours is clearly the best entry in response to my "Hardfire Modeling Challenge." With a little more work we should be able to use this model to make some pretty good predictions and learn something, and that is the hallmark of a good model.

I fully agree that more documentation is required, though I would not agree to implications of 'burying' calculations. The calculations are quite necessarily inter-dependant but have been made as clear as possible by sheet separation and the inclusion of the parameters sheet.

I fully intend on documenting each and every calculation used, though until then it is of course simply a matter using the dependancy tools in excel for those curious about such matters.

Explanation relating to how users may change sources values they disagree with would seem to cover most of your concerns, although of course, we may disagree on the values individuals may choose to actually use.

As far as I can see, focussed discussion can only result in improvement to the current model, so that's more than fine by me.

We will obviously have totally different opinions on the cause of the actual events, but I assume you will accept that biased discussion and inference should play no part in discussion of the model.

 

[femr2]

Factors which work in favour of continued collaspe:

1) The 'Cap' mass is modelled as a single, non-deformable, solid, and the entire mass is applied to subsequent collision calculations.
2) All impacts are currently inelastic, ignoring the significant levels of energy absorbed by the structure in a more accurate elastic collision impact simulation. This will be addressed in a future update.
3) All descent calculations, including the initial drop, are calculated using standard vacuum free-fall equations. The initial drop and every subsequent drop after the impact of each floor, is performed as a free-fall descent. In reality this obviously was not the case, as the descent involved continual deformation of the building structure. Inclusion of inelastic collisions and steel structure deformation mechanics will help address this assumption.
4) The time implications of the deformation of the structure are not included.
5) The energy expenditure for the crushing of all materials other than concrete and steel collapse energy is not included. This includes dry wall, fixtures and fitting such as tables, computers, filing cabinets, and specifically the energy required to collapse each floor truss structure. (An almost endless list 'could' be compiled)
6) The effect of Cap tilting is not included, as this would require a finite element analysis rather than what is intended to be a physics/math model.
7) The initial drop assumes all core and external columns for the failure floor have 'vanished'. This works VERY much in favour of continued descent (Paper)
8) If energy expenditure during any impact exceeds available energy, the simulation is artificially allowed to continue, with the initial cap velocity being zero. The collapse failure floor is recorded and displayed on the Parameters sheet, and deficiencies are included on the videos by the value being highlighted in red. If this situation occurred in reality, collapse would stop. This behaviour is included to enable users to see the effect on energy deficiencies upon collapse time.
9) The maximum rated loads are used on all floors. No inclusion of energy expended in deformation of any of the live or construction dead load materials is included.

Sorry if anyone thinks I'm spamming. Just wanted the 'conversation' up to date...
That's all for now

 

[R.Mackey]

... and now you see why a forum discussion is inappropriate. That's the short list of my concerns.

I also need to point out that the responses are mostly wrong. In particular, the first one -- the assumption built into this model is that the kinetic energy loss after inelastic impact is all dissipated somehow without any of it going into deformation or destruction of materials. This is, to put it mildly, dead wrong.

But, adjusted correctly, this model will predict the right answer -- about 15 seconds to collapse. It actually supports the so-called Official Theory.

It'll be interesting to see if its creator ever realizes this, however.

By the way, the JREF forbids reprinting e-mails without permission. According to you, femr2, we never conversed before, so I never gave you that permission. I'll do so anyway now, though, just to keep you out of trouble. No hard feelings.

 

[femr2]

... and now you see why a forum discussion is inappropriate. That's the short list of my concerns.

I also need to point out that the responses are mostly wrong. In particular, the first one -- the assumption built into this model is that the kinetic energy loss after inelastic impact is all dissipated somehow without any of it going into deformation or destruction of materials. This is, to put it mildly, dead wrong.

I see a forum as no different to an ongoing email conversation other that the public/private factor.

Most of my responses provided clarity and detail about the mechanisms available, and cannot BE wrong. They are simply statements of fact.

The first, conservation of momentum...

Energy loss due to conservation of momentum is indeed treated separately to other energy sinks such as the 'destruction of materials'.

This is not incorrect.

Consider the virtual model of the tower as virtual slabs of titanium held aloft with no support (floating). A titanium mass is dropped and upon each impact an inelastic collision occurs and the impacted mass 'welded' as per the normal construct for the crush-down models. Conservation of momentum applied to this model will result in kinetic energy 'consumption', even though there is no resistance, no supporting structure to deform, and no destruction of the titanium slabs themselves.

The kinetic energy sink is into heat, sound and (primarily) acceleration of the additional mass, not destruction of materials.

There is no destruction of materials in the model just described.

(Thanks for your permission btw)

 

[R.Mackey]

You're welcome.

Yes, it's incorrect. This is the same fundamental error that Gordon Ross made. Also see here for a humorous, but correct, illustration of the problem (just watch out for the obvious typo). You can also look up the classic ballistic pendulum experiment. The loss of energy primarily goes into deformation of materials.

This really isn't negotiable.

 

[femr2]

You're welcome.

Yes, it's incorrect. This is the same fundamental error that Gordon Ross made. Also see here for a humorous, but correct, illustration of the problem (just watch out for the obvious typo). You can also look up the classic ballistic pendulum experiment. The loss of energy primarily goes into deformation of materials.

This really isn't negotiable.

Let's make it very simple, and use actual values.

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

If you want to apply that to deformation of materials, then the energy required to accelerate the impacted mass to 7.96m/s must be separated.

Energy required to accelerate 2.47e6 kg to 7.96m/s = 78.32e6 J

I'll quite happily separate that sink. Heat and sound will also then be specified separately.

I note you have ignored the model mentioned previously which includes no deformation at all.

 

[R.Mackey]

Let's make it very simple, and use actual values.

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

If you want to apply that to deformation of materials, then the energy required to accelerate the impacted mass to 7.96m/s must be separated.

Energy required to accelerate 2.47e6 kg to 7.96m/s = 78.32e6 J

Looks like you're missing the point, as expected.

Your numbers above are fine, but the 83.33 MJ is all energy that goes into deformation (and heat, which is primarily caused by friction, and sound which is almost negligible). But don't confuse yourself.

System before has 38 Gg x 0.5 x (8.5 m/s)² = 1.4 GJ of energy.

System after has 41 Gg x 0.5 x (8.0 m/s)² = 1.3 GJ of energy -- and that includes the energy to accelerate the previously stationary mass. That mass is now moving in the calculation above.

But energy is conserved. We've accounted for everything moving before and after. So that means the difference, in my rounded calculation ~ 0.1 GJ and as you calculated to higher precision correctly = 83 MJ, all goes into deformation (etc.). You do not subtract from this to accelerate the previously stationary mass. You've already accounted for that.

I note you have ignored the model mentioned previously which includes no deformation at all.

I've ignored nothing, I only want to do this one step at a time, for reasons which should be obvious from the above.

 

[femr2]

Looks like you're missing the point, as expected.

I've ignored nothing, I only want to do this one step at a time, for reasons which should be obvious from the above.

The point is, Ryan, that in the model described, AND in the calculation just presented (by me), there is NO material to deform. There is no resisting structure, there is no deformation of materials. The energy usage is still exactly the same.

So tell me, Ryan, given that there is no resisting structure, and no deformation of materials, where are you suggesting that the energy actually goes ?

 

[R.Mackey]

If there's nothing to deform, you are unlikely to have an inelastic collision.

You could construct a model that has no resisting structure and no deformation of materials, but I fail to see its relevance. This also does not excuse your reasoning error above.

 

[femr2]

If there's nothing to deform, you are unlikely to have an inelastic collision.

You could construct a model that has no resisting structure and no deformation of materials, but I fail to see its relevance. This also does not excuse your reasoning error above.

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.

 

[alienentity]

Good evening Gentlemen. Hadn't checked the thread for a little while, and I was surprised that Femr2 has made an appearance.

A few weeks ago Femr2 was advertising one of his videos about Lateral Debris Ejection, and upon checking his channel I discovered he didn't allow comments. I thought this rather annoying so I made a clone of the video and put some critical comments as annotations on it.

He posted a few times in defense, but I found that he was not prepared to accept any answer I offered, and was very evasive. I suspected he would keep that game going indefinitely so I moved on, not wanting to waste time trying to prove something to him which he would never accept anyway.

I feel Ryan has correctly identified Femr2 as a deceptive provocateur rather than a sincere student of the truth and fact. But that's just my opinion.

That's my two cents. I know only a fraction of the physics that Mackey does, I would be very, very interested if he and Femr2 were to engage in a serious discussion. I would learn something. Perhaps that is still possible.....

goodnight

 

[R.Mackey]

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.

You're just tapdancing, already. I wish I was surprised.

The simple fact is, in your collapse model, you incorrectly handle the flow of kinetic energy. In your example above, you incorrectly treat conservation of momentum and energy. Now you're trying to concoct a situation where things are totally rigid, yet collide inelastically. This kind of philosophical argument has no bearing on the system we're actually trying to study.

I suggest you start by trying to understand your mistakes above. They are quite glaring.

 

[femr2]

You're just tapdancing, already. I wish I was surprised.

The simple fact is, in your collapse model, you incorrectly handle the flow of kinetic energy. In your example above, you incorrectly treat conservation of momentum and energy. Now you're trying to concoct a situation where things are totally rigid, yet collide inelastically. This kind of philosophical argument has no bearing on the system we're actually trying to study.

I suggest you start by trying to understand your mistakes above. They are quite glaring.

There is no mistake. The model as described is exactly the same as you outline in your video, without any resisting support structure. The laws of physics quite happily cope with 'floating' masses, which when impacted are still governed by the laws of physics.

That you seem unable to comprehend this virtual scenario does not bode well for future discussion.

I have described that model, and presented the calculations.

Conservation of momentum calculations result in the stated energy expenditure with zero resistance and zero deformation, totally negating your incorrect statement. The energy is not used in deformation of materials, because there IS no deformation of materials. Simple physics Ryan.

 

[R.Mackey]

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?

 

[alienentity]

BTW Ryan, I am in the midst of reading your DRG paper, and appreciated the observations you made regarding large debris that was ejected far from the towers. In particular the invalidation of explosives as the cause of this ejecta because of the lack of 'shrapnel'.

 

[R.Mackey]

BTW Ryan, I am in the midst of reading your DRG paper, and appreciated the observations you made regarding large debris that was ejected far from the towers. In particular the invalidation of explosives as the cause of this ejecta because of the lack of 'shrapnel'.

You're welcome. I haven't had any comments on that paper for a while, certainly none from Dr. Griffin himself...

This effect is quite clear to anyone who knows anything about explosives. You don't move heavy things by planting bombs next to them, and shrapnel is extremely dangerous. This is why glass windows are removed from real controlled demolitions, even the ones that use weakened columns and minimal charges. One with bombs big enough to toss heavy columns around would have inflicted incredible death and destruction through shrapnel.

 

[femr2]

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?

Turn the model on it's side and make it two wooden blocks on a frictionless surface, one traveling at 8m/s and the other stationary. They impact and are 'welded' as per normal inelastic collision mechanics. Same process.

No deformation of materials, no resistance. Same energy usage.

I'm afraid if this is as far as we can go, then I'll leave it to others to continue this extremely simple process.

The bottom line is that I most certainly will not be accepting that the majority of the energy sink goes into deformation of materials. All models just described have no deformation of materials, but still 'consume' the exact same levels of energy. Full stop.

Will be back when you've thrashed it around for a while.

 

[R.Mackey]

Turn the model on it's side and make it two wooden blocks on a frictionless surface, one traveling at 8m/s and the other stationary. They impact and are 'welded' as per normal inelastic collision mechanics. Same process.

How are they 'welded' together? Without some capture mechanism, this simply will not happen. Instead, the situation you describe will result in an elastic impact. This is why first-year physics students do experiments with weights and air tracks.

Like I said, this discussion is nothing but empty philosophy as you try to cover your mistake. Post #887 above is your example -- you created it, no prompting from me at all -- and you solved it wrong. You tried to double-count the energy to accelerate the stationary mass. This is true no matter what excuses you make.

This is also an extremely simple physics problem, high school level in fact. I'm increasingly concerned that you can't grasp it.

The bottom line is that I most certainly will not be accepting that the majority of the energy sink goes into deformation of materials. All models just described have no deformation of materials, but still 'consume' the exact same levels of energy. Full stop.

Then you fail. I believe we've just found your Irreducible Delusion.

The problem you have, I am unlikely to solve through forum posting, or even the e-mail you strangely avoid (which would have saved you some embarrassment). I can only recommend consultation with an educator, or perhaps courses in physics.

 

[GlennB]

Turn the model on it's side and make it two wooden blocks on a frictionless surface, one traveling at 8m/s and the other stationary. They impact and are 'welded' as per normal inelastic collision mechanics. Same process.

No deformation of materials, no resistance. Same energy usage.

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?