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  1. jaydeehess

    jaydeehess Senior Member

    Thanks Jeffrey, Mick. I was rushed at the time with little time to check out the thread in full.
     
  2. jaydeehess

    jaydeehess Senior Member

    I put it as that Gage's cardboard box demonstration is as relevant to the collapse of the towers as is the colour of the WTC tower lobbys.
     
  3. Mick West

    Mick West Administrator Staff Member

    Still waiting for the magnets, but managed to get together an 8 floor model with thin walls on both sides.

     
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  4. Mick West

    Mick West Administrator Staff Member

    20160328-115101-qg9rg.

    My 200 magnets arrived, time for a little building (and collapsing)
     
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  5. NoParty

    NoParty Senior Member

    Okay...I may get dinged for being off-topic...
    but damn, Mick, these little experiments are so cool...
    they're a significant part of why I pay the big bucks to be on this site!!

    Screen Shot 2016-03-28 at 1.26.48 PM.

    (Can't wait to see how the magnets work...)
     
    Last edited: Mar 28, 2016
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  6. tinkertailor

    tinkertailor Senior Member

    They're beautiful.
     
  7. Oystein

    Oystein Active Member

    Until about 5 minutes prior to starting this Reply, I was convinced that having floor heights of only a foot or so instead of the original 12 feet is a serious scaling error - but it seems floor height doesn't matter - floor number does.

    That's when I went back to an Open Office spreadsheet I did a while ago to compute the fall time of a pancaking collapse. In my model, I disregard completely any structural resistance at the connections, columns play no role. Imagine that each floor is held up just barely, and is let loose at the exact instance that the falling stack of floors impacts it. Floors have zero thickness, and at perfectly inelastic collision, velocity changes instantaneously to conserve momentum.

    My spreadsheet computes, floor by floor,
    • accumulated mass (input parameters are mass of the initial falling debris as a multiple of single floor mass, e.g. 15 for 15 floors initially falling; and the mass of one floor, generally set to 1),
    • fall distance (a multiple of single floor height, which is an input parameter)
    • elapsed time (starting with t0 = 0)
    • velocity before the floor starts moving (initial mass starts at v0 = 0) and after momentum has been transfered in an inelastic collision
    • velocity after the fall through 1 story at g (g is a global input parameter, which I set to 9.805 m/s^2)
    • average local acceleration resulting from 1 fall and 1 momentum transfer
    • kinetic energy before and after fall and momentum transfer, and KE thus dissipated
    • average acceleration total from initial release to floor n.
    For example: If I drop an initial 15 floors, then 95 floors later (for a total mass of 110 floors) the local acceleration is 3.30 m/s^2, and the overall average acceleration has been 4.47 m/s^2, to result in a total collapse time of 11.56 s (About 3 s, or 35%, longer than freefall) - this for a floor height of 3.77 m (415 m / 110 floors).

    Then I change the height of each floor to just 1 foot
    (0.305 m), expecting that acceleration would be much lower, perhaps even dropping below 0 after not too many floors - but, surprise: After 110 floors, I got the exact same average acceleration!

    This result confuses me, but I have confidence in my spreadsheet formulas.
    How does this jive with your intuition? Or can you show analytically that floor height cancels out in the derivation of average acceleration?

    Note that, while acceleration stays the same, total collapse time of course decreases, to 3.29 s, because the 95+1 floors drop is only 29.28 m for 1 ft stories instead of 361.92 m for 3.77 m stories.


    Short story: It doesn't matter what the story height is in your model. If you measure acceleration for N stories in your model, this ought to be comparable to N stories in the real WTC if you start out with the right initial mass and drop height - and manage to tune the relevant other factors right (e.g.: Make collisions sufficiently inelastic).
     
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  8. Jeffrey Orling

    Jeffrey Orling Active Member

    He's not interested in demonstration acceleration or G or a fraction of G. The model was apparently created to demonstrate progressive runaway floor collapse/destruction and resulting column instability.
     
  9. Mick West

    Mick West Administrator Staff Member

    Absent any resistance, acceleration is the same over any height - it's just g. Each floor hit by a falling mass m will offer a certain amount of resisting force (f), so it seems like the net acceleration { calculated as height fallen / (0.5*t*t) } will be g-f/m

    EDIT: m is the falling mass, not the individual floor mass, so will vary through the fall, but still not affected by floor height.
     
    Last edited: Mar 30, 2016
  10. Mick West

    Mick West Administrator Staff Member

    Actually I am interested in demonstrating acceleration. If the initial mass only falls a distance of one floor before taking out the first floor, then acceleration is essentially inevitable, even if the mass did not increase (which it does).

    Some of the other models (like Cole's) require a starting drop height of several floors to overcome the fact that their models are rigged with a >g deceleration required to break each floor.
     
  11. Oystein

    Oystein Active Member

    The resisting force increases with velocity (I think). So a larger free drop distance ends with a larger v and thus with a larger f - and that may cancel out. Gotta meditate about that.

    (My model implies an infinite f actually, as delta-v occurs instantaneously; f/m would be infinite, too. Infininte force however applies along a distance of zero - the work done / energy dissipated turns out to be finite)

    My intuition told me that there would be more collisions per second, hence more f/m per second, more delta-v per second; and that this would decelerate more. But again, with smaller delta-v's, that apparantly cancels out exactly.
     
  12. Mick West

    Mick West Administrator Staff Member

    Ah right, I was thinking only of the energy required to break the connections. The floor still needs to be accelerated up to account for conservation of momentum.
     
  13. Mick West

    Mick West Administrator Staff Member

    So I built an 8', 12 floor model:
    20160330-172725-nup2w.

    But I'm having a lot of problems with the falling mass bouncing outside the plane of the building

    20160330-172825-533nd.

    I did it twice, the second time was bit better


    20160330-172955-hsflk.

    Half a collapse:
    20160330-173046-m7ucm.

    So need to regroup. Maybe do a simple constraint to keep things inside. Maybe even some kind of side wall.
     
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  14. Spectrar Ghost

    Spectrar Ghost Senior Member

    Plexi side panels?
     
  15. Steve Funk

    Steve Funk Active Member

  16. Mick West

    Mick West Administrator Staff Member

  17. Mick West

    Mick West Administrator Staff Member

    Because of scale. You can't build a scale model that will behave like the real thing, you can only demonstrate certain aspects of the collapse. Here the material of the columns and floors is irrelevant - it's essentially infinitely strong at this scale. What I'm really modeling is the strength of the connections relative to the weight of a floor, and to do that I need to make weak connections using magnets.

    The primary thing being demonstrated is that a stable structure can progressively collapse from the top down, and accelerate. Anything else that happens to match is a bonus.
     
  18. deirdre

    deirdre Moderator Staff Member

    you really only need 1 side. :)

    take the 12th floor off. hold the 3 wood planks (without magnets) at floor 12 (going in the same direction as the floors..not perpendicular) and drop. ??
     
  19. Oystein

    Oystein Active Member

    A couple of bamboo sticks on either side might suffice to funnel.
     
  20. Mick West

    Mick West Administrator Staff Member

    Yes, I was thinking something along those lines too, some skinny poles just to guide things inwards. A simple option is to have the back against the wall, so I would only need two poles.
     
  21. Mick West

    Mick West Administrator Staff Member

    Yes, I think having them parallel would probably work better, as more of the mass starts out in the plane, and there's less likelihood of pivoting.

    None of these things are actual problems in a full sized WTC tower, just aspects of the scale and limited dimensions.
     
  22. Jeffrey Orling

    Jeffrey Orling Active Member

    Mick.... I know this will make you a magnet magnet.... but what about having the floor made in perhaps 3 sections connected by the same weak magnets... What would the collapse look like? same initial falling mass... ht and so on..???
     
  23. Efftup

    Efftup Senior Member

    Well what you have demonstrated so far is the mechanism by which a progressive collapse is not just possible, but inevitable UNLESS more than a floors worth of material is ejected for each floor fallen.

    In order to more accurately show how the towers were built though you MAY have to have plexiglass walls replacing the current mdf ones, but ALSO build up a floor out of separate sections of wood with STRONGER connections than the floor connectors, so that you have FOUR walls and each floor is completely surrounding the column. Obviously this will take a lot ore building and I don't think I am explaining myself very well. I may have to try and do a drawing to add.
     
  24. Mick West

    Mick West Administrator Staff Member

    I think that would be more realistic in some aspects, allowing the floors to break up, I did do a quick test along those lines, but with a double width floor:


    Three 6' sections would probably be about right

    Making the entire thing out of those sections would be WAY too much work, but I might try to make maybe three floors on one side, just to illustrate floors breaking
     
  25. aka

    aka Member

    [red emphasis mine]

    Apologies for chiming in with a little OT question, but may I inquire where in the computational model you account for the possibility of, or why you would reasonably expect, a collapse arrest (a ≤ 0), if structural resistance at the connections and columns are completely disregarded and thus, an even infinitesimally small impulse suffices to "break loose" the next floor slab, which will then, inevitably, fall -- accelerated by gravity -- without any resistance or deceleration at all? How could collapse possibly arrest here?

    Allow me to point out you say the model computes
    However, in the computational model you describe, all the PE is converted into KE, and just the right amount of KE transferred (in an ideally inelastic collision) to the next floor slab - it does not become clear how any energy is dissipated, i.e. transformed in an "irreversible process" into "heat" (friction, deformation....). Without dissipation, no net deceleration, without net deceleration, no arrest.

    But all hope is not lost: a simple addition to the spreadsheet should allow the model to account for friction, dissipation of energy (be it a fall through air, olive oil or a steel structure) and thus allow for the possibility of an arrest.
    ...whereas the Metabunk Model faces two challenges at once: it must be "rigged" with a >g deceleration for it to stand up in the first place, and, simultaneously, also be "rigged" with a <g deceleration to ensure inevitability of progression once initiated; and this is where Archimedes' Law of the Lever comes in.

    Am I merely pointing out the obvious? Then I shall fall silent again, you'll find me back in my thread on Bazants and NISTs claim of "inevitability" in Rambles.
     
    Last edited: Mar 31, 2016
  26. Mick West

    Mick West Administrator Staff Member

    In an inelastic collision, momentum is conserved, but energy is lost. It's a rather unintuitive result.

    So, lets say one floor falls at V, hits another, if we ignore the supports then the result after impact is something between the top floor stopping with the lower floor continuing at V, and both floors continuing falling at V/2. In both cases all the momentum (m*v -> m*v or 2*m*v/2) is conserved. But only in the perfectly elastic collision is the kinetic energy (0.5*m*v*v) conserved, in a fully inelastic collision half the energy goes into internal friction and destruction.

    I'm not sure how this relates to my physical model though. The wood collision are relatively inelastic, it might be woth seeing if adding a layer of soft tape or cardboard makes any difference. Minor point though.
     
  27. Mick West

    Mick West Administrator Staff Member

    That does not make any sense to me. It's "rigged" to model slender sectional columns braced by relatively wide floors that can support a dynamic load of six times their own weight. The fact that this results in a very stable structure that is susceptible to progressive collapse in extraordinary circumstances (to scale) is a consequence of these factors.
     
  28. aka

    aka Member

    My educators took great care to make sure I never make statements like "energy is created" or "energy is lost", as it is unscientific. Energy is converted from one form into another. Where is energy "lost" to anything in an ideally inelastic collision? It isn't. The KE is partially transferred to the next floor slab.

    https://en.wikipedia.org/wiki/Energy
    It relates to Oysteins spreadsheet model of ideally inelastic collisions.
     
  29. Mick West

    Mick West Administrator Staff Member

    The kinetic energy is "lost" as heat. It's no longer kinetic energy, it's heat energy.

    Do you understand that in an inelastic collision, kinetic energy is always converted to heat energy?

    https://en.wikipedia.org/wiki/Inelastic_collision
    Bringing this from the abstract to the reality of the model, the degree of elasticity in a collision is a factor to consider here - both in what actually happens in the model (which I can vary by using more or less elastic materials), and in how the model differs from the original WTC towers.

    The real towers floors collision are inelastic in a much more complicated way than my simple blocks-and-magnets. They bend, they crack, concrete is shattered and crushed, metal is twisted and ripped apart. This is all part of the inelasticity of the collision.

    In my physical model there's really only three factors relating to inelasticity:
    1. The simple coefficient of restitution of wood for the floors (somewhere under 0.5)
    2. The magnet/friction "seat" connections.
    3. The bending of the thin outer walls.
    There are all kind of interesting issues here, but this thread is about the physical model. I think the overall response of a floor to collision is not unrealistically far off the mark.
     
  30. Oystein

    Oystein Active Member

    I expected this because, well, because.... my brain farte... OH LOOK, THAT BEAUTIFUL BIRD! *red face*
    I was a bit tired when I composed that post, I mixed in a consideration of structural strength of the floor seats with some factor of safety - my expectation is wrong, thanks for pointing this out.

    Yes, all PE is converted to KE.
    An ideal inelastic collision has all the mass travelling at the same velocity after impact. If you do the math, you will find that this results in KE after collision being less than KE before collision: Some KE is lost. What is it lost to? Yes, heat and material deformation. My model is agnostic to the question how energy is dissipated, I just note that it actually is, because KE in fact decreases. Do you need to see the math?
    I bet my sweet arse that, in the real WTC event, >50% of the energy dissipation is material deformation (concrete crumbling, rebar bending, truss seats shearing...). My model is agnostic here, and I see no problem with that.

    You are confusing what happens in the two phases of each floor collapse in my model:
    I have alternating phases of floor stack falling freely through the height of 1 story, at g, without any resistance at all, and inelastic collisions through a distance of 0 where Conservation of Momentum results, inevitable, in a loss of KE, that is dissipated. I need not introduce a term to add this dissipation - it is a result of the model!

    Ok, here is the math.
    Suppose we have a mass of n floors moving down at velocity v0 as the collision with a floor of mass 1, that is at rest, begins.
    KE(before) = 1/2*n*v0^2 + 0
    M (momentum) = n*v0 + 0
    After the collision, the combined mass n+1 continues at a new velocity v1, such that Momentum is the same:
    M = (n+1)*v1
    <=> v1 = M/(n+1) | substitute M = n*v0
    <=> v1 = v0 * n/(n+1)
    The new KE, after collision, is
    KE(after) = 1/2*(n+1)*v1^2
    = 1/2*(n+1)*(v0*n/(n+1))^2
    = 1/2*(n+1) * v0^2 * n^2 *1/(n+1)^2
    = 1/2* n^2/(n+1) * v0^2
    = 1/2*n*v0^2 * n/(n+1)

    If you compare KE(before) with KE(after), you will see that they contain the same term in blue, but KE(after) has an additional factor:
    KE(after) = KE(before) * n/(n+1)
    KE is lost!
    The difference, 1/(n+1) * KE(before) is dissipated, and we need not know immediately how.


    No, you are pointing out the fallacious. You equate static forces with dynamic response. A brick laying on your head vs. a brick falling on your head comes to mind. I hope this image points out the obvious.
     
  31. Oystein

    Oystein Active Member

    Note that, in the big WTC, material deformation (including material partition) clearly dominated over heating (increase of temperature), and also over elastic wave propagation. I don't have a reference handy, and can't prove this from first principles, so I am asking you to just believe this for the moment. Furthermore, the collapsing debris within the footprint loses some KE to material separated from the slabs and ejected or simply continuing to move beyond the footprint.
    Your model, in contrast, creates no permanent material deformation, and unless you lose whole slabs, you also don't shed mass.

    This is pretty irrelevant to your model at the current stage, but we should keep this in mind when we discuss the results and relate them back to the original. An important insight into the collapse mechanism is that
    - the KE lost in the collisions due to inelasticity is MUCH more than the energy required to fail the floor connections
    - a majority of the KE loss is available to crush the floor slabs - this explains adequately the degree to which the floor where ground to small fractions
    - the truss seat failures are a small part of the overall material deformation that dissipates KE - they are already included in the deceleration due to CoM and do not slow the fall further (unless their strength was at least the same order of magnitude that the collision forces were - but it wasn't).
     
  32. Efftup

    Efftup Senior Member

    you meant STATIC, didn't you?
     
  33. aka

    aka Member

    Ah, I understand now. The difference between static load and dynamic load never occurred to me.

    Thanks for the clarifications, everyone!
    I was obviously not speaking about the energy dissipation that is already implied by computationally modelling the collisions as perfectly inelastic ones. That's a given. That should be clear to everyone. "The difference, 1/(n+1) * KE(before) is dissipated, and we need not know immediately how." - but we do know how: to model it as a perfectly inelastic collision!

    I was talking about the energy that is, as you would say around here, "lost" to deformation to unseat the floors in a real tower in the real, physical world, be it in Mick's garage or in Lower Manhattan, where the floors do not magically float mid-air, and where ripping them off their columns requires (or should require) a little energy too, slowing down the progression, even if only a little, and it can easily be added as a factor to the computational model Oystein proposed..

    Anyways, I only had a question about Oystein's beautiful bird, and did by no means intend to distract from the ongoing experimentation of the physical model. Now that the math is known, the computation of "kinetic energy before and after fall and momentum transfer, and KE thus dissipated" actually makes sense.

    Only one small clarification about the >g and <g issue:
    That requires a deceleration >g.
    And this requires a net deceleration <g.

    That's all I was trying to say :)

    Carry on!
     
    Last edited: Mar 31, 2016
  34. Mick West

    Mick West Administrator Staff Member

    No, I mean dynamic - a load suddenly applied, just not dropped very far. Gradual loading is very hard at this scale (microscopic motions can be huge when scaled up), very careful application can get it up to 10.
     
  35. Mick West

    Mick West Administrator Staff Member

    No, it requires the system to be in equilibrium, if it were >g then it would fly off into space. This isn't something you have to carefully fiddle to get the "deceleration" correct. If you place one girder on top of the other then you've got a system in equilibrium.

    Which is trivially arrived at for the floors by dropping other floors on them, which then puts the whole system out of equilibrium.
     
  36. aka

    aka Member

    Are Cole's and @psikeyhackr's models on their way to Sirius A or what?
     
  37. econ41

    econ41 Active Member

    No. Heating is not relevant to this part of the collapse mechanism.

    The model is a visual demonstration of part of the "progression stage" mechanism. Heating of steel was not a factor for progression - what happened AFTER the Top Block started to fall.

    Heat is the key factor in the "initiation" stage. - the mechanism that caused/allowed the top Block to start falling.
     
  38. Mick West

    Mick West Administrator Staff Member

    I think this confusion is perhaps the results of inexact terminology. It's tempting to think of the towers as a homogenous 110 floors, each floor stacked upon the other. Hence each floor is supporting all the floors above it being pulled down at g, and since there's a safety factor in there, then each floor is capable of supporting all the floors above it with a force >g (which, absent an increase in gravity, effectively means having the upper floors dropped on them from a very short distance, like an inch or so)

    But in discussing my model I've used the term interchangeably as a measure of a step (a level) in the height of the building and as the description of the entire system between two levels, and as indicating a floor slab (i.e. the piece of the building that is between the columns.).

    Now in my building, as in the WTC towers themselves, a floor (in the sense of the entire floor system on one level) is perfectly capable of supporting the entire building above it at accelerations of g and >g.

    A floor slab on the other hand, is not capable of supporting the entire building above it, even at significantly less than g. The floor slabs are designed to support their own weight plus a dynamic live load, plus a considerable safety margin. A floor slab can support six other floor slabs of weight being suddenly applied. But the weight of the upper floor levels is not supported by the floor slabs, it's supported by the columns.

    The problem with @psikeyhackr's and Coles models (a problem I'm attempting to remedy with my model) is that they don't model this. The floors in some models are infinitely strong (like the washers, or these tile squares)

    20160401-105343-hum4r.

    Cole's other model is actually much better, in that the floors actually break.
    20160401-100742-lmpuz.

    But he's using floor slabs that are again incredibly strong. They weigh under a pound and he's using a 20 lb weight dropped from a height of about 20 levels (his levels are only about 4" apart).

    So what I'm attempting to do is something that is closer to the actual situation.
    • Floor slabs that can support a sudden application of 6 times their own weight, but will fail at 12.
    • A falling mass that's just a the weight of a few floors.
    • An initial drop height that's just 1 or 2 floor levels
    20160401-103157-f3u66.

    Look at the numbers in my setup, and compare them to Cole's "better" model. I'd originally though that particular drop test was about 4 floor slabs worth of weight for the four bits of wood, and about six for the large single piece. But it turns out they were about the same, both under 2.25 the weight of a slab. The problem was that my drop pieces were of much lighter wood than the floor slabs. So I'm actually being incredibly conservative here. And yet the dropping of the 670g piece of wood on the 300g floor slabs, from a height of just two levels, resulted in a total collapse of that side.

    20160401-104326-ftytm.

    So I think in terms of modeling the actual situation here, I'm being very conservative (the falling mass was much greater in the real situation). I'm also dealing with a greater shedding problem due to my 2.5D model. And yet it still collapses.
     
    Last edited: Apr 1, 2016
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  39. Svartbjørn

    Svartbjørn Senior Member

    I know you're looking at this in the terms of re-usability.. but would it be a worthwhile experiment to actually light the thing on fire and let the damage caused help re-create the collapse? Is that even feasible?
     
  40. Mick West

    Mick West Administrator Staff Member

    Not really, I'm really modeling the progression of the collapse in the portions of the building that were not damaged by fire. The fire relates to initiation, not progression.

    And scale rears its head again here. The fire was burning from jet fuel and office contents, and it damaged non-flammable steel in a very localized manner (i.e. over a few floors). Here I have no office contents to burn, and setting the floors and columns on fire would just engulf the entire thing in fire before any substantive damage could be done.
     
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