C1 being slowed down and A97 being accelerated, while all the supports are being compressed? Yes, damping occurs, the collision is VERY elastic, but all of A97-A1 are then being accelerated for a short while, like all of C1-C13 are being slowed down for a short while ... first to zero velocity and then with an upward velocity, when/if bouncing takes place. Where do you get C1 being slowed only by half from? Do you think little C can one way crush down big A. Sorry, does not happen! How are you getting along with your design to prove the opposite?
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Proposed Design for Progressive Collapse Demonstration
- Thread starter Myriad
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C1 being slowed down and A97 being accelerated, while all the supports are being compressed? Yes, damping occurs, the collision is VERY elastic, but all of A97-A1 are then being accelerated for a short while, like all of C1-C13 are being slowed down for a short while ... first to zero velocity and then with an upward velocity, when/if bouncing takes place. Where do you get C1 being slowed only by half from? Do you think little C can one way crush down big A. Sorry, does not happen! How are you getting along with your design to prove the opposite?
bill smith
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FineWine can you honestly say with your hand on your heart that you don't see any of the things that the narrator in this video describes- as he describes it ? What DO you see ?
http://www.youtube.com/watch?v=dtx_GcFCs6c&feature=channel_page
Dave Rogers
Bandaged ice that stampedes inexpensively through
There's a useful concept called "the elastic limit" that modifies this situation rather critically anywhere other than Heiwa-world. I'm not sure whether Heiwa's decided that elastic limits don't exist, or decided that the impact forces aren't great enough to exceed it, but if it's the second option it might be worth pointing out that he had to use the smallest possible spring constant, divide the potential energy by two for no reason, and mis-calculate the elastic compression of the lower block, to come up with that result; if you get it right, there's enough PE from the fall of the upper block through a single storey to exceed the elastic limit of every column in the lower structure, and that means that there is no bounce.
Dave
I'm using your "funny m" model. A falling stack of masses "C" connected by springs lands on another stack of masses "A" connected by springs. Here's a quiz for you: at the moment the top of A and the bottom of C meet, how much are the springs in C being compressed? How much force is being applied to decelerate the floors above? I'll give you a hint: it's a very round number!
How much are the top springs in A being compressed? It's k times the weight of the top floor, plus some value that is the contribution from the collision. What is that additional value at the moment the floors meet? Hint: it's another very round number!
The floors collide inelastically C is moving at some velocity v, the bottom floor of C is slowed to v/2, the top floor of A is accelerated to v/2. Half the kinetic energy of the bottom floor of C is lost in the process. Want them to collide elastically instead? That's even worse, then less energy is absorbed and the top floor of C is moving downward even faster.
Springs don't transfer any force until they are already compressed. That happens after the floors collide. By then it's too late. The next floors in line are already on collision course. No way for the springs on either side to apply enough force over enough distance to stop them. The equation is a = -kx / m. Once the spring compression x is big enough it can bring the moving mass to a stop. But oops, making x big enough is what makes the supports break. Then you have another story to fall. Nothing comes to rest. Nothing bounces.
The only way to save your building is the way you did it in the funny m model: change your mind and build it out of rubber! Build it so that it sags under its designed static load by 3%, and can compress 10% of its height without breaking! That's easy for an inflatable building. The kids will love the bouncy floors, especially when the wind blows. But your grown-up tenants might complain, why do their filing cabinets keep tipping over? Why don't the elevators ever work? But you're right, it is hard to make an inflatable building progressively collapse. Unless there's a leak. Is your oil tanker design so safe because it's inflatable too? Tell the captain to watch out for narwhals. ("Too late, he got us! All hands to the emergency bicycle pumps!")
My demonstration design is coming along well, thanks. Since you changed your mind about offering a prize (and you keep dodging the question of what odds you want for a wager), I have to scale down. But analyzing your funny m model is showing me how to do that. It shows that even though greater h and m do make progressive collapse more likely (so large models are better), it can still be shown in a smaller model if the stress-strain curve of the materials is right. Something close to your original funny m model parameters, the realistic values for a skyscraper, .001h for the static load strain, .003h for the elastic limit, .005h for the breaking strain. Not the parameters you switched to when you realized realistic ones give you progressive collapse when you drop 2 floors onto 22 floors, or 14 onto 96. Not the rubber building pizza box values. It's hard to match the realistic skyscraper properties on a smaller scale, but I'll manage. I'll find something with the right properties at the smaller scale.
I'm developing some computer models to test various cases. It's bringing back memories of my high school computer science projects in numerical analysis. Heun's Method! The 4th order Runge-Kutta algorithm! Computers were slow back then, I'm sure I can do a lot more now!
But with the prize offer turning out to be false, I have to do all this in my spare time.
To give you something to read while you're waiting, I'll soon post further analysis of your funny m model (no computer simulations yet, just math and physics) proving that progressive collapse occurs for the n = 22 (2 dropped on 20) scenario.
Respectfully,
Myriad
BigAl
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More claims of silent demolition ?
Actually the springs in part A are always compressed a little to keep the horizontal elements in position.
The springs in part C during free fall are not compressed at all as no forces are acting on them (all masses in C are falling).
When bottom C m contacts top A m (assuming no springs are interfering between) energy is applied to all springs and forces develop in all springs.
To avoid just a bounce, the energy is assumed to be sufficient to produce a failure in the structure! So what element fails first and where? A spring or a connection of a spring to a horizontal m element?
In either case, you must ensure that all springs between horizontal elements or all connections between springs and one horizontal element fail simultaneously by gravity, otherwise the local failures will be unsymmetrical and the elements will displace sideways; further loads may then be directed outside the structure and just applied on ground, i.e. are lost to assist in the one-way crush down.
In CD it is easy to ensure that elements/connections fail simultanously and excessively to ensure CD. With only gravity available it is not so easy to ensure a one-way collapse. Gravity forces have a tendency to take the easiest route down, e.g. slide off an element in the way in lieu of crushing it.
Look forward to your final design.
Oh dear oh dear oh dear.
No, the forces do exactly what they are destined to do according to the laws of physics.
BigAl
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As someone who thinks in terms of momentum, vectors and cascading failure, the above makes no sense whatsoever.
Good, put your momentums, vectors and cascading failures together in a real structure A+C (C = 1/10 A and is carried by C) and C is dropped on A. If your part C structure then manages to make sense, i.e. to one-way crush down the A structure, i.e. break it into pieces, you are a genious. Such a structure has not yet been designed ... even if Myriad is trying hard with my assistance and encouragement.
You see, gravity and associated reaction forces applied on a structure do exactly what they are supposed to do when failures (or a part C drop) occur. They, e.g. forces applied by A on C, damage and stop part C. Part C has no chance to crush down part A. Part C gravity forces applied on part A are wasted on local failures, friction, or just applied on the ground, etc.
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BigAl
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To the degree that any of the above is valid, it depends of the values of the coefficients and parameters and generic handwaving isn't very interesting.
People that say "must" and can't" without showing their math are frequently making it up as the go in my experience.
bill smith
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Hey Al....Do me a favour.Here's a question. Tell me where 'Smith's Law' is wrong. ?
Smith's Law
''Whatever downwards force the moving body exerts on the stationary body of identical construction fixed in the ground is reciprocated by the stationary body equally and oppositely. After that it depends which body is used up first.''
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BigAl
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I hate to break this to you, that is nonsense.
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If I might interject?
It's the "used up" part where you're going wrong. Wreckage from A becomes part of the falling mass, C.
Debris from A in the collapse zone is not "used up" unless all of it is ejected. It isn't. It accumulates, accelerates and adds to the subsequent destruction of the remains of A.
Hope this helps.
alexi_drago
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Are you trying to suggest that many falling masses would have the same energy as one single large falling mass that has a mass equal to the sum of the many?
Someone who doesn't understand anything about physics is trying to formulate a universal law. The irony is lost on you.
The thirteen collapsing floors hit ONE floor, not the entire building. Next, fourteen collapsing floors hit ONE floor. And so on.
bill smith
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Hi FineWine. Have you seen the video I specially selected for you above ? I would be very interested in your considered judgement on it in the context of 'what DO you see ' ?. Please- a real answer.
BigAl
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The question is, why do you think it is anything but collapse induced by impact damage and fire and lack of firefighting?
Heiwa's "Funny M" Model Confirms Progressive Collapse
Heiwa has posted an essay on this Web site, describing a model of a structure comprised of floors of equal mass m, each supported on four vertical springs of relaxed length (height) h connecting with the next floor down (or the ground, for the lowest floor). The masses of the springs/columns are considered to be negligible. The number of floors is not specified, but all the examples in the essay use 22 floors so that is what I will use. The elements are numbered from the top down, so for instance M1 refers to the top floor and S1 refers to the springs supporting M1 above the next floor down.
Most of the model's parameters are left unspecified. The description does not specify the mass of the floors, the height of the floors, or the spring constants of the springs. We must simply refer to the uncompressed length of each spring as h, essentially making h our unit of measure (for instance, the total height of the structure with 22 floors, not accounting for compression of the springs under the static load, is 22 * h). Similarly we make m, the mass of one floor, our unit of measure of mass. For instance, the total mass of the structure with 22 floors is 22 * m.
Unlike the masses and (uncompressed) heights, the spring constants of each floor are different. However, they are all multiples of the spring constant of the top springs (S1). Considering the springs of S1 as a single spring with a constant four times the constants of each of the four springs, as per the rule of springs in parallel, I'll use k to designate that constant. (To indicate the spring constant in generic equations, I'll use capital K instead.) The spring constant of S2, for instance, I'll refer to as k2 and its value is 2*k, The spring constant of k3 is 3*k, and so forth.
k15 = 15*k
kn = n*k
Similarly, the notation xn will refer to the compression distance of a specific spring below floor n. To refer to the compression of a spring in generic equations, I'll use capital X instead.
The total potential energy of the structure with 22 floors, not accounting for spring compression under the static load, is (1 + 2 + ... 22) * h * m * g, or 253*mgh. Note that we don't know the actual value of g, because g would be earth's gravity expressed in our h units (and any convenient time unit such as seconds). But we do know that g is a constant, so we can compare any coefficient of mgh with any other. In other words, the unknown but constant magnitude of mgh can serve as our energy unit.
Similarly, the unknown but constant magnitude of mg can serve as our unit of weight and other forces.
The Model's Parameters
The only quantitative values given in the model are three dimensionless constants related to the performance of the springs/columns under load. The first, which I'll call C1, is the fraction of h that each floor of the structure compresses under the static weight of the floors above. The second, which I'll call C2, is the fraction of h that any of the model's springs can be compressed before it reaches its elastic limit. The third, which I'll call C3, is the fraction of h that any of the model's springs can be compressed before it reaches its plastic deformation limit and breaks.
Heiwa has provided two sets of these constants.
Original publication:
C1 = 0.001
C2 = 0.003
C3 = 0.005
Revised values that were posted later:
C1 = 0.03
C2 = 0.09
C3 = 0.1
At first glance, it might appear that there are too many unknowns in the model to determine much of anything about it, let alone whether it would progressively collapse. But as it turns out, there is enough information to prove that the model structure will progressively collapse, when evaluated with the original values of the parameters C1, C2, and C3.
The suggested parameter values in the current revised version of the paper can be disregarded. No real conventional structures, and certainly not the World Trade Center towers, sag anywhere close to 3% of their height under their own weight, or can be compressed by 10% of their height before they would break. (By "conventional" structures I mean structures built of ordinary building materials such as wood, concrete, and steel, omitting such special cases as inflatable structures, straw-bale or sod construction, sand castles, and so forth.) For instance, if the load-bearing walls of wood-frame houses compressed 3% under their expected working loads, a vertical 8-foot framing stud pulled out of a wall would spring back (lengthen) 2.88 inches. Imagine trying to squeeze a two-by-four hard enough to shorten it 2.88 inches, to get it back in there.
So, we'll work with the sensible original values of C1, C2, and C3.
Spring Constants
The first thing to do is derive the spring constants and other information related to the springs in terms of our unit m and h values. The equation for the relation between force and spring displacement is
F = -K * X where x is the distance the spring is shortened (or elongated). (eq. 1; Hooke's Law)
Parameter C1 tells us that the weight of n floors (equalling n * m * g), loading the nth spring down (Sn), whose spring constant is n*k, by .001*h. This results in each spring (and the entire structure as a whole) being shortened by ,001 of its height. Putting these values into eq. 1, we have
mg*n = -k * n * -.001h
k = 1000*mg/h
(Note that henceforth, I'll ignore both of the negative signs, and treat downward/compressive forces on the springs, and the change in length of the springs when compressed, as both positive in sign. The results will not be affected.)
The elastic strain energy stored in a static compressed spring, within its elastic limit, is
0.5*K*X2 (eq. 2)
So, for spring Sn, whose displacement x * h is less than or equal to C2 * h, the energy stored is
0.5 * n * k * x2 * h2
and substituting our known value for k we have
strain energy = n * 500 * x2 * mgh (eq. 3)
We can use this to determine how much elastic strain energy is in all the springs, when the structure is static. We know that x = C1 in that case.
Total elastic strain energy =
sum from n = 1 to 22 of [n * 500 * C12] * mgh (eq. 4)
= (1 + 2 + ... + 22) * 500 * C12 * mgh
= 0.126 * mgh
Remember that 1 * mgh is the potential energy from one floor mass falling one uncompressed floor height. The total elastic strain energy at rest is about 1/8 of that, or 1/2000th of the total gravitational potential energy of the structure (which is 253 * mgh).
We can also determine the reduction in the gravitational potential energy of the masses from the compression of the springs. The springs shorten by .001 * h, so each mass loses .001 of its total height off the ground. So, the potential energy loss is .001 * mgh , or 1/1000th of the total gravitational potential energy of the structure.
Why doesn't the decrease in potential energy equal the strain energy? That is, why isn't all the lost gravitational potential energy from the sagging of the masses stored as elastic strain energy in the strings? Because we're describing the structure at rest, which means that half the gravitational potential energy from the dropping masses must have been lost to damping effects (such as heating of the springs and moving the surrounding air) in order to bring it to rest. If we raised all the masses in the structure up to their full h and let them go and there were no damping losses, the (assumed perfect) springs would just keep oscillating forever. To damp it to a static condition, energy is lost. (If it weren't for the lost energy, the elastic strain energy could just turn back into gravitational potential energy by having the floors spring back up again, which is what happens when it's oscillating.) Of course, in reality, that damping would occur a little bit at a time as the structure was built.
Strain Energy Capacity of the Springs
We can also derive two more energy values: the maximum amount of elastic strain energy that a given spring can have before it goes beyond the elastic limit, and the additional amount of energy of plastic deformation a spring can absorb before it breaks.
The maximum elastic strain energy that a spring Sn can have is
0.5 * n * k * C22 * h2 (eq. 5)
substituting our value for for k
= n * 500 * C22 * mgh
= n * .0045 * mgh
To derive the maximum energy a spring can have before it breaks, we have to make some reasonable assumptions for how it behaves at compression distances between C2 and C3. (The model omits this information.) One reasonable assumption is that the resistive force remains constant at the value it reaches at C2, which is
n * k * C2 * h
= n * 1000 * C2 * mg
= n * 3 * mg
... so the total energy absorbed in plastic deformation is that force times the distance C2 to C3, or
n * 3 * mg * (.005 - .003) * h
= n * .006 * mgh
But just to make sure we're not underestimating any energy sinks, let's instead assume that the springs are made of super tough stuff that continues to resist deformation with its characteristic Hooke's law force (that is, the force with which it resists further deformation = -kx with the same k as for its elastic region) as it strains plastically, right up until it breaks. (This also makes some of the subsequent calculation easier.) So the energy absorbed in plastic deformation is instead
0.5 * n * k * (C32 - C22) * h2
= n * ,008 * mgh
The total amount of strain energy to break a spring is the sum of the elastic strain energy plus the plastic deformation energy. So,
Energy to break spring Sn = n * (.045 + .008) * mgh = n * .0125 * mgh (eq. 6)
Note that this makes the assumption, in favor of collapse arrest, that the elastic strain energy in the springs is "lost," when it is actually a form of potential energy that is still available to do work on the masses it's connected to. We will nevertheless assume that when a spring breaks, its elastic strain energy is expended in a way that does not help contribute to further collapse, such as propelling the broken spring parts sideways away from the structure.
The total amount of energy required to break all the springs is:
(1 + 2 + ... 22) * .0125 * mgh
= 3.16 * mgh
That's more energy than the gravitational potential energy of two floors' mass falling one floor's height (which is 2 * mgh). So does that mean Heiwa is correct? If we remove the springs S2 and allow M1 and M2 to fall onto M3, will the springs absorb all the energy and the falling floors just bounce off?
Not so fast. The problem with that conclusion is something Heiwa himself has taken pains to point out repeatedly: the energy doesn't get precisely distributed to all the springs in direct proportion to their strain energy capacity. Instead, the weaker springs tend to fail first!
Also note that while the total energy required to break all the springs is greater than the gravitational potential energy converted by two floors falling a distance h, it's less than the total gravitational potential energy of the structure by a factor of 80. That suggests that while two floors dropping cannot cause an instantaneous global failure of the structure, a progressive failure resulting ultimately in global failure remains a likely possibility.
What we have to consider is, how much downward-directed kinetic energy applied to M3 can the lower 20 floors absorb before one of its springs breaks, and which spring breaks first?
How much energy from downward force acting over a distance can the lower structure absorb in strain energy?
To compute with any precision how much energy the springs can actually absorb under the impact of falling masses, we'd need more complete descriptive data than the model includes, and then to perform fairly complex dynamic calculations or use a computer model. However, we can more easily compute an upper limit for that value, that is, the amount of strain energy the lower structure can absorb under the very best possible conditions for absorbing the most energy possible.
This approach is a bit like testing the weight of cargo that, when carefully loaded a little at a time, will cause a ship to sink while motionless in glassy calm water, in order to show that the same ship cannot carry more than that weight of cargo without sinking if the load were dropped into the hold from a helicopter in one big crate while out at sea during a hurricane. Obviously, it's going to be a significant overestimate of the actual capacity.
In the case at hand, the best possible conditions are created by applying a slowly increasing force gradually compressing the structure (that is, acting over a distance), slowly enough that all oscillations can be assumed to be damped and slowly enough that the inertia of the floors has no effect on the distribution of strain.
It is easy to see that both of those conditions are necessary for storing the most strain energy possible in the springs. If the blow from above is faster so that inertia comes into effect, the inertia of the upper floors limits the rate at which strain can be transmitted to the lower springs, because doing so requires the floors to be accelerated. So if we find that the upper springs fail first under the ideal conditions of very slowly applied force, we can be certain that they would have an even greater tendency to fail first with a more rapidly applied force. And if oscillations are not damped, oscillations will cause springs to be compressed to their failure points sooner than would occur when everything stays damped to steady state. The damping itself is an energy sink.
(As an example showing why this is the case, consider spring S1 on a rigid base, for which K1 = k, which is then suddenly loaded with a mass of 3 * m. We know that in a steady state condition after damping, if the spring were compressed by a force of 3*mg, the spring would be compressed just to its elastic limit at x = C2 * h. But if that weight were loaded onto the spring all at once with no damping, the mass would still be moving as x reaches C2 * h, and its inertia would keep it in motion, further compressing the spring until x reached 2 * C2 * h. (If all the strain were elastic, it would then continue to oscillate sinusoidally between x = 0 and x = 2 * C2 * h.) But since C3 < (2 * C2), the spring breaks first. So while the spring can support a weight of 3*mg steady state without any plastic deformation, it deforms and then breaks if that much weight is "dropped" onto it from a height of zero.)
Properties of the Lower Structure after Removing M1 and M2
In the drop test, we would begin by holding M2 in place and removing spring S2. This leaves us with a "lower structure" of 20 floors, each of mass m, connected by springs with spring constants ranging from 3*k for S3 (now the top springs of the lower structure) to 22*k for S22.
The lower structure now has 2*m*g less load on it before, so the springs relax somewhat from their initial static state of being each compressed by C1 * h. The masses rise a little as a result. We can calculate the new total strain energy of the lower structure as the sum
strain energy = sum for n=3 to 22 of [0.5 * n * k * xn2] * mgh (eq. 7)
where xn = h * (n - 2) / (n * k)
= 500 * sum for n=3 to 22 of [n * ((n - 2) / (1000 * n))2] * mgh (eq. 8) which results in .0894 * mgh That's about 29% less than the original structure, because we've removed two springs from consideration entirely, and reduced the load on each of the other ones by 2 * mg. In the original structure, M3 is at a height of 20 * h * (1 - C1) = 19.98 * h. With the top two floors removed, the new neutral position for M3 is 20 * h - sum for n=3 to 22 of [h * (n - 2) / (n * k)] (eq. 9) = 20 * h - sum for n = 3 to 22 of h * (1/k - (1/n * 2/k)) = 19.9844 * h The displacement of M3 from the ideal position of 20 * h is .0156 * h, about 22% less than in the full 22-floor tower. Maximum Strain Energy that the Lower Structure Can Absorb Now, for our best-case spring energy test, we begin applying a slowly increasing downward force on M3. This causes the entire structure to displace downwards until something breaks. However, the different M's do not displace equally, nor in proportion to their height off the ground, because each has a different amount of force pressing it down and each has a different spring constant for the spring holding it up. We have to consider what's happening to each spring. Call the extra force applied to M3 at any given moment F * mg (so that F = 1 represents the weight of one floor). For our best-case test, everything is slow and damped, so we can evaluate it as series of steady-state conditions. At any given moment, the force on m3 equals F * mg. The force on spring S3 equals F plus the weight of M3, or (F + 1) * mg. The force on spring S4 equals F plus the weight of M3 plus the weight of M4, or (F + 2) * mg. In general, the force on Sn = (F + n - 2) * mg. The spring constant kn = k * n = n * 1000 * mg/h The quotient of force over spring constant gives us the strain (compression distance) of each spring, which is xn = h * (F + n - 2) / (n * 1000). This can also be expressed as xn = h * (F - 2) / (n * 1000) + .001. (eq. 10) Note that when F = 2, equivalent to slowly loading the top two masses back on, this gives us a strain of .001 * h on every spring, just as we would expect given the original starting conditions for the 22-floor structure. We can see in eq. 10 that for all values of F>= 2, the larger n is, the smaller the strain xn. So, we can now be certain that despite the extra weight pushing down on the lower springs, under these ideal conditions the top spring of the lower structure, S3 will reach x = C3 * h first, and will break first. The total amount of strain energy that the springs can contain when that happens is the absolute theoretical limit, under ideal conditions and all assumptions favorable to collapse arrest, of the amount of energy that the springs can absorb in strain before S3 breaks. Solving eq. 10 for F when n = 3 and x = C3 = .005, F = 14 * mg So S3 breaks under the additional load of 14 * mg (the weight of 14 floors), applied to the top of M3 (or a total load of 15 *mg, the weight of 15 floors). We can now calculate and sum the total strain and total strain energy of all the springs when F = 14 * mg and a steady state exists. The total strain energy is the sum for n=3 to 22 of [0.5 * n * k * xn2] * mgh (eq. 11a) where xn = (F + n - 2) / (k * n) (eq. 11b) and k = 1000 The sum calculates out to 0.5227 * mgh. But before applying any additional F to the 20-floor lower structure, we started with .0894 * mgh of strain energy already in the springs, so the additional amount that can be absorbed, under these absolute best-case conditions, is .5227 - .0894 = .4333 * mgh. Dropping M2 Alone on the Lower Structure We can now easily show that dropping only M2 onto the lower structure will cause springs S3 to break. That's enough to prove that dropping the entire upper unit M1 + M2 + S1 must also cause springs S3 to break. We don't yet know exactly what M1 and S1 will do (we'll consider that later) but even if they magically evaporate into weightless dust the moment M2 touches M3, contributing no energy at all to the collision or subsequent events, S3 still breaks and a progressive collapse still ensues. Because of the readjustment of the springs under load, the distance from M2 to M3 is a little less than h, it's about .995 * h. That gives it .995 mgh of potential energy converted into kinetic energy before it reaches M2. We assume a fully inelastic collision between M2 and M3. This is another worst-case assumption against progressive collapse. It means half of the kinetic energy is immediately expended in inelastic deformation of M2 and M3 (whereas a more elastic collision would transfer more energy to M3 and a fully elastic collision would transfer all the energy to M3). That leaves .498 * mgh of kinetic energy. As established above, the maximum amount of strain energy the springs in the lower structure can absorb is .433 * mgh. Not enough. S3 must break. Note that the collision can be "spread out over time" due to imperfect contact, deformation of the M's during the inelastic collision, or any other reason, and it doesn't matter. The determination of the maximum energy the lower structure can absorb already assumes an ideal case where the application of force is spread out over essentially infinite time and everything is perfectly damped. Since the actual collision is faster and less damped than that, the actual amount of the energy the springs can absorb will be less. Also, that's without any contribution from M1. But in fact, when M2 collides and slows down, the momentum of M1 keeps M1 moving toward M2, compressing springs S1 which transfer additional energy to M2. Exactly how much doesn't matter, because M2 by itself already had enough kinetic energy to break S3. What happens then? That first impact seemed kind of a close call. (At least, until we remember that we way overestimated the ability of the lower structure to absorb strain, and that we ignored the contribution from M1.) There's not much energy left over. But now M2 and M3, having collided inelastically, fall together onto M4. Assuming they have no kinetic energy left over when S3 breaks, they still fall onto M5 with close to 2 * mgh of kinetic energy after falling another distance just short of h. This time the inelastic collision with M5 only absorbs 1/3 of that kinetic energy instead of 1/2, leaving 1.33 * mgh that would need to be absorbed by the lower springs. Those same lower springs have already absorbed most of the energy of the first collision, already bringing M4 fairly close to its breaking point. (You can consider the possibility that the springs below may, depending on the actual values of h and m, have had time to spring back up, but in that case the energy of those oscillations is still in play and must still be considered, and any plastic deformation is permanent). Also remember that it only takes 3.16 * mgh to break all the springs in the entire original structure.) Repeat the methods and calculations above could determine the exact values, but it's clear that M4 is going to break. This time there will be plenty of energy left over to contribute to the next collision with M5. It can end only in global progressive collapse. We still don't know what happened to M1 and S1. It doesn't actually matter. But let's examine the question next. Does S1 Break? To address this question, we'll have to depart from the realm of best-case energy transfer calculations, and evaluate the dynamics of the event. The problem is, while the M1 - S1 - M2 system is fairly straightforward, two falling masses connected by a spring, the lower structure, of 20 masses connected by 20 springs of different coefficients, Equations for the dynamics of the whole system, even if they could be fully derived, would be rather opaque and not likely to help in intuitive understanding of why the system behaves as it does. Instead I'm going to again examine a key limiting case. Immediately before the collision of M2 and M3, M1 and M2 is falling at a velocity we will call v (which happens to be sqrt(mgh)), and M3 is stationary. Immediately after the collision, before large forces have had time to develop in the springs and before the spring forces have had significant time to act on the masses to change their velocities, M1 is still moving downward at v, while M2 and M3 are moving downward at v/2. They are moving closer together at v/2, which will compress S1. For S1 to survive, their relative movement toward one another must stop and reverse before the spring is compressed beyond C3 * h. M1 gets accelerated upward by the spring force of S1 and downward by gravity. M2 gets accelerated downward by the spring force of S1 and by gravity, but upward by the spring force of S3, at least until S3 breaks. Clearly any force contributed by S3 adds upward acceleration to M3-M2 which helps compress S1. So we can examine a limiting case in which S3 makes no contribution, as if it disappeared at the moment of the collision of the floors. If S1 breaks under that ideal starting condition, it must also break with S3 present (regardless of when S3 breaks). Without S3, we have a much simpler initial condition: two masses in freefall, connected by a spring, with the top one moving downward at v and the other moving downward at v/2. Because gravity accelerates both equally, we can ignore it, and we can shift the system into the reference frame of the center of gravity of the two masses. In that reference frame, we have a simple harmonic oscillator with a fixed center of gravity, starting in mid-cycle (x = 0), with M1 moving toward the center of gravity at v/3 and M2-M3 moving toward the center of gravity at v/6. This, it turns out, is way too high an amplitude (or way more kinetic energy, however you choose to calculate it) than S1 can withstand, by a factor of more than 13. So, now we know: S1 does break, within a moment before or after S3 does. Note, however, that this result has not been shown to apply generally to the initial collisions in other cases (such as dropping 14 floors onto 96), although it probably does; and it has not been shown to apply generally to subsequent collisions in the progressive collapse, and it probably does not. As the collapse front is slowed less by conservation of momentum in subsequent collisions, the relative speed of any remaining uncrushed upper floors decreases, and in a case such as 14 floors dropping on 96, the ratios of the spring constants in upper and lower blocks are not as great as for the 3:1 disparity between S3 and S1. To establish for certain whether or not that is the case, a computer model is now in development, using numerical methods to deal with the n simultaneous equations of motion for a funny m system and show its evolution over time in various collapse scenarios. This will aid in the development of the physical model, which as designed so far resembles the characteristics of the funny m model in some respects (thought it does not accurate represent the exact characteristics of the wtc towers, or the details of the wtc's most likely collapse mechanisms in which separation of floors from columns generally precedes fracture of the columns). Conclusions This analysis refutes any notion that "1/11 of a uniformly constructed structure cannot destroy the other 10/11" is a valid universal principle. Even if there is objection to the features of the model or the specific parameters chosen, and/or doubt that they accurately describe the WTC towers, there is no doubt that with those particular features and parameters in place, progressive collapse can occur and indeed cannot be avoided. For instance, the funny m model does not specify any ratio of width to height, so any objection that collapse must in all cases be arrested by torquing due to asymmetrical breaking of the springs or the loss of mass over the side can be refuted by referring to a funny m structure that is ten times wider than its total height. It has also been clearly established that imperfect collisions that "spread the force out over time" cannot be assumed capable of increasing the energy dissipation via elastic strain enough to make a difference in the outcome, since in the case evaluated, even the best-case dispersion of strain energy into the lower structure given unlimited time and perfect damping was not enough to arrest collapse. No doubt there are some uniformly composed structures for which 1/11 (or 1/10) of the structure falling one story's height cannot cause progressive collapse of the other 10/11 (or 9/10). But the proposition that this must be true for all uniformly composed structures is now disproven by the very model that that hypothesis's most vocal proponent has put forward in an attempt to support it. Respectfully, Myriad
Heiwa has posted an essay on this Web site, describing a model of a structure comprised of floors of equal mass m, each supported on four vertical springs of relaxed length (height) h connecting with the next floor down (or the ground, for the lowest floor). The masses of the springs/columns are considered to be negligible. The number of floors is not specified, but all the examples in the essay use 22 floors so that is what I will use. The elements are numbered from the top down, so for instance M1 refers to the top floor and S1 refers to the springs supporting M1 above the next floor down.
Most of the model's parameters are left unspecified. The description does not specify the mass of the floors, the height of the floors, or the spring constants of the springs. We must simply refer to the uncompressed length of each spring as h, essentially making h our unit of measure (for instance, the total height of the structure with 22 floors, not accounting for compression of the springs under the static load, is 22 * h). Similarly we make m, the mass of one floor, our unit of measure of mass. For instance, the total mass of the structure with 22 floors is 22 * m.
Unlike the masses and (uncompressed) heights, the spring constants of each floor are different. However, they are all multiples of the spring constant of the top springs (S1). Considering the springs of S1 as a single spring with a constant four times the constants of each of the four springs, as per the rule of springs in parallel, I'll use k to designate that constant. (To indicate the spring constant in generic equations, I'll use capital K instead.) The spring constant of S2, for instance, I'll refer to as k2 and its value is 2*k, The spring constant of k3 is 3*k, and so forth.
k15 = 15*k
kn = n*k
Similarly, the notation xn will refer to the compression distance of a specific spring below floor n. To refer to the compression of a spring in generic equations, I'll use capital X instead.
The total potential energy of the structure with 22 floors, not accounting for spring compression under the static load, is (1 + 2 + ... 22) * h * m * g, or 253*mgh. Note that we don't know the actual value of g, because g would be earth's gravity expressed in our h units (and any convenient time unit such as seconds). But we do know that g is a constant, so we can compare any coefficient of mgh with any other. In other words, the unknown but constant magnitude of mgh can serve as our energy unit.
Similarly, the unknown but constant magnitude of mg can serve as our unit of weight and other forces.
The Model's Parameters
The only quantitative values given in the model are three dimensionless constants related to the performance of the springs/columns under load. The first, which I'll call C1, is the fraction of h that each floor of the structure compresses under the static weight of the floors above. The second, which I'll call C2, is the fraction of h that any of the model's springs can be compressed before it reaches its elastic limit. The third, which I'll call C3, is the fraction of h that any of the model's springs can be compressed before it reaches its plastic deformation limit and breaks.
Heiwa has provided two sets of these constants.
Original publication:
C1 = 0.001
C2 = 0.003
C3 = 0.005
Revised values that were posted later:
C1 = 0.03
C2 = 0.09
C3 = 0.1
At first glance, it might appear that there are too many unknowns in the model to determine much of anything about it, let alone whether it would progressively collapse. But as it turns out, there is enough information to prove that the model structure will progressively collapse, when evaluated with the original values of the parameters C1, C2, and C3.
The suggested parameter values in the current revised version of the paper can be disregarded. No real conventional structures, and certainly not the World Trade Center towers, sag anywhere close to 3% of their height under their own weight, or can be compressed by 10% of their height before they would break. (By "conventional" structures I mean structures built of ordinary building materials such as wood, concrete, and steel, omitting such special cases as inflatable structures, straw-bale or sod construction, sand castles, and so forth.) For instance, if the load-bearing walls of wood-frame houses compressed 3% under their expected working loads, a vertical 8-foot framing stud pulled out of a wall would spring back (lengthen) 2.88 inches. Imagine trying to squeeze a two-by-four hard enough to shorten it 2.88 inches, to get it back in there.
So, we'll work with the sensible original values of C1, C2, and C3.
Spring Constants
The first thing to do is derive the spring constants and other information related to the springs in terms of our unit m and h values. The equation for the relation between force and spring displacement is
F = -K * X where x is the distance the spring is shortened (or elongated). (eq. 1; Hooke's Law)
Parameter C1 tells us that the weight of n floors (equalling n * m * g), loading the nth spring down (Sn), whose spring constant is n*k, by .001*h. This results in each spring (and the entire structure as a whole) being shortened by ,001 of its height. Putting these values into eq. 1, we have
mg*n = -k * n * -.001h
k = 1000*mg/h
(Note that henceforth, I'll ignore both of the negative signs, and treat downward/compressive forces on the springs, and the change in length of the springs when compressed, as both positive in sign. The results will not be affected.)
The elastic strain energy stored in a static compressed spring, within its elastic limit, is
0.5*K*X2 (eq. 2)
So, for spring Sn, whose displacement x * h is less than or equal to C2 * h, the energy stored is
0.5 * n * k * x2 * h2
and substituting our known value for k we have
strain energy = n * 500 * x2 * mgh (eq. 3)
We can use this to determine how much elastic strain energy is in all the springs, when the structure is static. We know that x = C1 in that case.
Total elastic strain energy =
sum from n = 1 to 22 of [n * 500 * C12] * mgh (eq. 4)
= (1 + 2 + ... + 22) * 500 * C12 * mgh
= 0.126 * mgh
Remember that 1 * mgh is the potential energy from one floor mass falling one uncompressed floor height. The total elastic strain energy at rest is about 1/8 of that, or 1/2000th of the total gravitational potential energy of the structure (which is 253 * mgh).
We can also determine the reduction in the gravitational potential energy of the masses from the compression of the springs. The springs shorten by .001 * h, so each mass loses .001 of its total height off the ground. So, the potential energy loss is .001 * mgh , or 1/1000th of the total gravitational potential energy of the structure.
Why doesn't the decrease in potential energy equal the strain energy? That is, why isn't all the lost gravitational potential energy from the sagging of the masses stored as elastic strain energy in the strings? Because we're describing the structure at rest, which means that half the gravitational potential energy from the dropping masses must have been lost to damping effects (such as heating of the springs and moving the surrounding air) in order to bring it to rest. If we raised all the masses in the structure up to their full h and let them go and there were no damping losses, the (assumed perfect) springs would just keep oscillating forever. To damp it to a static condition, energy is lost. (If it weren't for the lost energy, the elastic strain energy could just turn back into gravitational potential energy by having the floors spring back up again, which is what happens when it's oscillating.) Of course, in reality, that damping would occur a little bit at a time as the structure was built.
Strain Energy Capacity of the Springs
We can also derive two more energy values: the maximum amount of elastic strain energy that a given spring can have before it goes beyond the elastic limit, and the additional amount of energy of plastic deformation a spring can absorb before it breaks.
The maximum elastic strain energy that a spring Sn can have is
0.5 * n * k * C22 * h2 (eq. 5)
substituting our value for for k
= n * 500 * C22 * mgh
= n * .0045 * mgh
To derive the maximum energy a spring can have before it breaks, we have to make some reasonable assumptions for how it behaves at compression distances between C2 and C3. (The model omits this information.) One reasonable assumption is that the resistive force remains constant at the value it reaches at C2, which is
n * k * C2 * h
= n * 1000 * C2 * mg
= n * 3 * mg
... so the total energy absorbed in plastic deformation is that force times the distance C2 to C3, or
n * 3 * mg * (.005 - .003) * h
= n * .006 * mgh
But just to make sure we're not underestimating any energy sinks, let's instead assume that the springs are made of super tough stuff that continues to resist deformation with its characteristic Hooke's law force (that is, the force with which it resists further deformation = -kx with the same k as for its elastic region) as it strains plastically, right up until it breaks. (This also makes some of the subsequent calculation easier.) So the energy absorbed in plastic deformation is instead
0.5 * n * k * (C32 - C22) * h2
= n * ,008 * mgh
The total amount of strain energy to break a spring is the sum of the elastic strain energy plus the plastic deformation energy. So,
Energy to break spring Sn = n * (.045 + .008) * mgh = n * .0125 * mgh (eq. 6)
Note that this makes the assumption, in favor of collapse arrest, that the elastic strain energy in the springs is "lost," when it is actually a form of potential energy that is still available to do work on the masses it's connected to. We will nevertheless assume that when a spring breaks, its elastic strain energy is expended in a way that does not help contribute to further collapse, such as propelling the broken spring parts sideways away from the structure.
The total amount of energy required to break all the springs is:
(1 + 2 + ... 22) * .0125 * mgh
= 3.16 * mgh
That's more energy than the gravitational potential energy of two floors' mass falling one floor's height (which is 2 * mgh). So does that mean Heiwa is correct? If we remove the springs S2 and allow M1 and M2 to fall onto M3, will the springs absorb all the energy and the falling floors just bounce off?
Not so fast. The problem with that conclusion is something Heiwa himself has taken pains to point out repeatedly: the energy doesn't get precisely distributed to all the springs in direct proportion to their strain energy capacity. Instead, the weaker springs tend to fail first!
Also note that while the total energy required to break all the springs is greater than the gravitational potential energy converted by two floors falling a distance h, it's less than the total gravitational potential energy of the structure by a factor of 80. That suggests that while two floors dropping cannot cause an instantaneous global failure of the structure, a progressive failure resulting ultimately in global failure remains a likely possibility.
What we have to consider is, how much downward-directed kinetic energy applied to M3 can the lower 20 floors absorb before one of its springs breaks, and which spring breaks first?
How much energy from downward force acting over a distance can the lower structure absorb in strain energy?
To compute with any precision how much energy the springs can actually absorb under the impact of falling masses, we'd need more complete descriptive data than the model includes, and then to perform fairly complex dynamic calculations or use a computer model. However, we can more easily compute an upper limit for that value, that is, the amount of strain energy the lower structure can absorb under the very best possible conditions for absorbing the most energy possible.
This approach is a bit like testing the weight of cargo that, when carefully loaded a little at a time, will cause a ship to sink while motionless in glassy calm water, in order to show that the same ship cannot carry more than that weight of cargo without sinking if the load were dropped into the hold from a helicopter in one big crate while out at sea during a hurricane. Obviously, it's going to be a significant overestimate of the actual capacity.
In the case at hand, the best possible conditions are created by applying a slowly increasing force gradually compressing the structure (that is, acting over a distance), slowly enough that all oscillations can be assumed to be damped and slowly enough that the inertia of the floors has no effect on the distribution of strain.
It is easy to see that both of those conditions are necessary for storing the most strain energy possible in the springs. If the blow from above is faster so that inertia comes into effect, the inertia of the upper floors limits the rate at which strain can be transmitted to the lower springs, because doing so requires the floors to be accelerated. So if we find that the upper springs fail first under the ideal conditions of very slowly applied force, we can be certain that they would have an even greater tendency to fail first with a more rapidly applied force. And if oscillations are not damped, oscillations will cause springs to be compressed to their failure points sooner than would occur when everything stays damped to steady state. The damping itself is an energy sink.
(As an example showing why this is the case, consider spring S1 on a rigid base, for which K1 = k, which is then suddenly loaded with a mass of 3 * m. We know that in a steady state condition after damping, if the spring were compressed by a force of 3*mg, the spring would be compressed just to its elastic limit at x = C2 * h. But if that weight were loaded onto the spring all at once with no damping, the mass would still be moving as x reaches C2 * h, and its inertia would keep it in motion, further compressing the spring until x reached 2 * C2 * h. (If all the strain were elastic, it would then continue to oscillate sinusoidally between x = 0 and x = 2 * C2 * h.) But since C3 < (2 * C2), the spring breaks first. So while the spring can support a weight of 3*mg steady state without any plastic deformation, it deforms and then breaks if that much weight is "dropped" onto it from a height of zero.)
Properties of the Lower Structure after Removing M1 and M2
In the drop test, we would begin by holding M2 in place and removing spring S2. This leaves us with a "lower structure" of 20 floors, each of mass m, connected by springs with spring constants ranging from 3*k for S3 (now the top springs of the lower structure) to 22*k for S22.
The lower structure now has 2*m*g less load on it before, so the springs relax somewhat from their initial static state of being each compressed by C1 * h. The masses rise a little as a result. We can calculate the new total strain energy of the lower structure as the sum
strain energy = sum for n=3 to 22 of [0.5 * n * k * xn2] * mgh (eq. 7)
where xn = h * (n - 2) / (n * k)
= 500 * sum for n=3 to 22 of [n * ((n - 2) / (1000 * n))2] * mgh (eq. 8) which results in .0894 * mgh That's about 29% less than the original structure, because we've removed two springs from consideration entirely, and reduced the load on each of the other ones by 2 * mg. In the original structure, M3 is at a height of 20 * h * (1 - C1) = 19.98 * h. With the top two floors removed, the new neutral position for M3 is 20 * h - sum for n=3 to 22 of [h * (n - 2) / (n * k)] (eq. 9) = 20 * h - sum for n = 3 to 22 of h * (1/k - (1/n * 2/k)) = 19.9844 * h The displacement of M3 from the ideal position of 20 * h is .0156 * h, about 22% less than in the full 22-floor tower. Maximum Strain Energy that the Lower Structure Can Absorb Now, for our best-case spring energy test, we begin applying a slowly increasing downward force on M3. This causes the entire structure to displace downwards until something breaks. However, the different M's do not displace equally, nor in proportion to their height off the ground, because each has a different amount of force pressing it down and each has a different spring constant for the spring holding it up. We have to consider what's happening to each spring. Call the extra force applied to M3 at any given moment F * mg (so that F = 1 represents the weight of one floor). For our best-case test, everything is slow and damped, so we can evaluate it as series of steady-state conditions. At any given moment, the force on m3 equals F * mg. The force on spring S3 equals F plus the weight of M3, or (F + 1) * mg. The force on spring S4 equals F plus the weight of M3 plus the weight of M4, or (F + 2) * mg. In general, the force on Sn = (F + n - 2) * mg. The spring constant kn = k * n = n * 1000 * mg/h The quotient of force over spring constant gives us the strain (compression distance) of each spring, which is xn = h * (F + n - 2) / (n * 1000). This can also be expressed as xn = h * (F - 2) / (n * 1000) + .001. (eq. 10) Note that when F = 2, equivalent to slowly loading the top two masses back on, this gives us a strain of .001 * h on every spring, just as we would expect given the original starting conditions for the 22-floor structure. We can see in eq. 10 that for all values of F>= 2, the larger n is, the smaller the strain xn. So, we can now be certain that despite the extra weight pushing down on the lower springs, under these ideal conditions the top spring of the lower structure, S3 will reach x = C3 * h first, and will break first. The total amount of strain energy that the springs can contain when that happens is the absolute theoretical limit, under ideal conditions and all assumptions favorable to collapse arrest, of the amount of energy that the springs can absorb in strain before S3 breaks. Solving eq. 10 for F when n = 3 and x = C3 = .005, F = 14 * mg So S3 breaks under the additional load of 14 * mg (the weight of 14 floors), applied to the top of M3 (or a total load of 15 *mg, the weight of 15 floors). We can now calculate and sum the total strain and total strain energy of all the springs when F = 14 * mg and a steady state exists. The total strain energy is the sum for n=3 to 22 of [0.5 * n * k * xn2] * mgh (eq. 11a) where xn = (F + n - 2) / (k * n) (eq. 11b) and k = 1000 The sum calculates out to 0.5227 * mgh. But before applying any additional F to the 20-floor lower structure, we started with .0894 * mgh of strain energy already in the springs, so the additional amount that can be absorbed, under these absolute best-case conditions, is .5227 - .0894 = .4333 * mgh. Dropping M2 Alone on the Lower Structure We can now easily show that dropping only M2 onto the lower structure will cause springs S3 to break. That's enough to prove that dropping the entire upper unit M1 + M2 + S1 must also cause springs S3 to break. We don't yet know exactly what M1 and S1 will do (we'll consider that later) but even if they magically evaporate into weightless dust the moment M2 touches M3, contributing no energy at all to the collision or subsequent events, S3 still breaks and a progressive collapse still ensues. Because of the readjustment of the springs under load, the distance from M2 to M3 is a little less than h, it's about .995 * h. That gives it .995 mgh of potential energy converted into kinetic energy before it reaches M2. We assume a fully inelastic collision between M2 and M3. This is another worst-case assumption against progressive collapse. It means half of the kinetic energy is immediately expended in inelastic deformation of M2 and M3 (whereas a more elastic collision would transfer more energy to M3 and a fully elastic collision would transfer all the energy to M3). That leaves .498 * mgh of kinetic energy. As established above, the maximum amount of strain energy the springs in the lower structure can absorb is .433 * mgh. Not enough. S3 must break. Note that the collision can be "spread out over time" due to imperfect contact, deformation of the M's during the inelastic collision, or any other reason, and it doesn't matter. The determination of the maximum energy the lower structure can absorb already assumes an ideal case where the application of force is spread out over essentially infinite time and everything is perfectly damped. Since the actual collision is faster and less damped than that, the actual amount of the energy the springs can absorb will be less. Also, that's without any contribution from M1. But in fact, when M2 collides and slows down, the momentum of M1 keeps M1 moving toward M2, compressing springs S1 which transfer additional energy to M2. Exactly how much doesn't matter, because M2 by itself already had enough kinetic energy to break S3. What happens then? That first impact seemed kind of a close call. (At least, until we remember that we way overestimated the ability of the lower structure to absorb strain, and that we ignored the contribution from M1.) There's not much energy left over. But now M2 and M3, having collided inelastically, fall together onto M4. Assuming they have no kinetic energy left over when S3 breaks, they still fall onto M5 with close to 2 * mgh of kinetic energy after falling another distance just short of h. This time the inelastic collision with M5 only absorbs 1/3 of that kinetic energy instead of 1/2, leaving 1.33 * mgh that would need to be absorbed by the lower springs. Those same lower springs have already absorbed most of the energy of the first collision, already bringing M4 fairly close to its breaking point. (You can consider the possibility that the springs below may, depending on the actual values of h and m, have had time to spring back up, but in that case the energy of those oscillations is still in play and must still be considered, and any plastic deformation is permanent). Also remember that it only takes 3.16 * mgh to break all the springs in the entire original structure.) Repeat the methods and calculations above could determine the exact values, but it's clear that M4 is going to break. This time there will be plenty of energy left over to contribute to the next collision with M5. It can end only in global progressive collapse. We still don't know what happened to M1 and S1. It doesn't actually matter. But let's examine the question next. Does S1 Break? To address this question, we'll have to depart from the realm of best-case energy transfer calculations, and evaluate the dynamics of the event. The problem is, while the M1 - S1 - M2 system is fairly straightforward, two falling masses connected by a spring, the lower structure, of 20 masses connected by 20 springs of different coefficients, Equations for the dynamics of the whole system, even if they could be fully derived, would be rather opaque and not likely to help in intuitive understanding of why the system behaves as it does. Instead I'm going to again examine a key limiting case. Immediately before the collision of M2 and M3, M1 and M2 is falling at a velocity we will call v (which happens to be sqrt(mgh)), and M3 is stationary. Immediately after the collision, before large forces have had time to develop in the springs and before the spring forces have had significant time to act on the masses to change their velocities, M1 is still moving downward at v, while M2 and M3 are moving downward at v/2. They are moving closer together at v/2, which will compress S1. For S1 to survive, their relative movement toward one another must stop and reverse before the spring is compressed beyond C3 * h. M1 gets accelerated upward by the spring force of S1 and downward by gravity. M2 gets accelerated downward by the spring force of S1 and by gravity, but upward by the spring force of S3, at least until S3 breaks. Clearly any force contributed by S3 adds upward acceleration to M3-M2 which helps compress S1. So we can examine a limiting case in which S3 makes no contribution, as if it disappeared at the moment of the collision of the floors. If S1 breaks under that ideal starting condition, it must also break with S3 present (regardless of when S3 breaks). Without S3, we have a much simpler initial condition: two masses in freefall, connected by a spring, with the top one moving downward at v and the other moving downward at v/2. Because gravity accelerates both equally, we can ignore it, and we can shift the system into the reference frame of the center of gravity of the two masses. In that reference frame, we have a simple harmonic oscillator with a fixed center of gravity, starting in mid-cycle (x = 0), with M1 moving toward the center of gravity at v/3 and M2-M3 moving toward the center of gravity at v/6. This, it turns out, is way too high an amplitude (or way more kinetic energy, however you choose to calculate it) than S1 can withstand, by a factor of more than 13. So, now we know: S1 does break, within a moment before or after S3 does. Note, however, that this result has not been shown to apply generally to the initial collisions in other cases (such as dropping 14 floors onto 96), although it probably does; and it has not been shown to apply generally to subsequent collisions in the progressive collapse, and it probably does not. As the collapse front is slowed less by conservation of momentum in subsequent collisions, the relative speed of any remaining uncrushed upper floors decreases, and in a case such as 14 floors dropping on 96, the ratios of the spring constants in upper and lower blocks are not as great as for the 3:1 disparity between S3 and S1. To establish for certain whether or not that is the case, a computer model is now in development, using numerical methods to deal with the n simultaneous equations of motion for a funny m system and show its evolution over time in various collapse scenarios. This will aid in the development of the physical model, which as designed so far resembles the characteristics of the funny m model in some respects (thought it does not accurate represent the exact characteristics of the wtc towers, or the details of the wtc's most likely collapse mechanisms in which separation of floors from columns generally precedes fracture of the columns). Conclusions This analysis refutes any notion that "1/11 of a uniformly constructed structure cannot destroy the other 10/11" is a valid universal principle. Even if there is objection to the features of the model or the specific parameters chosen, and/or doubt that they accurately describe the WTC towers, there is no doubt that with those particular features and parameters in place, progressive collapse can occur and indeed cannot be avoided. For instance, the funny m model does not specify any ratio of width to height, so any objection that collapse must in all cases be arrested by torquing due to asymmetrical breaking of the springs or the loss of mass over the side can be refuted by referring to a funny m structure that is ten times wider than its total height. It has also been clearly established that imperfect collisions that "spread the force out over time" cannot be assumed capable of increasing the energy dissipation via elastic strain enough to make a difference in the outcome, since in the case evaluated, even the best-case dispersion of strain energy into the lower structure given unlimited time and perfect damping was not enough to arrest collapse. No doubt there are some uniformly composed structures for which 1/11 (or 1/10) of the structure falling one story's height cannot cause progressive collapse of the other 10/11 (or 9/10). But the proposition that this must be true for all uniformly composed structures is now disproven by the very model that that hypothesis's most vocal proponent has put forward in an attempt to support it. Respectfully, Myriad
I see an ignorant fraud dishonestly describing the debris falling from a collapsing building as "explosions." They don't look like explosions and they aren't explosions.
Thanks for your analysis and conclusions. Let's assume that S1 does break before S3, which I consider likely. This means that M1 is disconnected from the remainder of the structure for a certain time and that thus the structure below can decompress a little before M1 connects (impacts) again (it has to drop h). In my opinion this means that S3 does not break.
Furthermore, does all four springs S1 break simultaneously? Let's assume they vary a little and that one break before the others. This means that M1 cannot impact the structure below perfectly; it is tilting.
Thus the Funny m model will not one-way collapse, reason being that upper part breaks apart first.