The 9/11 Forum

I was careful to say "relatively" constant because the observed tilts follow the calculated tilts only to some approximate degree; in fact I do see what look like periodic "wobbles" in the data which may be due to variable resistance to the rotational motion; or could be artefacts in the data!

Relative constant resistance torque with small wobbles (if any) sounds like a signature of specific types of collisions. It would be hinging upper block vs intact, connected flooring below floor 98.


I'd look at the angles theta= arctan (13n/207) where n=1, 2, 3... for slight wobbles of increased resistance as the south wall upper block perimeter makes contact with flooring.


If so, interesting.

Don't want to jump to conclusions without more detailed study of the data.


Dr G, consider what type of resistance you'd expect if the tilt occured as a result of 3 or 6 floor perimeter buckling along the south face. And what type of collision might we expect after the buckle (3 to 6) floors totally folds in and the upper rotation meets stiffer resistance again?



Now lets consider anticipated resistance if a demo team initiates tilt by dividing upper and lower perimeter into two sheets (along bolt connections if you wish) and kicks the lower sheet inside the upper, or the upper inside the lower.

Now column bucking is not a factor in torque resistance. It's just spandrel plates hitting outer floor connections. This would look like relatively constant torque of resistance with little wobbles.



Hmmmmm.

Major_Tom wrote:Femr, in your GIF of eastern antenna movement a few posts ago...

It seems as if there is sharply increased NW corner resistance about 6-7 stories into the fall.

A soft collision. Hat truss vs lower NW corner (or fl 98 vs fl 92, or fl 106 vs fl 98???) after 6-7 stories of relatively smooth descent.


Hmmmmm.

As a slight aside, have been discussing floor layouts on another thread here, and though this may be relevant...



Image flips between floor 82 and 77. 82 is the sparse frame.

Not an ideal choice of floors, but the transition from many lift shafts to much less is at floor 81.

Given the huge structures put in place at the top of lift shafts, and the increased support members, thought you may have something to say...

I can't see it as being from way down there.

Something which makes a solid structure with the hat truss has to be responsible for the collision.

So I am seeing something from floor 104?, 105?, 106?.....100? colliding with something below after a 6 to 8 story fall.

82 seems too low.


Even if there was a solid upper block (which there wasn't), the lowest we could go is floor 98 colliding with floor 92 to 90.

Using the tilt rate formula provided by Dr. G, here are best fit angles (c01s) to compare to the measurements used (b01):

Code: Select all

  Estimated angles
t= -3.637 b01=  0.000
t= -1.549             c01=  0.050
t= -0.300 b01=  0.681 c01=  1.228
t=  0.033 b01=  1.080 c01=  1.807
t=  0.367 b01=  1.738 c01=  2.499
t=  0.701 b01=  2.939 c01=  3.301
t=  1.034 b01=  5.308 c01=  4.215
t=  1.368 b01=  5.948 c01=  5.241
t=  1.702 b01=  7.464 c01=  6.378
t=  2.035 b01=  8.088 c01=  7.626
t=  2.369 b01=  8.545 c01=  8.984
t=  2.536 b01=  9.067 c01=  9.704
t=  2.703 b01= 10.173 c01= 10.453
t=  2.869 b01=  9.106 c01= 11.228

This image help to see the Eastward tilt, if watched for a while...

femr2 wrote:This image help to see the Eastward tilt, if watched for a while...

The video shows that the antenna tilts and falls to the southeast.

I finally did what should have been done right away; plotting the angles va time. This immediately suggested a cubic polynomial fit to the data. That worked so well that I also tried a quartic polynomial fit. Which is better?

One means of answering is to use AICc from
http://en.wikipedia.org/wiki/Akaike_inf ... _criterion

Aha, then the difference in AICc values for the two fits is only less than 0.2 AICc units; essentially the same. So neither is better on those grounds. While BIC penalizes extra parameters more heavily, I didn't bother to check the BIC values; instead I choose the cubic polynomial on the general grounds of parsimony and the fact that the angle reversal at the end does not appear in the cubic polynomial fit, but does in the quartic polynomial fit.

Here is the cubic polynomial fit (c01s) to the b01 angles.

Code: Select all

  Estimated angles
t= -3.637 b01=  0.000 c01=  0.190
t= -0.300 b01=  0.681 c01=  0.205
t=  0.033 b01=  1.080 c01=  1.276
t=  0.367 b01=  1.738 c01=  2.415
t=  0.701 b01=  2.939 c01=  3.589
t=  1.034 b01=  5.308 c01=  4.767
t=  1.368 b01=  5.948 c01=  5.918
t=  1.702 b01=  7.464 c01=  7.010
t=  2.035 b01=  8.088 c01=  8.010
t=  2.369 b01=  8.545 c01=  8.887
t=  2.536 b01=  9.067 c01=  9.270
t=  2.703 b01= 10.173 c01=  9.610
t=  2.870 b01=  9.106 c01=  9.903

femr2 wrote:This image help to see the Eastward tilt, if watched for a while...

It shows also quite clearly how upper part C is destroyed (blown apart) prior it allegedly impacts lower part A and supposedly compacts part A into compressed rubble part B in a one-way crush down.

Heiwa --- No it does not.

But more important, you are yet once again off topic on this thread. Start another for your pontifications.

DDB - according the BLGB theory that you have developed there shall be no tilt (topic) or rotation of any kind, etc., but the complete upper part C shall remain intact and displace down as one rigid block at more or less constant acceleration (like a vertical avalanche!) impacting/compressing/crushing lower part A (it becomes smaller) into rubble part B (which becomes bigger). No tilt or rotation of upper part C is permitted in this funny theory. If they are observed, your theory is no longer valid.

Heiwa --- A one dimensional model is obviously an approximation. But good enough for some purposes.

Understanding the tilt is of value to accurately calculate meters vertical drop from the pixel angular measure.

David B. Benson wrote:Heiwa --- A one dimensional model is obviously an approximation. But good enough for some purposes.

Understanding the tilt is of value to accurately calculate meters vertical drop from the pixel angular measure.

The one dimensional BLGB model - a rigid point, C, with infinite 1-D density dropping on and applying gravity force/energy on a line, A, representing the bottom structure of WTC 1 with finite, linear, homogeneous density x, which then is compressed into a line B, with linear density 5x, by C, crush down, until there is no line A left but a line B that is 1/5th of the original line A, which finally destroys the point C in a crush up, is evidently a little crude to understand the tilt of point C.

However, the tilt shows that C cannot be regarded as neither a point nor anything with infinite density but something else, e.g. a weak top structure, C, of a sligtly stronger structure below, A. The tilt shows that something happens to C that is not considered by the BLGB approximation.

I actually analyze the matter in my discussion paper (about BLGB) to be published in ASCE Journal of Engineering Mechanics soon (I have been told) to which you have been requested to reply. I look forward to that.

Heiwa:

So if you think you have a better model, LET'S SEE IT!

Does your model account for the tilt?

Is it a Sqrt{sin(theta)} function?

Have you looked at the difference between the rotational work, potential energy loss and kinetic energy gain from purely vertical descent?

Dr. G wrote:Heiwa:

So if you think you have a better model, LET'S SEE IT!

A better model may be a classical system of n material points constrained to move without friction along a straight line. Each material point i is characterized by a constant mass charge m_i and by a position x_i relative to an origin O (with infinite inertial mass) at position 0. Suppose two adjacent material points are linked by a potential (e.g. a spring), thereby forming a chain. The chain is said to be “fixed”, if the potential also links O and mass 1, and “floating”, if only the masses are linked together.

The potential U_i (x_i-x_(i-1) ) describes the displacements of the ends of each spring until it breaks.

Adding up all potentials yields a Lagrangian equation that in turn gives the Euler-Lagrange equations of motion for the chain.

Assume that all material points are initially located close to their “normal” equilibrium.
If the masses are initially at rest, each mass i will remain at rest at x_i.
A small impulse (in the unconstrained direction) on one of the masses can be modelled by giving the mass a small non-zero initial velocity. This will start the propagation of a longitudinal elastic wave. Each mass will display periodic motion around its equilibrium position at x_i.
A large impulse can be modelled by giving a mass (or group of masses) a large initial velocity. This may be sufficient for a local potential to fail.

A dissipative term must be introduced in the equation of motion. This term should seek to oppose the relative motion of adjacent masses, for example through a Newtonian frictional viscosity term with a damping coefficient. The system will then exhibit viscoelastic behaviour.

Finally, in the presence of coupling g between the mass charge m_i and a uniform gravitational field (in the direction of the origin), a gravitational term -m_i g must also be added.

Thus a complete model for all masses and potentials in the chain system in a gravitational field may be created.

The model should consist of two chains A and C. A consists of, e.g. 97 material points and potentials and is fixed (to ground). C consists of, e.g. 13 material points and is floating. The potentials are all different, e.g. the bottom potential of A is 110 times ‘stronger’ than the top potential of C.

Prior impact all A potentials are slightly compressed and equally stressed due to gravity. The C potentials are not compressed.

C is then is dropped by gravity on A and, at a given velocity, gives A an impuls (impact). At the impact two longitudinal elastic waves are formed; one in A and one in C and the material points displace due to the potentials deforming. If the impuls is great enough one potential will fail and a new drop by gravity will occur.

Using above model you can try to break all A potentials while C remains intact and you have verified BLGB! However, it would appear that only C potentials will break at impact and that then the destruction is arrested due to lack of energy.

Using the cubic polynomial fitted to my tilt data, I ran my version of the B&V crush-down equation for the first 2.805 seconds of the collapse. As predicted, vaf continues to win out over the BLGB forces, this time by 13.642 AIC units. That is enough to definitely reject the latter forces in favor of vaf.