crush.dia CRUSH PILE DYNAMICS
The remaining building weight (ctr = cement tons remaining) rests on the slab,
which travels, at our first estimate, 2 slab thicknesses (slabt) through the
inside wall rubble pile before the slab breaks up near ^.
ctr
!
| _ _
| / \ | |
|______/_____\_______________|___________________ 2slabt
slabt |_________________________________________:______ |_ tps
/------------\ | ^
/ \ | Breaking
hpi - / \ hpi Point
2slabt / trib trib \ |
/_____________|______________\ |
The remaining detached slab at : travels hpi-2 slabt with weight of
rat x tps where rat is ratio of the area stl-perimeter to the inside wall
slab area and tps is tons/slab. Here, stl is the stretched length of the
squashed rubble spread out to the break near :. The cross sectional area
of rubble volume is the same but with height = slabt. With the rubble pile
height reduced from hpi -> 0 in most cases. This is a minimal dust
contribution because of the broken slab's reduced weight crushing the
rubble.
: . , . ,.. : ' . . '' .
. '. '. . . . ., . .. ' .. . '. . '
. ' ' . , . , . . ' ' ' .
|_______________________________________ ______
rat*tps |_______________________________________:______ slabt
<-------------- stl=cac/slabt --------->
Below, we show the slab base crushing downward at various levels to 2
slab thicknesses. At the top, there are not many particles to prevent
the horizontal components <---> so there is much lateral movement of
these particles. As all these particles together try to act like a
fluid, the pressure is being supplied in all directions. So the least
resistance is in the horizontal. This is a sort of slow bubble burst
resulting in a decrease in pressure in all directions, including the
downward vertical component, once the rubble moves horizontally <--->.
The flattened cone head area becomes larger so the psi decreases.
As the slab-building progresses downward, more particles so more
frictional resistance so less horizontal component vectors. Also with
an increasing surface area of the flattening rubble against the slab,
we have the weight of the slab-building pushing down in a broader range,
resulting in the PSI decreasing. Also, stacking more vertically aligned
particles that won't slide, thus causing pressure columns to further
stop horizontal movement. With these 2-3 effects, hardly any more bubble
bursts are possible. The total vertical weight component from the slab
building is not diminished but it's weight is applied to the pile until
the slab breaks because the momentum of the slab-building encountering
a more resistance pile. This slab encounter finally breaks away from it's
wall attachment and the rest of itself at say 2 slab thicknesses vertically.
weight _________________________________________________________ _
ctr ! /"\ ! |
<-------------------/*#**$\------------------->
....................../.........\........................ 2
/#**$*#*%*#**~\
<-------------/*$*#+*?*&&*#**#% \-------------> s
................/.....................\.................. l
/*~##*%$$*?*&*?+*%**#%#*#*\ a
<-------/+*#@*%&$~*?*$*??*%*&#%#*##*$ \-------> b
........../.................................\............ t
/*~##*%$$*?*&*?+*%*+*#@*%&$~*?*$*??*% \
<-/&&+*#@*%&$~*?*$*??*%*&#%#*##*$*?+*%**#^? \-> |
____/_____________________________________________\______ _
/<---------------------- br --------------------->\
/ hr*trib | hr*trib \
/ \
/ Large and smaller /_\ in proportion (~) so height ratio, \
/ hr=2*slabt/hpi and whole base area br=2*hr*trib. If psi \
inversely ~ to ctr/br, it's inversely ~ to 1/hr, by similar triangles.
Since ctr is constant and br increases, so is the lateral pressure for
rubble particles to escape. Now frictional force Fr also increases with
increases area or ~ to length br. So total downward vertical pressure
increases by a factor say Pv=f(br) ~br^2 = (2*hr*trib)^2 = a*hr^2, where a=2*trib.
To feel this in the real world, we might say "Go pound sand". Make a sand pyramid
and slam the top with a board or sea shell harder and harder each time. By integration
INT, the area of a*hr^2 is INT a*hr^2 = a*INT hr^2 = a[hr2^3/3-hr1^3/3] with limits hr1=0,
hr2=2*slabt. The ratio of the INT to the AREA xy = (hr2-hr1)*a(hr2^2-hr1^2) is
a/3[hr2^3-hr1^3]/[a(hr^2-hr^1)*(hr2-hr1)] = [hr2^3-hr1^3]/[3(hr2^2-hr1^2)*(hr2-hr1)].
With hr1=0, we have INT=a[hr2^3/3] and AREA xy = (hr2)*a(hr2^2) = a(hr2^3) so INT/AREA xy
= a[hr2^3/3]/a(hr2^3) = 1/3. Note, this is the same as the rubble /_\ to the rubble + 2
horizontal components inverted \/ triangle areas.
This assumes that all visible horizontal components end at 2 slabs. Using the 1/3 correction
in run with DUSTBM.c model run produced a max of near 7" with 83% of dust deposited within 2 miles.
This amount of tons of dust is reasonable but the less crushing caused a particle density shift to larger
particles that were deposited closer to WTC. The horizontal components should end at # of slabs to cause
the slab to break. Experimenting, we found that 3.6 x slab thickness rather than 2, and the occurrences of
0 <----> is at 1/1.4 instead of 1/3 correlates better with the observed WTC street dust photos from the web.
That is, we correct the main work done slabt*ctr by 2/3, but now better correlated with 3.6/1.4 with max dust
2.74" @ .147 mi and 38.8% of dust was deposited within 2 miles with 62.9% of WTC's to dust.