VShear2.dia = Vector Shear Diagram 2
A Better Perspective
t7 V7=14
> levl 7
\
\ Particle Size o
\ layer 7
\
b7 \t6 V6=12
> levl 6
\ \
\ \ parallelogram Each layer has it's own parallelogram shape here. If visible,
\ \ layer 6 we could see the top dust cylinder (20 mi diameter) tilt more
\ tilt \ & more to 12 miles in descent to the surface layer before
\b6<>t6 \t5 V5=10 collapsing out to 13 mi on the surface.
> levl 5
\ \
\ \ parallelogram
\ \ layer 5
\ shear \
\b5<>t5 \ t4 V4=8
> levl 4
\ \
\ \
\ \
\ \
\b4<> \t3 V3=6
> levl 3
V2=4
> levl 2
t V1=2
> levl 1



 shear
___________b<>t vo=0
~~~~~~~~~~~~~~~~~~~~~~\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ levl 0
 \
~~~~~~~~~~~~~~~b'~~~~~~~~t'~~~~~~~~~~~~~~~~~~~~~~~~~~~
b t
> 6
This over simplified wind profile has equal layer thicknesses and a linear increase of wind
with height. We have a layer of particle size o starting it's descent from the top layer. This
top layer is denoted by 'layer 7' with the top by 't' and the bottom by 'b' which can represent
the center of the cylinder's top & base respectively. Initially, we have a vertical cylinder
with center t aligned vertically with b. All the levels below are indicated by it's equal
thicknesses and decreasing east wind component by arrow length (V?). Both particles t & b move at
their adjacent ambient winds. However, t starts out with higher winds at it's level than b.
When t arrives at levl 6, b arrives at levl 5 if layer 7 equals layer 6 thickness which we do here.
The difference between t and b is the position difference or cylinder tilt levels of top vs. bottom
shown by b<>t. Continuing, t moves to levl 5, and b moves to levl 4. During this time, t moves
more than b again because of more shear. Therefore, the cylinder tilt increases.
From there down to the surface ~~~~~~~ layer (levl 0), t is at the top (1) and b is at the ground
level. Our final tilt by shear is b<>t @ levl 0. Now b has landed and is stationary as seen by b'
but t goes through the surface and lands at t'. The final shear, but no tilt because now they both have
landed, appears as the same difference @ levl 6. That is, if b's position is held constant, t will have
a downwind lead on b, which we can say, is t's starting position compared to b. Now only shown in
layer 6, they follow parallel paths but not at the same time. The parallel paths goes all the
way down to the surface and would coincide if t started at level 6 as b did. That's obvious, the
same particle size, same position, same winds and same path. So simple but this is the best way to
visualize. However, t is displaced further east because of it's travel through layer 7. If b had waited
for t, then they both took off, we would have parallel and simultaneous paths, therefore like a
parallelogram, the adjacent pair of sides are parallel so the tilt by shear at levl 6 is the same at
ground level. This has to be since they both took the same path from levl 6 down so the only difference
is the initial separation that occurred in t's 1st descent. That is tt(V7+V6)/2.
We care about this shearing tilt for the overlapping or stretched cylinder calculations for non
overlapping cylinders. However, if the tilt were visible, it would be t's (V7+V6)/2*tt  b's
(V6+V5)/2*tt. So the difference of these is their difference of their respective positions
(mean layer velocity * tt) top and bottom of the cylinder or tt[(V7+V6)/2(V6+V5)/2] = tt[V7+V6V6V5]/2
= tt[V7V5]/2. That is 13  11 = 2 or
(1410)/2 = 4/2 = 2 for 1st and last exprtession respectively.
With the same layer thickness, tt is the same for both layers. Now, our previous procedure was that we
summed all 1/2 of the vector differences for the cumulated shear. Looking at this example, and do this
for each term, do we arrive at the same value?
tt{[V7+V6V6V5]/2 (14+121210)/2 = 4/2 = 2
+ [V6+V5V5V4]/2 (12+10108)/2 = 4/2 = 2
+ [V5+V4V4V3]/2 (10+886)/2 = 4/2 = 2
+ [V4+V3V3V2]/2 (8+664)/2 = 4/2 = 2
+ [V3+V2V2V1]/2 (6+442)/2 = 4/2 = 2
+ [V2+V1V1V0]/2 (4+220)/2 = 4/2 = 2
+ [V1+V0]/2} (2+0)/2 = 2/2 = 1
___
= tt[V7+V6]/2 (14+12)/2 = 26/2= 13
Even if the layer thickness or particle size vary, we would have different travel times tt7, tt6, etc.
for each layer but again, parallel paths and different tt's with each pair of V's would still cancel each
other. Also, regardless of the values of the vectors V's could vary and even the directions < >
but the parallel paths still apply. Therefore, we don't need to keep track of each layer, just the top
for the shear  tilt. This will reduce the size and CPU time processing pttc.cor. Since we are using
trajectories and with a new format, we now call this pttct.cor.