VShear1.dia = Vector Shear Diagram 1
A review from Fall Vect.dia is seen here with a brief explanation. The average speed of
15 mph is shown by the underline which is represented at 50'. That is, assuming a linear
(dx/dy = 10 mph/100') plot of speed verses height, the mid point is half the thickness
representing half the speed. The particle fell through 5 levels before it reached 20 mph
but averaged 15 mph.
H'
10
100 *
12
80 *
14 mph
60 *
      +      50'
 16
40  *


 18
20  *


 20
0  *
15mph
__________________________________________
0 Distance or Sec
Now the fall velocity (fps) determines the travel time T by H/fps, that is ft/(ft/sec)=sec or T.
Then the distance D=RT, as we know, where R is given above each particle *, the particle
is at base level 0 at Distance RT or 15T. Recall, this applies to a wind vector with the
same direction, we must also have to apply it to a vector shift in direction also.
This is vector translation we need the coordinates W_E and N_S so we break the vector into
it's components by deg=270dir or deg=(90+dir) and dx=vel*tt*cos(deg) and dy=vel*tt*sin(deg).
By the above diagram, we have vel=(V1+V2)/2 where V1=10 and V2=20 in our example and tt=travel
time T. The bottom of this layer becomes the starting position of the next layer beneath. We average
the next layer vector components vel=(V1+V2)/2 again. For direction also, we would have E and N
components for a layer, de=(velb*cos(degb)+velt*cos(degt))/2, dn=(velb*sin(degb)+velt*sin(degt))/2.
The t & b refers to the vectors at the top, and bottom respectively.
As we referenced a surface cylinder depositing as the top level travels finally reaches the ground,
we have an elongated grounded cylinder. Actually, we could consider the surface as a wind layer that
has a wind shear opposite to the surface layer? Why, we have brought the translating particles to 0.
So in the above diagram, we could represent it as a wind layer <== 20 mph. As the particles near the
base move ==> 20 mph, we have a fictitious wind of <== 20 mph since ==> 20 + <== 20 = <> 0.
Why bring this abstract meaning here? Simply, then as the wind layers below a level adds to the
translation, the difference in the wind vectors add to the distortion of the vertical dust cylinder by
the shearing effect just as the surface does. This means a cylinder will change it's shape by it's
decent through all the wind levels. By the formula for de,dn components, we would now have
des=(velb*cos(degb)velt*cos(degt))/2, dns=(velb*sin(degb)velt*sin(degt))/2 where we have component
delta e,n by shear des,dns. That is, translation is half the vector sums, shear is half the vector
differences.
To illustrate, lets look at the layers in condensed size and just show the top, bottom and mean
wind vectors of some layers. We centered the dust cylinders through the wind layers after the
translation below so we can see the effects of the wind shear (ds) better.
____________________ 0
  Initial o top dust layer  assume no wind ds=0
 0 ==> 
____________________ 0
:
ds
____________________ 20 Falls through 1st wind layer.
\  \ Trans Position = 25tt
> \  25 ==>  \ ds=10tt to the east between top & bottom layer
\_________________\ 30
____________________ 30
\   \ Falls through 2nd wind layer.
\  35 ==>  \ Trans Position = 25tt + 35tt = 60tt or tt=TT/2 30TT
> \___________________\ 40 New ds = Old ds (10tt) + this ds (10tt) = 20tt
ds
ds :
t _____________________40 Caught results before entering through 3rd wind layer or assume
\   \ 3rd wind layer has no wind shear as seen below. Ds still = 20tt.
\  40 ==>  \ After passing through here, the last tilted cylinder above would
> b \__________________\ have the same shape but translated further downwind @ 40 mph.
40 ds
:
Now suppose we pass through a west wind layer that decreases in speed from 40 > 20 mph. Showing
____________________40 this before this dust layer would enter, passing the initial o through it would deform it as shown.
/ / The ds is now 20tt to the west. So passing the above layer through it, we have the final
/ 30 ==> /  distortion below decreasing the shear ds. Even though we are still translating eastward, the
/____________________/> cylinder orientation has shifted back to a more upright position because ds=ds20tt.
20 ds The top particle has averaged 40 mph east while the bottom particle has slowed from 40 mph > 20 mph.
>
                                                                   
Above, notice that the layers passing through each other all had the same thickness.
That is the wind height levels h2h1 were the same. This is hardly the case. Let's
look at cases that would be most likely to occur. To make it simple, we start out
with an the original layer o falling below another layer again.
____________________ 0
  Initial o dust layer  assume no wind
 0 ==> 
____________________ 0
:
ds
________________________ 10 This wind shear layer is twice as thick
\  \ as the dust layer above passing through it.
> \   \ Then we divide it into n layers, 2 in
\   \ this case as seen below.
\   \
\   \
\______________________\ 20
ds
________________________ 10 The initial o dust layer is seen twice.
\  \ In the first layer, we have an average of
> \  12.5 ==>  \ 12.5 mph, and 17.5 mph for the second.
\______________________\ 15 So we average 12.5 & 17.5 mph = 15 mph.
\   \ Notice the shear is +5 at each level.
\  17.5 ==>  \ This is the same as considering the whole
\______________________\ 20 layer, above, as we did in equal thickness
showing translation is averaging 10,20=15
and the shear is 2010 or 10. However, the
tilt is 1/2 the shear. So for the bottom
layer, the tilt is 17.5  12.5 = 5.
____________________ 20
  Double thickness of the initial o dust layer passing
  through a wind layer 1/2 it's size. In this case, let's
 20 ==>  pass 1/2 of the initial o dust layer through one at a
  time by dividing it into 2 equal layers.
 
____________________20
____________________ 20 t = 1/2 top initial o dust layer assume no wind shear.
 
 20 ==> 
____________________20 b = 1/2 bottom initial o dust layer assume no wind shear.
 
 20 ==> 
____________________20
:
_________________________10 With the same thickness, b passes through as seen here. Now t does the
\  \ same because it's identical to b. So for the whole layer t + b, we have
\  15 ==>  \ 2 identical shears and translation and assuming b does not move into the next
\______________________\ 20 layer below yet, or assume that we have a layer with no shear (20 mph @ top &
ds bottom), b is translated to b' while t falls to t'. This way we examine the
layer passing through before being sheared by the next layer. In all the previous
ds cases above, the layer passing through was less than or equal to the new shearing
_________________________10 layer below it. In this cases, the shearing layer is only half as thick so it's travel
\  \ time tt to react on the falling layer is half the falling layer if it were the same
t' \  \ thickness. This half effect is the ___ line halfway down to the left. While b' was
\ ____________________ \ still being translated by the lower half of the shearing layer, t' was added to the
\   \ top of b' to complete the original (t + b) dust cylinder thickness within this 1st
b' \  \ wind layer. So for the total shear, we have 2 x 1/2 = 1 or the shear is still 2010=+10,
\______________________\20 the same as the layer it was passing through but twice the thickness or time to do so.
ds
The factor that links this all together is the travel time tt which is a particle's
fall velocity through the lower shearing layer. Now holding the fall velocity constant
through the lower shearing layer, the travel time depends on the shearing layer's
thickness, which translates the upper portion at a different rate and distance
than the lower portion. Thinking of a single particle at the bottom (b) of an original
layer passing through all the levels below it to the ground arrives at b', then the
particle at the top of an original layer (t) will arrive at the top of the ground layer's
cylinder at t', which will be somewhat upwind of it's more downwind partner b' because of
the accumulated +, shear components of all the layers and t' has yet to fall through the
last (surface) layer and may become more downwind than b', depending on the tilt of this
last dust cylinder layer as seen below.

t  t

==>  ===>


b t' b'  b b' t'
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
ds > > > >
initial final initial final
On the left ^, the lighter winds ==> causes t descent to b to still be upwind of the permanent b'.
On the right, stronger winds ===> with less initial shear (ds) causes t' to be displaced downwind
from b'.