FallVect.dia
==> 10 mph V1
ft
10
100 *
12
80 *
14 mph
60 *
             50'
 16
40  *


 18
20  *


 20
0  *

_____________+_____________________________
0 15 20 mph 40
====> 20 mph V2
In this case, we ignore the directional shear and assume that it is the same direction
at the top and bottom of a wind layer 100' thick. This layer has a particle * falling through
it @ 10 mph at the top and 20 mph at the bottom. The particle has a constant fall velocity,
but not needed here because the x axis is in units of speed (mph) and independent of the
fall velocity. Regardless of it's fall velocity, the particle passes through increasing wind
velocities showing it's horizontal downwind speed, labeled above the particle, at the bottom
of each 20' descent.
The average speed of 15 mph is shown by the dashed line 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 the average speed. The particle fell through 5 levels before it
reached 20 mph but averaged 15 mph.
ft
10
100 *
12
80 *
14 mph
60 *
             50'
 16
40  *


 18
20  *


 20
0  *

__________________________________________
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, not 20T. Unit check RT = (mi/sec)*sec = mi or
ft converted.
As this applies to a wind vector with the same direction, we also have to apply it
to a vector shift in direction also. Do we average the directions separately? No.
This is important when we average wind direction for hourly reports for NOAA,
Air Force & Navy hourly wind direction observations for aircraft and in air pollution models,
but not here. We must average vectors, which includes direction & velocity as we are
doing a trajectory analysis, keeping track of it's path position, or complete path of a
particle.
One of the coordinate transformation to be made is the math world to the map world.
In the math world, 0 degrees points to the right or East. In the map world, 0 degrees
points North, or 90 degrees cc to the map world. 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 but here, we stopped at the
surface and would start applying our cylinder deposition. However now that this surface
cylinder is not vertical but assumed shifted by direction and speed, we subdivide
this tilted cylinder into say 5 sub cylinders that are assumed vertical but treat
all it's different center coordinates as a unique cylinder to apply our new method.