Non Overlapping Cylinder Method
vol = cylinder h pi (d/2)^2 = h x Area
vel = mean wind velocity
Cylinder h = cylinder thickness
Side View tt = travel time h/fps
No Shear dis = downwind distance
.<---------------- Twice this distance --------------->.<------------==>
by wind translation
_______ dis = vel x tt
| | dis = vel x h/fps
| |
| |
| | 8
| | h
| |
| |
|_______|
d .
_______ Here we have h/2, tt=h/2/fps _______
| | so dis = vel x h/2/fps or | |
| | half the original distance. | |
| | 4 vol = 2(h/2) area = h x area. | | 4
|_______| |_______|
_______ _______ _______ _______
| | 2 | | | | | | 2
|_______| |_______| |_______| |_______|
Now we have h/4, tt=h/4/fps
so dis = vel x h/4/fps or a
quarter of the original distance.
vol = 4(h/4) area = h x area.
_______ _______ _______ _______ _______ _______ _______ _______
1 |_______|_______|_______|_______|_______|_______|_______|_______| 1
Now we have h/8, tt=h/8/fps
so dis = vel x h/8/fps or an
eighth the original distance.
vol = 8(h/8) area = h x area.
________________________________________________________________
1 /_______/_______/_______/_______/_______/_______/_______/_______/ 1
Wind Shear (/) Added
In this diagram, we forced the issue of subdividing the cylinders until
all the sub cylinders came end to end to match how far the original cylinder
would have traveled or the top layer subdivision. How did we force this issue?
We chose a dummy wind velocity to make the eight layers come out end to end.
An increase in wind velocity would increase the downwind distance traveled
and we wouldn't have the last 2 end to end sketches above end to end.
If we subdivide further, can't show it graphically here, say 1/2 again,
the distance traveled would be 1/2 again and there is no more room to stack end
to end. What happens, we will be stacking them on top of each other or back to
the overlap. So we have double the new height which is 1/2 again which means the
overlap height remains constant. So n # of subdivisions matches n # of stacks,
h/n x n = h, which remains constant.
Notice in all the diagrams above, the total volume is always the same as the original
and better be. Since original cylinder vol = h x base area, we have n subdivisions
x (h/n) x area, n(h/n) x area = h x area, the original volume. So regardless of the
# subdivisions, the cumulative vol is constant. However, a unique subdivision can be reached
that the sub cylinders can be end to end with no more increase in height h by stacking. This
eliminates the over prediction lumps and again smoothes the data within the process.
Instead of 8 or any height h, we set the scale as h=100% or 1.0 then fractional
subdivisions of this is the reduced % or another type of probability as we did with
the overlap stacked cylinders, shoved and pressed pancakes demo diagram.
All we need to calculate is how many cylinder diameters d can be stacked to the
downwind travel distance from the top of the cylinder, then this is the # of subdivisions
required or h/n. We will use equivalent square areas (s) instead of circles (d)
to simplify. We subdivide the square by original top level travel n=dis/d or dis/s.
Then rat=1/n is the % probability. Actually, rat = s/dis directly.