Cell Size.dia
Depending on how many points in a tile, which vary in size, we determine that a
data point represents an area cs^2 of the grid. However, if we use the raw data, we
would have many holes. We have allowed for 2 holes between data points to be filled
in by interpolation. This will also change the grid size. As an example of all this,
we look at New Orleans' .las data 468.las to make a 468.grf grid.
1. Raw Data Points: = npr = 4948963 from NEWGS.c
minx=777346 maxx=783555 miny=3315337 maxy=3322419 from 468.xml
dxm = 6209 The # of data points, if distributed equally 1 m^2
3322419 ------------------ over the 468.grf tile grid, is the area (dxm*dmy).
| | However, we divide by the # of data points npr=4948963.
| cs | So (6209 x 7082)/4948963 = 8.8851. However, since it
| _ | isn't a square grid such as MIA & PHL, we have dmy > dxm
| |.|c | 7 or a wider range of meters along the y-axis than x.
d | |_|s | 0 So if equally space along x, rmbpx=rmbp*dxm*dym=7.7898 &
l y | | 8 if larger equally space along y, rmbpy=rmbp*dym*dxm=10.1344.
y m | | 2
| | However, we assume that the lidar receptor tries for samples
| | in fixed distances in both x & y sweeps, ignoring direction.
| | For that case, we stay with rmbp=8.8851.
| |
| | Then the range of the grid for this case is lx=dxm/rmbp=699
3315337 |__________________| & ly=dym/rmbp=797 because dmy > dxm. Thus if the data points
777346 783555 are equally space distributed, we would have a data point for
lx every m along x & y (no holes but also no overlaps). Since we
allow for 2 holes, we could expand the lx,ly grid for more
accuracy in case 2 points may overlap so we separate them into
an adjacent grid space. That is, the original cs along an axis
is now divided into thirds so our new cs=cs/3=2.96m, then the
corresponding tripled lx=dxm/cs=2096 & ly=dym/cs=2391. Finally,
the cell size cs in feet (fpc) is 9.717' for MIA @ cityl.dat.
This is close enough to the PHL fpc of 10'.