[GRASS-user] EPSG 32633 and spatial resolution for hourly LST product from Copernicus

[Markus Metz]

On Mon, Oct 1, 2018 at 4:28 PM Nikos Alexandris <nik at nikosalexandris.net>
wrote:
>
> Nikos wrote:
>
> >> Import using r.in.gdal, _without_ any of `-l` or `-a` and then I get
the
> >> closest to the reported spatial resolution. Else, with `-a`, for
> >> example, the spatial resolution is not as close to the "original" one.
> >> Makes sense?
>
> >what would be the output resolution and extends with r.in.gdal -a?
> >Generally r.in.gdal -a provides the best results.
>
> Importing with/out `-a` and `-al`:
>
> for MAP in $(g.list raster pattern=import*) ;do echo $MAP && r.info -g
$MAP && echo ;done
>
> import
> 360 degree EW extent is exceeded by 0.00019226 cells
> north=80.0223214291667
> south=-79.9776823855556
> east=179.977687153889
> west=-180.022321429167
> nsres=0.0446428582072328
> ewres=0.0446428582072241
> rows=3584
> cols=8064
> cells=28901376
> datatype=CELL
> ncats=0
>
> import_with_a
> north=80.0223333333333
> south=-79.9634444444444

something went wrong here:
> east=179.945666666667
east should be 180 - ew_res / 2

just like
> west=-180.022333333333
i.e west = -180 - ew_res / 2

I will investigate

> nsres=0.0446388888888889
> ewres=0.0446388888888889
> rows=3584
> cols=8064
> cells=28901376
> datatype=CELL
> ncats=0
>
> import_with_al
> north=80.0223333333333
> south=-79.9634444444444
> east=179.968
> west=-180
> nsres=0.0446388888888889
> ewres=0.0446388888888889
> rows=3584
> cols=8064
> cells=28901376
> datatype=CELL
> ncats=0
>
> The first "import" map's ns/ew resolution is closest to the original.
>
> >>
> >> Better to cut off the west side (?):
> >> ```
> >> g.region raster=g2_BIOPAR_LST_201606220100_GLOBE_GEO_V1.2.nc_LST -pag
> >w=-180 e=180
> >
> >this shifts the grid by half a cell to the east. The -a flag does not
make
> >sense because 1) you want to force new extends, 2) there is no resolution
> >given for use with the -a flag.
>
> More precise, this shifts the computational's region grid by half a
> cell to the east. Right?

it shifts the raster map to the east, not the computational region.
>
> If the default of `-a` is "to align the region resolution to match the
> region boundaries",

?
from the manual:
"With the *-a* flag all four boundaries are adjusted to be even multiples
of the resolution, aligning the region to the resolution supplied by the
user."

i.e. with the -a flag, the region boundaries are modified and the
resolution is kept. The default (without -a) is to align the region
resolution to match the region boundaries.

> won't the above command try to modify the spatial
> resolution, of the computational region, so as to perfectly fit inside
> the currently set, of user provided, boundaries?
>
> If, say, the region's boundaries are n=80 s=-80 w=-180 e=180 and the
> raster map's rows and columns 3584 and 8064 respectively, won't the
> above command try to adjust the region's resolution so as to fit these
> cells inside these boundaries?

If you want to "adjust the region's resolution so as to fit these cells
inside these boundaries", you must not use the -a flag.

Markus M
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.osgeo.org/pipermail/grass-user/attachments/20181001/9923141c/attachment.html>