[Proj] To handle geographic info from HDF5 formated Satellite Data

[Aydın Gürol Ertürk]

Frank,

Thank you for response.
if I use your suggestion, for example pixel (1000,1000);

longitude = -25 + 0.560648 * 1000 = 5581.48
latitude = 75 - 0.560648 * 1000 = -5531.48

So, it is not possible.

I do not have problem converting projected space to geographic space using
proj lib. My problem is how can calculate the distance from the corner using

below parameters than I can convert those to lat/lon.

Best Regards,
Aydin


2008/5/7, Frank Warmerdam <warmerdam at pobox.com>:
>
> Aydin wrote:
>
> > I try to handle geographic coordinates (lat/lon) from a sattellite data
> > in HDF5 format. You can find necessary information below.
> > Could you please tell me how can initialize proj4 program and how can
> > obtain lat/lon values for one pixel.
> >
> > detailed info is below:
> >
> > geo_number_rows = 4500
> > geo_number_columns = 3165
> > geo_pixel_size_x = 0.560648
> > geo_pixel_size_y = -0.56215554
> > geo_row_offset = -4759.928
> > geo_column_offset = -1636.3636
> > geo_dim_pixel = KM, KM
> > geo_pixel_def = CENTRE
> > geo_product_corners = -25.0, 25.0, -25.0, 75.0, 45.0, 75.0, 45.0, 25.0
> > geo_product_center = 9.719999, 50.47374
> >
> > projection_name = EQUIDISTANT_CYLINDRICAL
> > projection_proj4_params = +proj=eqc +lat_ts = 0 +lon_0=0.000000
> >
>
> Aydin,
>
> PROJ.4 does not address transformation from pixel/line space to
> georeferenced
> space.  Just projected space to geographic space (and back).  The info
> above
> does not indicate an ellipsoid, but we might assume WGS84 as a guess.
> If the corners are in long/lat ordering we can assume the top left corner
> is 25W, 75N.  This can be converted to equidistant cylindrical coordinates
> like this:
>
> proj +proj=eqc +lat_ts=0 +lon_0=0.000000 +ellps=WGS84
> -25 75
> -2782987.27     8348961.81
>
> So, the top left corner is -2782987.27mE, 8348961.81mN.
>
> Actually, treating this as a projection issue is rather silly since
> the projection is eqc which is just a scaling of geographic space.
> And since all the locations and pixel sizes are expressed in decimal
> degrees you might as well just ignore that aspect, and treat this as
> a WGS84 image with the given corners.
>
> So, if the top left corner is -25,75 and the pixel size is
> 0.560648, 0.56215554 you could compute pixel (20in from left,15 down
> from top) as:
>
> longitude = -25 + 0.560548 * 20
> latitude = 75 - 0.560648 * 15
>
> etc
>
> Best regards,
> --
>
> ---------------------------------------+--------------------------------------
> I set the clouds in motion - turn up   | Frank Warmerdam,
> warmerdam at pobox.com
> light and sound - activate the windows | http://pobox.com/~warmerdam
> and watch the world go round - Rush    | President OSGeo, http://osgeo.org
>
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