I have processed the *.tif-file from https://astrogeology.usgs.gov/search/map/Mars/GlobalSurveyor/MOLA/Mars_MGS_MOLA_DEM_mosaic_global_463m as gdalbuildvrt mars.vrt <filename>, and now gdalinfo mars.vrt gives

Driver: VRT/Virtual Raster
Files: mars.vrt
Size is 46080, 23040
Coordinate System is:
PROJCS["Equirectangular Mars",
Origin = (-10669675.197320545092225,5334837.598660272546113)
Pixel Size = (463.093541550370901,-463.093541550370901)
Corner Coordinates:
Upper Left  (-10669675.197, 5334837.599) 
Lower Left  (-10669675.197,-5334837.599) 
Upper Right (10669675.197, 5334837.599) 
Lower Right (10669675.197,-5334837.599) 
Center      (   0.0000000,   0.0000000) (  0d 0' 0.01"E,  0d 0' 0.01"N)
Band 1 Block=128x128 Type=Int16, ColorInterp=Gray
  NoData Value=-32768

My questions (as I am a newbie in GIS) are the following: 1)does this all mean that the Mars surface have been projected to a 2-dimensional plane, in particular to a rectangle occupying the space between (-10669675.197,-5334837.599) and (10669675.197, 5334837.599)? In other words, does this mean I do not have to reproject it using e.g. gdalwarp?

I query the data with e.g. gdallocationinfo -geoloc mars.vrt x y to get the "height" of the point (x,y). Here x,y lie in the above rectangle. For example, when I type, say, dallocationinfo mars.vrt -geoloc 10 10, I get

  Location: (23040P,11519L)
  Band 1:
    Value: -1512

i.e. the numbers are different from the ones supplied. Also, the "height" is negative, but I presume it is normal since what is measured is below the "sea-level".

I read the man-page of gdallocationinfo more carefully and I seel that P-and L-suffixes of the above numbers come from pixel/line space.

  • Please take the Tour, which emphasizes the importance of asking One question per Question. – Vince Apr 3 at 12:22
  • Thanks -- updated! – Ilonpilaaja Apr 3 at 14:28
  • Why are you building a vrt? The only data in that vrt is coming from the tif itself. You should get equivalent gdalinfo output from running on the tif. The tif is a raster in that projection - ie approximately 463x463m squares (at the equator, and squeezed at the poles if I understand the equirectangular projection). – Spacedman Apr 3 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.