1

I tried using GDAL and got:

# gdalinfo nicholas_map_2.tif 
Driver: GTiff/GeoTIFF
Files: nicholas_map_2.tif
       nicholas_map_2.tfw
Size is 28799, 27875
Coordinate System is:
GEOGCS["WGS 84",
    DATUM["WGS_1984",
        SPHEROID["WGS 84",6378137,298.257223563,
            AUTHORITY["EPSG","7030"]],
        AUTHORITY["EPSG","6326"]],
    PRIMEM["Greenwich",0],
    UNIT["degree",0.0174532925199433],
    AUTHORITY["EPSG","4326"]]
Origin = (-122.325695999999994,37.876846000000000)
Pixel Size = (0.000000441883000,-0.000000350276000)
Metadata:
  AREA_OR_POINT=Area
Image Structure Metadata:
  COMPRESSION=LZW
  INTERLEAVE=PIXEL
Corner Coordinates:
Upper Left  (-122.3256960,  37.8768460) (122d19'32.51"W, 37d52'36.65"N)
Lower Left  (-122.3256960,  37.8670821) (122d19'32.51"W, 37d52' 1.50"N)
Upper Right (-122.3129702,  37.8768460) (122d18'46.69"W, 37d52'36.65"N)
Lower Right (-122.3129702,  37.8670821) (122d18'46.69"W, 37d52' 1.50"N)
Center      (-122.3193331,  37.8719640) (122d19' 9.60"W, 37d52'19.07"N)
Band 1 Block=28799x32 Type=Byte, ColorInterp=Red
  Mask Flags: PER_DATASET ALPHA 
Band 2 Block=28799x32 Type=Byte, ColorInterp=Green
  Mask Flags: PER_DATASET ALPHA 
Band 3 Block=28799x32 Type=Byte, ColorInterp=Blue
  Mask Flags: PER_DATASET ALPHA 
Band 4 Block=28799x32 Type=Byte, ColorInterp=Alpha

I assume pixel size is what I'm looking for? But is that in degrees?

Nikos Alexandris
  • 1,350
  • 16
  • 31
Nicholas Pilkington
  • 41
  • 2
  • 3
  • 1
    That would be in degrees. You could use this estimation that one degree is approximatly 111,111m. That means your pixels are approximately 5cm x 4cm. Does that seem right to you? That's pretty high resolution. I would follow @Ryan's advice and convert to a projected coordinate system. I am a bit dubious of the negative pixel size, however. – Fezter Jan 17 '14 at 03:52
  • actually Y pixels are often negative in size to allow for the different origins used in geography and graphics – Ian Turton Jan 17 '14 at 09:26
  • Compute the lengths of the side of the image using a geodesic calculator. This gives about 1120 m E/W by 1084 m N/S and a resolution of 0.03885 m squared. The negative Y pixel size is totally standard. It just reflects the normal storage order for images, i.e., top down. This is in the negative sense for latitudes (and northings). – cffk Jan 18 '14 at 01:33

1 Answers1

1

You could convert the raster to a projected coordinate system (ie: UTM). The pixel value will be represented in meters. Then with simple math you can calculate the resolution in cm (1 = 100cm, 0.5 = 50cm, 0.25 = 25cm, etc.)

Ryan Garnett
  • 9,479
  • 8
  • 61
  • 106