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Restated, how do I convert arc degrees to meters? For example, I have an elevation data raster that has the following metadata:

WEST LONGITUDE=64.97798789° E
NORTH LATITUDE=33.02003415° N
EAST LONGITUDE=66.03338707° E
SOUTH LATITUDE=31.98030163° N
PROJ_DESC=Geographic (Latitude/Longitude) / WGS84 / arc degrees
PROJ_DATUM=WGS84
PROJ_UNITS=arc degrees
EPSG_CODE=4326
NUM COLUMNS=9001
NUM ROWS=9001
PIXEL WIDTH=0.0001111 arc degrees
PIXEL HEIGHT=0.0001111 arc degrees

I can compute by hand the pixel width in arc degrees as follows:

(EAST LONGITUDE - WEST LONGITUDE) / NUM COLUMNS

Similarly, I can compute by hand the pixel height in arc degrees as follows:

(NORTH LATITUDE - SOUTH LATITUDE) / NUM ROWS

My question is how to compute the pixel width and height in meters.

| improve this question | |
  • Do you have access to GIS software? You could try and project it using meters as the unit of measurement. – djq Nov 17 '10 at 22:05
  • Thanks for the suggestion. I know how to use Global Mapper, which is where I obtained the sample data above. Through a sequence of steps I am able to use Global Mapper to see the pixel width and height as meters, but it's a hassle, and I want to be able to verify by hand what Global Mapper says. – Sipp Nov 18 '10 at 16:02
  • I'm sure you're aware that the length of a degree of longitude varies with latitude (from infinitesimally small at the poles to roughly 111 km at the equator). I would have thought that to calculate cell dimensions in meters by hand would be more of a hassle than with e.g. Global Mapper. – jbaums Nov 18 '10 at 20:12
  • That said, take a look at Wikipedia's Great-circle Distance page. en.m.wikipedia.org/wiki/Great-circle_distance – jbaums Nov 18 '10 at 20:42
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You can use the quick-and-dirty (yet fairly accurate) conversion described here. As an example,

  • Pixel height will be essentially constant (it is constant on a sphere and approximately so, to a fraction of a percent variation, on the WGS84 ellipsoid). 0.00011111 degrees = 1/9000 degrees = (approximately) 111111/9000 meters = 12.35 meters.

  • Pixel width will be the cosine of the latitude times the height. At the top of your grid, cos(33.02003415) = 0.8385, whence the width will be 10.35 meters. At the bottom of your grid, cos(31.98030163) = 0.8482 for a width of 10.47 meters. The variation is so small you can safely linearly interpolate between these to estimate widths at other locations in the grid.

| improve this answer | |
  • Thanks! Quick and dirty is exactly what I was looking for...just something to get a figure of resolution in meters per pixel when Global Mapper reports it to me in arc degrees. – Sipp Nov 20 '10 at 16:54

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