Take the 2-minute tour ×
Geographic Information Systems Stack Exchange is a question and answer site for cartographers, geographers and GIS professionals. It's 100% free, no registration required.

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.

share|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
add comment

1 Answer 1

up vote 2 down vote accepted

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.

share|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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.