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My question is related to this:

http://stackoverflow.com/questions/7927863/location-data-format-for-adehabitat-package

I have a relational database with 'world tiles'. Each tile has a side length of 0.25 degrees (plus top left and lower right lat/long coordinates). This means I have 1036800 tiles (720 for latitudes and 1440 for longitudes). I would like to understand distortion a bit better (sorry about this stupid question). How do you compute the size of each tile? I understand that the distortion is largest at the poles and that the tiles are not necessarily square. Any links/feedback would be very much appreciated. Many thanks!

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2 Answers

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yosukesabai's answer works for general applications where a spherical Earth is an acceptable assumption. The calculations become much more complicated if one considers the Earth as an oblate spheroid that is not a true sphere. When I had this problem a few months ago, I needed to calculate with very exact precision each pixels area of a raster in a geographic coordinate system. I wrote an ArcPython script that made pixels as polygon ArcPy objects, projected them to UTM, and calculated their area. This was then output to a different raster. If you are interested, I can send you this code (it's on my home computer).

Another option to get a good understanding of the distortions caused by map projections is to look at Tissot's Indicatrix (wikipedia),(GIS.se tissot topics). Each dot represents a perfect circle on the actual face of the Earth. The degree to which the circles are squashed and stretched on a map should give you an idea for how that map projection incurs distortion.

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These are good points. Another interesting consideration is this: what do the edges of the cells represent? Are they geodesics connecting the cell vertices, or are the top and bottom edges actually intended to lie along lines of latitude (which, as geodesic circles, are straight only at the equator). Also, when you projected to UTM, you automatically made an error that is typically about 0.04% and varies rapidly east to west. Arguably, then, assuming a spherical earth and using spherical geometry may have been a more accurate solution for your problem. – whuber♦ Nov 3 '11 at 17:45
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It may also be worth pointing out that @yosukesabai's answer can be made extremely accurate by first replacing the latitudes by their authalic latitude equivalents and then applying the spherical formula. Authalic latitudes can be computed (with some effort) but they can also be looked up in tables, interpolating as necessary, which is a straightforward task on a relational database platform. This solution will be far more accurate than any projection-based calculation. – whuber♦ Nov 3 '11 at 17:47
Thanks for that authalic latitude link...I never knew that transformation was possible. Also, when I did the UTM projection I made the column of cells right along the central meridian of the UTM zone, which should have even less distortion. – dmahr Nov 3 '11 at 18:37
I think csetzkorn is not that much into accuracy but problem is to finding a projection method which is (1) preserve area (2) preserve angle locally and (3) ideally cover entire globe at once, with only exception of inhabitable region near the polar regions. If you see the link near the top of original post, you see that he want to apply a model called ecological niche factor analysis (ENFA) which appears to expect location is specified by square grids. i am wondering what is a good map projection for such purpose. sorry for stepping into somebody else's thread. – yosukesabai Nov 3 '11 at 18:59
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Approximating the earth with a sphere gives you an idea without too much hassle in math.

http://en.wikipedia.org/wiki/Solid_angle#Latitude-longitude_rectangle

( sin(phi_N)- sin(phi_S) ) * ( theta_E - theta_W), where phi_N and phi_S are north and south lines of latitude and theta_E and theta_W are east and west lines of longitude

if you want the approximate area of these lat/lon rectangles, multiply 4*pi*r^2 by the formula above, where r is the radius of the earth, approximately 6371 km.

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+1. Just be sure to convert degrees to radians before applying the formula :-). – whuber♦ Nov 3 '11 at 14:17
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@whuber: Thank you for fixing grammar + spellings... – yosukesabai Nov 3 '11 at 15:18

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