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I have several coordinates provided in WGS 84 which I need to rasterize. For that purpose I simply interpret the lat,lon coordinates as x,y coordinates and I apply some scaling. So e.g. anything in the rectangle described by (54.3460005,10.1494873)(54.2984447,10.2002991) will be scaled to the rectangle (0,0)(10000,10000). In further processing steps this raster is translated back into the WGS 84 coordinate system.

I am aware of that this simple translation would be problematic for larger areas but I only look at smaller urban areas so I think the error introduced is acceptable. Further, visual inspection showed that the difference between my raster method translated back into WGS 84 coordinates and the original coordinates is practically neglectable. Yet I feel uncomfortable doing so without knowing the exact error I introduce by this method. At least I want to report it.

How can I calculate on a theoretic basis how large the error is which I introduce by simply interpreting lat,lon (WGS 84) values as x,y values?

EDIT: The error I am concerned about is the level of distortion of a pixel. While in the northern part of the map a pixel covers a smaller area (when being in the Northern hemisphere like my example), in the southern part the area covered is larger.

  • Depends what you mean by "error". Map projections (which is what you are doing by treating lat-long as cartesian coordinates) have errors in distances, directions, and areas depending on the projection. There's no error in a point location in the transformation of a single point (except maybe some computational precision). – Spacedman Apr 8 '17 at 7:54
  • As I will present the rasterized graphics as-is I am concerned about the area covered by k pixels in different sections of the map. One idea I had was to report the geographical distances in the northern and in the southern border of the map using the spherical geographical distance function D = 6371,009 * ((lat1-lat2)^2 + ((cos(lat1+lat2) / 2) * (lon1-lon2))^2)^0.5 taken from en.wikipedia.org/wiki/Geographical_distance – mkastner Apr 8 '17 at 8:16
  • (To an excellent approximation) the area distortion is proportional to the secant of the latitude, whence the relative distortion across the map is proportional to the ratio of secants of the latitudes closest to a pole and closest to the Equator. (There are other distortions inherent in your method: of aspect, orientation, and angle, but these you do not mention.) In the example you give this ratio amounts to 1,155 parts per million, or around 0.11%. – whuber Apr 11 '17 at 23:08
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One can use one of the geographical distance functions presented e.g. in https://en.wikipedia.org/wiki/Geographical_distance and compare the Northern with the Southern map border length.

When sticking to the example:

north = 54.3460005
west = 10.1494873
south = 54.2984447
east = 10.2002991

Using tools like http://www.movable-type.co.uk/scripts/latlong.html show that at lat=54.3460005 the distance from east to west is 3.293km while at lat=54.2984447 it is 3.297. So when you go 5.288km south, there is a difference of 4m already.

That website also gives a hint about taking care about radians and degrees.

Calculating the distance by hand assuming a spherical shape:

radius = 6371.009km

delta_east_west = 0.0508118° = 0.0008868332088649

D_north = radius * ((phi1-phi2)^2 + ((cos((phi1+phi2) / 2)) * (lat1-lat2) * pi/180)^2)^0.5 = 6371.009 * cos(54.3460005) * 0.0008868332088649 km = 3.2933m

D_south = 6371.009 * cos(54.2984447) * 0.0008868332088649 km = 3.2971m

More precise results can be achieved by assuming an ellipsoidal earth, improving the result by 0.3%, cf. How accurate is approximating the Earth as a sphere? .

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    Your further note is incorrect: interpreting lat,lon as Cartesian coordinates is called the "Plate Carree" projection. Except near the Equator it isn't remotely like a Mercator projection. – whuber Apr 11 '17 at 23:10
  • Thanks for your comment, in the Wikipedia description it looked like a Mercator projection can be rectangular as well but The Plate Carree projection is a by far better match. – mkastner Apr 13 '17 at 6:16

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