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I heard from one source 21km square, and from another 245km square...

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    360 / 1920 degrees width and 180 / 1080 degrees height. This cannot be converted into metres. – Michael Stimson Jun 16 '16 at 5:06
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    It depends on the map projection. – daniel.neumann Jun 16 '16 at 5:16
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    ... and where the pixel is on the globe. – PolyGeo Jun 16 '16 at 5:26
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Assuming you are talking about a map of the world in Mercator Projection, Something like this:enter image description here

Then, You should know that not all pixels will represent the same area. There is a large amount of distortion as you move towards the poles, the area represented by each pixel will decrease.

Tissot's indicatrices show a good visual representation of the distortion in the area: enter image description here

But the area and the distortion both depend on the projection of the Map, Unless it's a projection that preserves area, the Pixels will not represent the same area. Since we are trying to map 3d features in 2d, there will always be some sort of distortion in every projection.

Most projections try to best account for a specific type of distortion, for example, In the Mercator Projection linear scale is equal in all directions around any point, thus preserving the angles and the shapes of small objects (which makes the projection conformal), but the projection distorts the size of objects as the latitude increases from the Equator to the poles, where the scale becomes infinite. Projections will always have to make a compromise.

Take a look at this, and this might give you some insight on distortions and map projections.

PS: To answer your original question, if we assume that the pixel in question is at the equator, and there is no distortion (of area) at the equator (which is the case with Mercator Projection), Each pixel would be 20.85km wide ( Circumference at the equator [40,030 km] divided by pixels [1920] ).

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