I imagine an actual answer to this is impossible since WorldA doesn't really correspond to the world, but roughly (discounting borders/coastlines) what land area is represented by 1 pixel on WorldA?
This can be done either comparing the size of the map with the real proportions of the Earth employing the circunference or by aproximating it from the area.
Let's start with the circunference method. Discounting the two pixels that make the borders of the map, WorlDA is 1202px wide at the equator. Earth's
equatorial circunference is 40075.017 Km. So assuming the WorlDA shows the Earth's 360º of longitude, a pixel in the equator should be roughly equivalent to roughly
33.34 Km or 0.2995 degrees of longitude. Note that I calibrated the WorlDA so the center of the map alligns the center of the picture, thus making both hemispheres equal, not sure if current "standard" WorlDA does that or not, also pretty sure it does not reach 90ºN.
We could do the same for the
latitude. The WorlDA is centered on the Oslo Meridian, and again discounting the two pixels of the border (658 in total), and that Earth's polar circunference is 40007.86 Km (which we would have to divide by two since we go from pole to pole, not from one pole all the way around to the same place) would give a value of
30.40 Km per pixel or 0.2735 degrees of latitude.
However, one must take into account that while longitudinal variations do not affect the actual size of a degree, latitudinal variations do (degrees closer to the pole are shorter), and the fact that the projection itself is wrong.
This is no equal area projection,
so the value of 33.34 X 30.40 Km only works at the equator, pixel sizes closer to the poles are way bigger than they should be, to the point where at the last row of pixels before the pole (646 pixels across and 30.40 Km from the pole*) it is wrong by several orders of magnitude.
Another way to do it is
considering the map as an ellipse, and we have both semiaxes (601px and 329px) respectively, which gives us an area of 621184 square px. The area of the Earth is 510072000 square Km, so one pixel would correspond to
821.1287 square Km per pixel or a pixel of 28,6553 Km of side. Again, this number is also distorted by the projection of the map, same distortion as with the pixel count before.
Again, these are approximations and should not be taken seriously. The first only works at the Equator, while the the second only works at a hypothetical point where the distortion is somehow cancelled in the mid latitudes (if I knew the projection I could try to pinpoint it, but that's too much work).
