Toady has said that for minecart physics purposes, a tile was 2x2 meters, not 1x2. I don't think we have definitive dimensions in terms of SI units otherwise. I think you could safely make some other assumptions too, e.g. that the mantle and core of the planet are below hell. I.e. Dwarves can only dig about in the crust, which is only 5 (ocean) to 50km (mountains) thick on earth as well.

Each embark tile being 48x48 was known, although I'm not sure how well it's documented e.g. on the wiki.

I think it should be possible to calculate the radius (/circumference) of the planet even if you take the latitude of the hot side of the map as an unkown, as you still have a temperature gradient over a known distance. May involve calculus though, so is of course more complicated, and said latitude being 0 will indeed probably be a valid answer. Or at least a possible one.

edit: What did you assume/measure for the solar constant/how did you, in general, get from the temperature gradient to the radius of the planet? For a solar constant S

_{0} (skipping the subscript from here on), the amount of radiation reaching the surface at any given location is (at solar noon, averaged over the year, or during the equinoxes) is Q = S*cos(y), where y = latitude, north is positive. For 2 places on the map, Q

_{1}=S*cos(y

_{1}) and Q

_{2}=S*cos(y

_{2}) must be true, and thus you can get

Q

_{1}/cos(y

_{1}) = Q

_{2}/cos(y

_{2}), or

Q

_{1}/Q

_{2}= cos(y

_{1})/cos(y

_{2})

So hm. You can get ratios of the cosines of latitudes. Which you have done for the case of y

_{1} = 0 degrees and y

_{2} = 60 degrees (insolation half of the equator's). Note that I'm using latitudes, you were using angles of incidence. With no axial tilt, the sun's height above the horizon is 30 degrees at a latitude of 60, and sin=cos(Pi/2-a), or vice versa.But what if the bottom of the map is instead e.g. y

_{1} = 20 degrees? Then with Q

_{1}/Q

_{2}=2 as before, y = arccos(0,5*cos(20 degrees)) = 62.0 degrees (to 3 S.F.). Huh. So apparently y

_{2} doesn't react very quickly to changes in y

_{1}. Further testing shows that eg. y

_{1} = 60 degrees only gives y

_{2} = 75.5 degrees.

There's still the option of what if the hot end of the map is in the other hemisphere, but I don't feel like getting into that, and it does seem unlikely.

There's still lots of stuff that we're just assuming/can be artifacts of the time/space compression, especially in fort mode. I was about to remark that we don't really know if there even is a sun (like you commented about the moon earlier), but we do at least know that dwarves do get "irritated/nauseated by the sun" if cave adaptation has set in

. One thing that bugs me is that fort mode has no day/night cycle, for instance, so are the surface temperatures supposed to be the ones for noon, average daytime temperatures, or averaged over 24 hours?

Just hoping I got the math more or less right, it's been a couple of years since the climatology course where we did these kind of calculations.