関連記事

No Man's Sky planet size calculation

I’ve always wondered how big the planets in No Man’s Sky are. I already knew they were going to be small, but it turned out that they are tiny.

I placed two save beacons at a distance of 418u from each other, collected the coordinates using the analysis display, obtaining the coordinates A = -32.81 / +54.24 and B = -32.50 / +53.99.

Using the haversine formula and assuming that u = meter, I found that the distance between the two points on the test planet should be 41.67 km (25.89 mi) instead of 418 m (0.26 mi) if the planet had the diameter of the Earth.

The question soon arose of what the size of the planet should be for the result to be 418 m and it turns out that the planet should have a radius of 64 km (39.77 mi) or 2⁶ km (it may have been a coincidence, but I think it is likely that the radius of the planets is 2⁶ km due to how the calculations have to be done within the game). The surface area is 804 km² (310.43 mi²).

For comparative purposes, the test planet would be the size of Weywot, which is a natural satellite of a trans-Neptunian dwarf planet in the solar system. Almost the distance from Boston (Massachusetts) to Providence (Rhode Island), i.e. absolutely minuscule.

Calculation details in the comments.

(there may be grammar mistakes)

Edit: If the average surface area of ​​the planets in the game is 804 km² as originally calculated, then the total explorable surface area in the game is equivalent to 29 trillion Earth planets (29,075,048,491,006.6), which would be 72 times the amount of stars present in the Milky Way.

submitted by /u/brunnomenxa
[link] [comments]

続きを読む シェア
0

Community

Popular Posts