Making triangle faces of a world polyhedron. I will start with a world surface area and get the diameter, radius and circumference. The world div would then be a rectangle circumference/100 km pixels wide and circ/200 km pixels tall. Diameter will be the square root of surface area in m. km^2/pi, rounded to the nearest tenth of a kilometer.
479.8 m. km^2, 12358.1 km across, or 6179 km in radius. The world div is 388 pixels across and 194 tall.
39.6 m. km^2 of land with an inital side length of 6292.8 km
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