4 ms·
An image of earth at very roughly 4cmx4cm resolution? (If I've knocked the zero's off correctly)
by flir 11mo ago
An image of earth at very roughly 4cmx4cm resolution? (If I've knocked the zero's off correctly)
- aidenn0 11mo agoEach pixel would represent roughly 16cm^2 using a cylindrical equal-area projection. They would only be square at the equator though (representing less distance E-W and more distance N-S as you move away from the equator). No projection of a sphere on a rectangle can preserve both direction and area.
- flir 11mo agoI admit it, I was applying Cunningham’s Law. Disappointingly(?), you came to the same answer.
- aidenn0 10mo agoI admit I trusted your math; you seem to be off by a factor of 4: You have: 510.1e6km^2/1073741824/1073741824 You want: cm^2 * 4.4244122 / 0.22601872 Strangely enough, units lacks area_earth, so I used the number from https://iere.org/what-is-the-area-of-the-earth/ https://iere.org/what-is-the-area-of-the-earth/
- flir 10mo ago:D I was starting with the length of the equator and assuming a spherical cow^Hplanet.
- aidenn0 10mo agoDid you perhaps use the diameter of the earth rather than the radius? #PiIsWrong https://www.tauday.com/ https://www.tauday.com/ You have: (40075km/tau)^2*4*pi/1073741824/1073741824 You want: cm^2 * 4.434018 / 0.22552908