Question

If the Earth’s atmosphere is considered to extend from the surface to 20 km, the earth’s radius is 6,200 km, and the atmosphere is partitioned into cells spaced 0.5 km apart, how many floating‐point calculations will be required to run one iteration of a global weather simulation, assuming that each cell requires 200 floating‐point operations? If it requires 20 cycles (iterations) of simulation to calculate the weather over the next 24 hours, and a forecast is needed in 30 minutes, what must the FLOPS rating of the computing system be?

I know the answer is:

The volume of a sphere is 4/3rt3. In this case we have two spheres, the Earth and the Earth + 20 km (i.e., including the atmo

Can someone explain the math in the last section of that answer? Like " why If the sampling points are ½ km apart, there are 8 per km^3 for the entire atmosphere" and so on.

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Answer #1

0.5 Voals los so Now. See that in 1 km of length there are two cells. It will be best to take valye at mid points 0.5 yee you

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