Re: Re : Lunar Solar Power Stations vs. The Glaser Proposal

Jake McGuire wrote:

Integral of max(cos(theta), 0) from 0 to 2 pi, I think.

Exactly. The flat solar panel gives output proportional to sin
[elevation] for elevation angles from 0 to pi, and zero for elevation
from pi to 2pi. Integrate it out, and if the solar array is on the
equator, the average power is 2/2pi.

Less for solar arrays off the equator. (The moon has negligible axial
tilt, so "on the equator" and "on the subsolar track" are essentially
identical. Doesn't work quite so well on Earth.)

--
Geoffrey A. Landis
http://www.sff.net/people/geoffrey.landis

.

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