On Wed, Feb 24, 2010 at 03:27:42AM -0800, Chris LeBron wrote: >Problem is, I don't know how to do the math for it. Specifically, I'm >tripping over the inverse of "at point N from a light source you get X >amount of light, at point N2 you get 1/4X, and point N3 you get 1/9X, >etc." I get that one, but whet about when you go closer to a light >source from a known point? I know how powerful it would be from Earth >orbit, and I can calculate for the 'Belt, but what about Venus and >Mercury?
Just continue the progression. Here's an algorithm that'll work anywhere in the solar system. Assuming that the "in space" of the original description is at the distance from the sun of Earth's orbit, just: - work out the distance from the sun; [example: Venus averages 0.72 AU] - divide by the distance from the sun of Earth's orbit in the same units; [example: this is why I picked AU, because Earth's orbit is 1 AU. 0.72.] - square the result; [example: 0.52] - invert the result (i.e. divide 1 by it). [example: 1.9] That's the multiplier for sunlight intensity. Roger _______________________________________________ GurpsNet-L mailing list <[email protected]> http://mail.sjgames.com/mailman/listinfo/gurpsnet-l
