Anselmo, I might have an easy way to solve this in two dimensions. Start by imagining a three-dimensional ring-shaped object, which represents all of the points in the celestial sphere where the sun can be. It looks like part of a truncated globe wherein only the part between the tropics remains.
Okay, now look at this ring from the side (line of sight lies in the equatorial plane and tropic lines are horizontal). What you see is a rectangle with the short sides bowed out into arcs. Draw this shape on a piece of paper. Now bisect it horizontally (at the equator), and shade the part of your drawing below that line. What you have is a representation of the sunlight received at the north pole over the course of a year. Now if you draw a vertical bisection, you see how that sunlight is distributed on opposite sides of your hypothetical wall. Both sides get an equal amount of light. Set that drawing aside, and draw the bowed-out rectangle again, this time rotated 90 degrees (with the arcs at the top and bottom). Bisect horizontally. As before, above the line represents the sunlight, and below the line represents darkness. Bisect vertically. To the left represents sunlight falling on the north side of the wall, and to the right sunlight on the south. Again, both sides get equal light. Set that drawing aside, and draw one where the rectangle is tilted only 45 degrees.... You get the idea? >From here, it is a matter of algebra to find where is the highest ratio of north to south light. You'll need formulas for the area of a triangle, and the area of a segment, and you'll need to figure out where the formula is non-linear (i.e. where the corner of our rectangle crosses one of our bisection lines.) The only other thing to figure out is to what degree this 2D representation matches 3D reality? Rod Heil ~41.3 N 105.5 W __________________________________________________ Do You Yahoo!? Yahoo! Sports - sign up for Fantasy Baseball http://sports.yahoo.com -
