Rod, At the moment I sent my message about Anselmo's question I received your message. At the time I started with gnomonics, some more than 20 years ago, one of the things I did was making an instrument of simple plastic materials that exactly does what you described. That little instrument was a good help to me to understand what happens in the sky. The instrument still exists and of course I used it now for this problem. Indeed it is an easy method to solve the problem and the answer also is 45 degrees. Thanks for your attention to this.
Best wishes, Fer. Fer J. de Vries [EMAIL PROTECTED] http://www.iae.nl/users/ferdv/ Eindhoven, Netherlands lat. 51:30 N long. 5:30 E ----- Original Message ----- From: "Rod Heil" <[EMAIL PROTECTED]> To: <[email protected]> Sent: Sunday, March 03, 2002 8:16 PM Subject: Re: On northern vs. southern dials > 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 > - -
