Hello everybody! For a long time now I have wanted to make an "all business" scientific horizontal string sundial which shows declination lines with a longitude-corrected analemma offset from the meridian which shows 12:00 noon in Standard Mean Time. The dial plate of a similar dial is shown opposite pg. 200 in Mayall.
Traditionally, a nodus is placed somewhere along the style. Since my style is a cable, a little sphere attached to the cable would make a good nodus. The objective would be to locate the nodus as far north from the dial center as possible so that the distance between the winter and summer solstice declination lines is maximized and the reading of dates and the analemma would be more precise. But this distance can't be so far that the winter solstice line overlaps the hour numerals. Even on my large 40" dial this distance would be pretty small with limited precision. Now, what if instead of placing the nodus on the style, I use a tall vertical pointed rod at the dial center for the nodus? (taller than the one I am currently using) The space between the winter and summer declination lines could then be much greater than the space projected from a nodus on the style, the analemma would be larger and precision and legibility of the declination lines and analemma would be much greater. Have any of you ever seen a dial built like this? I know how to calculate and draw the declination lines for a vertical nodus projecting onto a horizontal face. But nowhere can I find a formula for an analemma projected onto a horizontal plane. In the new BSS June 1999 Bulletin there is a good article by Allan Mills (pg.62) which deals with the projection of analemmas on perpendicular, polar, south vertical, and even curved surfaces. But no mention is given to the projection onto a horizontal plane. Does anybody know what this formula is? Or better yet, would any of you with computer sundial generators be willing to sell me a computer generated drawing of the longitude corrected analemma and declination lines for Tucson AZ (lat.:32 13' 18" N.; long.:110 55' 33" W.) Thanks for any of your thoughts and help with this problem. John Carmichael http://www.azstarnet.com/~pappas
