Joel, Yes, it sure does sound like too much math, but now we're getting somewhere. I'll admit I don't know enough about the problem to realize it's complexity, and it isn't practical then to solve from ephemeris data. I do appreciate you taking the time to enlighten me. I gather a modern planetarium program will do the job, but I see that it would be used in an iterative manner as I suspected the calculations might be. There are probably many, but could you suggest an affordable one that is easy to learn and use? A free one would be nice as a learning tool even if it didn't have a lot of extra features. Thanks for your help and humorous mention of calculation accuracy as interpreted in the courts of the 1600's. Ed ------------------------------ Quoting Joel Stucki <joel.stucki@gmail.com>:
Gee that sounds too much like math.
So the first problem I see is your question is based on alt-azimuth coordinates (10 edge above horizon, how far from east) and ephemeris data is in declination-right ascension coordinates. This makes your problem much harder to solve. Ephemeris data is very hard to calculate which is why historically it was published in tables and in modern times it is done by computer. Computers have a pretty easy time going from ra-dec to alt-az for a given location. Ephemeris.com uses the nasa JPL data as its source. JPL does give instructions on calculating position using Kepler laws of planetary motion. http://ssd.jpl.nasa.gov/?planet_pos This is based on math from the 1600's and not as precise as modern computer models, but apparently was close enough to have Kepler's mother tried for witchcraft.
Using a modern astronomy planetarium program it's pretty simple to pick a date and location, display the alt-az for your object then step through time until it reaches the height your interested in.