Hello, Someone from Erie, PA recently asked how far above the ascending space shuttle would pass above their horizon. I tried using a formula that I had stored on my calculator, but it gave very ridiculous results. I'm probably missing a couple of parentheses or something like that. I have the formula as I have it below; perhaps someone can tell me the correct formula? DIST = R*(cos^-1 (cos (a)*R/(R+Height))/57.29578)-a*R/57.29578 where R is the radius of the earth (~ 6371 km), Height is the altitude of the satelllite in km, a is the height of the observer's artificial horizon in degrees, and DIST is the farthest distance and observer can be from the satellite's ground track and still have the satellite be at least a degrees in altitude. ------------------------------ Jonathan T. Wojack tlj18@juno.com 39.706d N 75.683d W http://www.angelfire.com/stars2/projectorion 5 hours behind UT (-5) ________________________________________________________________ GET INTERNET ACCESS FROM JUNO! Juno offers FREE or PREMIUM Internet access for less! Join Juno today! For your FREE software, visit: http://dl.www.juno.com/get/web/. ----------------------------------------------------------------- Unsubscribe from SeeSat-L by sending a message with 'unsubscribe' in the SUBJECT to SeeSat-L-request@lists.satellite.eu.org http://www.satellite.eu.org/seesat/seesatindex.html
This archive was generated by hypermail 2b29 : Wed Nov 28 2001 - 16:26:14 EST