At 18:15 24/03/01 , Tony Beresford wrote: > The answer is about 2 minutes >from 20 degrees elevation to 20 degrees elevation thru the zenith. A minor glitch its actually about 1 minute. My explanation of the calculation should perhaps be rewritten as >One just calculates the great circle distance to an object at an elevation >of 20 degrees and a height of 90 Km. double this answer, and calculate the fraction of the orbit that this represents, multiply by the orbital period ( 87 minutes say) to get the answer. The approximations involved can only possibly affect the answer by 1 in the 2nd significant figure. Tony Beresford ----------------------------------------------------------------- Unsubscribe from SeeSat-L by sending a message with 'unsubscribe' in the SUBJECT to SeeSat-L-request@lists.satellite.eu.org http://www2.satellite.eu.org/seesat/seesatindex.html
This archive was generated by hypermail 2b29 : Sat Mar 24 2001 - 01:19:38 PST