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Re: Local / Sidereal Time math problem

Started by tvansickle, 09/05/2004 03:40AM
Posted 09/05/2004 03:40AM | Edited 09/05/2004 03:42AM Opening Post
Steve, There is no simple formula. The sidereal day is about 4.25 minutes shorter than the solar day. That is caused by the fact that the Earth moves around our orbit during the day. If we measure the length of a day as the time required for the Sun to pass from the zenith to the zenith the next day you will find that time is exactly 24 hours. On the other hand, if you measure the time required for a star to pass from the zenith one day to the next, the time is less by about 4.25 minutes. This can all be figured out exactly from the geometry of our orbit.

Anyway, since the sidereal day is somewhat shorter than the solar day, you have to basically count the number of solar days from a known sidereal time and adjust the day count from solar days to sidereal days. These days are measured in fractions of a day which can then be converted to the appropriate time.

Just remember, the Local Sidereal Time--LST--is the Right Assention of the zenith at your current location. You will find some algorithms to calculate the LST in Meeus, Astronomical Algorithms.

It is a fun problem to try to work out on your computer.
Ted
Posted 09/09/2004 05:30PM #1
well, I'm not sure about the "fun" part, especially this is not a very new problem -- others have solved it for you:

http://tycho.usno.navy.mil/sidereal.html