**Previous message:**Tony Beresford: "Feb 22 observations correction"**In reply to:**Ed Cannon: "Calculating declination of geosat belt?"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]

Hey Ed, Here is what I came up with, "back of the envelope", with almost no error checking:) There are 2 constants and 1 input to the following equation to calculate the *declination* of the Clarke orbit *at the observers local meridian*. The two constants are: Re = The Earth's Equatorial radius = 6378 km Rg = The Geosats distance from Earth center = 42,241 km (Note: this is *not* the height of the geosat above the equator) The one input is: ThetaL = Observer's Latitude (Note: for Northern latitudes this is entered as a positive number and for Southern latitudes this is entered as a negative number. - Sorry Greg and Robert!) The one output is: ThetaD = Geosat's declination at the observer's local meridian The formula: ThetaD = arcsin (h/Rs) where: h = Re*sin(-1.0*ThetaL) Rs = sqrt(Re**2 + Rg**2 - 2*Re*Rg*cos(ThetaL)) Note that "h" is the observer's distance from the equatorial plane and that "Rs" is the distance from the observer to the Geosat. The math gets way to complicate for proper calculation of the declination off the local meridian. But it will be *smaller* (that is, closer to the celestial equator) the farther from the local meridian you get. Hope this is right! Regards, Jeff Umbarger Plano TX ------------------------------------------------------------------------- Subscribe/Unsubscribe info, Frequently Asked Questions, SeeSat-L archive: http://www.satobs.org/seesat/seesatindex.html

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