Re: Geostationary apparent declination

From: b_gimle@algonet.se
Date: Mon Oct 17 2005 - 04:45:47 EDT

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Jeff wrote in www.satobs.org/seesat/Feb-2005/0231.html:

> ... The math gets way to complicated for proper
> calculation of the declination off the local
> meridian. But it will be *smaller* ...

Not much - I find thinking in Cartesian coordinates easier.
Replace computation of Rs and ThetaD by:

TheD = ARCTAN(h/(SQRT(Xsat^2+Ysat^2)))
where:
Xsat = Rg*COS(Longi)-Re*COS(ThetaL)
Ysat = Rg*SIN(Longi)
Longi = Satellite longitude relative local meridian.

In Excel, with H2 (, H3, H4...) = Longi
(and Re,Rg,ThetaL,h in A2 to D2) this becomes:
=DEGREES(ATAN(\$D\$2/(SQRT(I2^2+J2^2))))
where X = I2 (, I3, I4...) :
and Y = J2 (, J3, J4 ...) :

This spreadsheet (I hope I did it right) is at
www.algonet.se/~b_gimle/programs/ibmpc/Clarke_Belt.xls
(in Excel 5.0 format to be compatible with more users).

and ThetaL your geocentric latitude, both of which are smaller.
At 60 degrees, the diffs are about -13 km, -0.2 degrees;
less near the equator.

see http://satobs.org/seesat/Mar-2005/0235.html

/Björn

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