Converting RA/Dec to earth centered coordinates?

In article ,
"W. eWatson" writes:
My xy plane is in the plane of the equator, and its projection into the
sky represents the celestial equator. Declination is measured +/- from
the celestial equator to each pole along great circle lines that pass
through each pole.

Somewhere on the celestial equator is point from where RA is measured.

To go from RA/Dec to an x,y,z unit vector is just simple trignometry:

x = cos(dec)*cos(RA)
y = cos(dec)*sin(RA)
z = sin (dec)

Make sure to convert RA/dec to degrees or radians (whatever units
your calculator or program uses). It's easy to forget to multiply
hours by 15 to get degrees.

The zero point of RA is the place where the ecliptic and celestial
equator intersect with the equinox heading north. Both this zero
point and the location of the celestial poles changes with respect to
the stars because of precession. If you want a _current_ x,y,z unit
vector, you need to start from current RA/Dec coordinates rather than
coordinates at a standard "ecliptic and equinox" (B1950 or J2000).
The Meeus book will tell you how to do that calculation.

Ah, the obliquity of the ecliptic (e) is what I need.

Don't see why you need that. Did you mean ecliptic coordinates
rather than celestial?

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