Re: Mixed Coordinate conversion - ra from long and dec

Dave Blake wrote:

Can any one help me with a bit of math.
I would like to get right ascension given the celestial longitude and
declination of an object. Yes, that's right I do mean mixed co-ordinate
systems.

I think this is easier to do in Cartesian coordinates.

The solution is the intersection of two circles. One is a great circle
of constant longitude. The latitude changes as you move along this
circle. The other is a (non-great) circle of declination. The right
ascension changes as you move along that one.

Each circle is embedded in a plane. The intersection of the circles is
on the line which is the intersection of the planes. You should be able
to find this as the vector cross-product of the plane normals, which are
also the pole vectors of the two circles.

I don't have time to work out what this would look like, but to give you
a start: you need to define both circles in the same coordinate system.
Choose the Cartesian system with its axes aligned with the plane and
pole of the equatorial (ra and dec) system. In this system, Z points to
the north celestial pole, and X and Y form the plane of the equator.

The center of the longitude circle is just (0, 0, 0). The center of the
declination circle is (0, 0, sin dec). The radius of the dec circle is
cos( dec ), and the radius of the longitude circle is 1.

The normal of the dec circle is (0, 0, 1). The normal of the longitude
circle is a vector pointing toward (lon + 90, 0). Convert that to ra
and dec in the usual way, then convert the result to a vector using

nx = cos( dec ) * sin( ra )
ny = cos( dec ) * cos( ra )
nz = sin( dec )

to get the components of the longitude circle's normal.

Now you have the two planes in point-normal form.

dec: P = (0, 0, sin dec) N = (0, 0, 1)
lon: P = (0, 0, 0) N = (nx, ny, nz)

Googling will reveal how to find the line representing the intersection
of these two planes. Then you just need to find the two points on this
line that intersect the dec circle.

All those 0's and 1's probably mean that this simplifies a great deal
once you work through it. Unfortunately the margins of my newsreader
are too small to contain the proof.

- Ernie http://home.comcast.net/~erniew

.

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