Stories of the Stars: Alphecca. Artwork by Frank Paul

If you really want to do the job right, you should convert the co-ordinates to galactic co-ords. This is where the X-Y plane is the plane of the Galaxy, which you might agree is a tad less parochial than the earth centered ones. The positive X-axis points at the galactic center. I will note in passing that I have yet to see a published 3-D starmap that used galactic co-ords. Lazy of them....  Late breaking news, the guidebook Proximity Zero contains galactic co-ordinates! Check it out! Some astronomers in galactic structure use +Y as the coreward axis, while +X points anti-spinward.

 Xg = -(0.0672 * X) - (0.8727 * Y) - (0.4835 * Z)

 Yg = (0.4927 * X) - (0.4504 * Y) + (0.7445 * Z)

 Zg = -(0.8676 * X) - (0.1884 * Y) + (0.4602 * Z)

 Example: If Tau Ceti is at 3.13, 1.49, -1.01, it's galactic co-ords will be -1.02, 0.12, -3.46

 These equations are from New Perspectives on Nearby Stars by Bruce Webster, in BYTE Magazine issue July 1985. They seem to work, I fed in the co-ordinates of the galactic center and got a point about 10 kiloparsecs on the positive X axis. (RA 17^h42^m4^s, DEC -28°55', DIST 27,000 light years, in case you were wondering.). Click on "July", look at READ.ME, and you'll find a list of the Macintosh HQX files to download containing the program.
 

Late breaking news!  The above equations are based on epoch 1950.0 star data. If your star data is in epoch 1950.0, use the above equations. (The Gliese data is epoch 1950.0, as far as I can tell.) However, new star data is being published as epoch 2000.0. If you are using this data, use the following equations:  Xg = -(0.0550*X) -(0.8732*Y) - (0.4839*Z)  Yg = (0.4940*X) - (0.4449*Y) + (0.7470*Z)  Zg = -(0.8677*X) - (0.1979*Y) + (0.4560*Z)

 These equations are based on the following co-ordinates:

 Galactic center: RA = 17^h45.6^m; Dec = -28°56.3'

 Galactic north pole: RA = 12^h51.4^m; Dec = +27°07.7'

 These equations are curtesy of William Sandberg. For those who are interested in the mathematical derivation of the equations, go here. The derivation was also independently worked out by Andrew Goddard, who has also written a nice set of astronomical online calcuators. (I would have posted Andrew's derivation, except I lost both it and his email address the last time my hard drive crashed.)

Not so fast, says Chris Lawson:

AN END TO HELIOCENTRIC CO-ORDINATES?

Our Sun lies roughly 50 light-years North of the galactic plane, and therefore a less heliocentric co-ordinate system will put Sol at (0, 0, 50) rather than (0, 0, 0). However, as non-heliocentric systems go, this is hardly better.

Three things define our current "galactic" co-ordinate system: our Sun, the galactic center, and the plane of the galaxy. The position of our Sun is crucial to this co-ordinate system. A truly galactic co-ordinate system would have the galactic center at (0, 0, 0) and our Sun would be at (x, y, 50) light-years. The precise value of x and y would depend on how the x and y axes are defined, but since the galactic center is 27,000 light-years away, then x^2 + y^2 = 27,000^2.

This still leaves us with the task of defining x and y axes. We could use a line from galactic center to Andromeda center, but this has the disadvantage of tilting the co-ordinate system away from the natural planes of the galaxy. We could use a line from galactic center to the Sun, an inversion of the current system, but this would still be heliocentric. My own suggestion is that we use the major and minor axes of the galaxy's elliptical shape. Winchell has pointed out that this is a non-trivial task given that the Milky Way is slightly irregular, has several arms, and is currently colliding with a dwarf galaxy and interacting with the two Magellanic Clouds. Still, I believe that this is the least heliocentric co-ordinate system that is likely to be used by a hypothetical galactic empire, even if it requires a few arbitrary decisions, such as exactly where the elliptical axes run and which way is positive on each axis.

Why bring up hypothetical galactic empires? Unless you happen to live in a galactic empire, then you have no need for "true" galactic co-ordinates. When you're stuck in a single system, as we are, or in a handful of local stars, then it makes perfect sense to define yourself as (0, 0, 0) so that you can put Barnard's Star at (4.96, 2.96, 1.44) instead of (15234.72, 22292.84, 51.44). The maths is a lot easier when you live at (0, 0, 0).

Anyway, for true non-helocentric co-ords, first you have to choose what defines the X-axis. For purposes of illustration, let's say that the negative X-axis points at the projection of Andromeda Galaxy onto the galactic plane. If you start with equitorial co-ords, and use the equations at the top of this page, you will have heliocentric galactic co-ords. (Sun at origin, x-y plane is galactic plane, positive X-axis points at galactic center)

Subtract 8,300 parsecs (27,000 light years) from each X co-ord and add 15 parsecs (50 light years) to each Z co-ord to get Sun indexed galactic co-ords. (Galactic center at origin, x-y plane is galactic plane, negative X-axis points at the Sun)

Xs = Xg - 8,300

Ys = Yg

Zs = Zg + 15

Say the Andromeda Galaxy is at Xa, Ya, Za. The projection of the Andromeda Galaxy on the x-y plane is at Xa, Ya, 0.

To transform all the co-ordinates so that the negative X-axis points at the Andromeda Galaxy's projection, do the following:

R = sqrt( Xa^2 + Ya^2)

X' = (-Xs * (Ya/R)) + (Ys * (Xa/R))

Y' = (-Xs * (Xa/R)) + (Ys * (Ya/R))

Z' = Zs

(thanks to Erich Schneider for the above equations)

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