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.

Some astronomers in galactic structure use +Y as the coreward axis, while +X points anti-spinward. See details here.

What you have to do is perform a coordinate transform to convert from equatorial to galactic co-ords. This means you'll take the equatorial x, y, and z coordinates and feed them into three equations. Out will pop the galactic x,y,z.

Epoch 1950.0

If your star data is in epoch 1950.0 (like the Gliese data), use the equations below from Bruce Webster's New Perspectives on Nearby Stars, published in July 1985 issue of BYTE Magazine.

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

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

Z_g = -(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.

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.)

Late breaking news, the latest (2019) measured distance to the center of the galaxy is 26,660 light-years. Specifically 8178 parsecs ± 13stat. ± 22sys.

Epoch 2000.0

If your star data is in epoch 2000.0, use the following equations:

X_g = -(0.0550*X) -(0.8732*Y) - (0.4839*Z)

Y_g = (0.4940*X) - (0.4449*Y) + (0.7470*Z)

Z_g = -(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 those 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.)

And if you really want to be anal-retentive, Sol is currently about 50 light years above the plane of the galaxy. So add 50 light years (or 15.3 parsecs) to the galactic Z co-ord. (a 1995 survey by Humphreys, Larsen, and Cohen gave values of 8, 15.5 and 14 parsecs depending on the wavelength used for the survey. Take your pick.)

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.)

X_s = X_g - 8,300

Y_s = Y_g

Z_s = Z_g + 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( X_a² + Y_a²)

X' = (-X_s * (Y_a/R)) + (Y_s * (X_a/R))

Y' = (-X_s * (X_a/R)) + (Y_s * (Y_a/R))

Z' = Z_s

(Thanks to Erich Schneider for the above equations.)

If you want to go into more mathematical detail, be sure to check out James Shuster's incredible SGC page. Very well done, and much more mathematically rigorous (translation: most of it is over my head...) He also has some notes on interstellar navigation.