What is the conversion formula that converts values FROM equatorial
celestial coordinates (RA/dec) to GALACTOCENTRIC galactic
coordinates (alt/az)?

Please note that I am not asking for a conversion to heliocentric
galactic coordinates, which is what is commonly implied when
articles talk about galactic coordinates.  GALACTOCENTRIC
galactic coordinates have an origin at the center of the
Milky Way Galaxy.

This does require knowledge of the Sun's galactocentric
coordinates, however the best I've been able to find is
a distance of r=8.5 kpc but no azimuth/altitude (for instance
the Sun may very well lie below or above the plane of the galaxy).

Thank you very much!

---

Clarification of Question by mjsmigel-ga on 15 Mar 2003 12:30 PST

VERY IMPORTANT ...... the galactocentric galactic coordinates
the conversion cranks out needs to be in the form ALT/AZ/DISTANCE,
not alt/az as I posted since alt/az doesn't define a point in 3-D
space.  I will also be willing to accept a X/Y/Z cartesian
coordinate system rather than alt/az/distance.  Thanks.

---

Clarification of Question by mjsmigel-ga on 15 Mar 2003 16:23 PST

Sorry, I am bungling this up -- the INPUT would be RA/dec/distance
and the OUTPUT would be alt/az/distance.  This will yield a 3D
answer from a 3D input.  Thanks to xarqi-ga for catching my error.

What you ask is impossible.  No conversion can create a 3D answer from
2D data.  Sorry.

See my clarification above.

I think the only other thing we need is the vector (RA/dec, or
whatever) from Earth to the galactic centre - somewhere in Saggitarius
isn't it?  After that, I think some fancy vector arithmetic will give
what you want.

Do I understand you want to define ie the location of one of the moons
of Mars with respect to the Galactic center? The moon to the galactic
center distance is about one trillion times greater than the moon to
Mars distance, so there appears to be little utility in such an
approach. Because of the motion of the moon, the average distance and
average angle is the same for the moon as for Mars plus or minus
perhaps one part per quadrillion = one million times a billion. Unless
I am missunderstanding, so many significant digits are required, as to
make computations impossible with most calculators. The motions are
also quite complex, so writing equations, that apply at any time in
the past or future would be like defining the movement of the
escapemement in a watch with respect to the hour hand in an old type
watch.  Neil

Neil:
My take on this Q is that a more likely scenario than one involving a
moon would be something like:
Given the RA/decl/distance of (say) Achenar relative to the Earth,
what is its Alt/Az/r from the centre of the galaxy?
More generally, how can common or garden "Earthling" co-ordinates for
stars be converted to universal (well - galactic) co-ordinates.

Is that right mjsmigel?

Xarqi - you've precisely hit the nail on the head.  Neilzero - I didn't say
anything about using trivial distances such as lunar or planetary ones.

Thanks for the clarification.  I know *how* to work out the method, I
just need to get a couple of bits of info, think a bit more, and push
some numbers around.  I'll comment again when I have more - meanwhile,
a ga-researcher (as opposed to myself, a commentator) may pick up this
interesting question and provide you with an answer sooner.

Strangely, this question doesn't seem to have drawn much attention
from the real experts here, so you may have to put up with me!
Here's where I'm at.
Given the vector to the target (that is, its direction specified in
RA/decl - not alt/az as this will make it earth time dependent - yuk,
and its distance), and the vector from Earth to the galactic centre,
it is a matter of fairly simple trigonometry (pythagoras actually) to
work out the vector from the galactic centre(GC) to the target, thus
effecting the shift in reference frame.  So that's the how.
The vector to the GC I'm sure can be found easily enough.
But - at some point there has to be a conversion to the galatic frame
coordinate system, that is, some assumptions need to be made - like -
Galactic "North" is the direction along the axis of rotation from
which the galaxy would appear to be rotating counter-clockwise.  This
gives us a galactic "z".  We also need "x" and y".  X could be defined
as the line between GC and the Earth, or perhaps more appropriately,
the direction of travel of the galaxy toward Virgo (or maybe toward
the great attractor).  Y could then be defined as being mutually
perpendicular to the other two axes, and signs ascribed arbitrarily.

I'm pretty sure all this has been done, and a galactic co-ordinate
system is defined somewhere.  My next step would be to track it down.

Then we just have to figure out how to handle vectors is degrees and
parsecs, and we're just about there.

I'll keep plugging away at it, but it will be a bit slow - sorry.  So
this is just an update and maybe a spur to some other researchers.

Galactic coordinate system:
http://cse.ssl.berkeley.edu/chips_epo/coordinates2.html

Thanks xarqi... appreciate the comments.  I think if this question
stagnates another week I will cancel it and move it over to the
mathematics section.  The only reason I posted it in astronomy
was in hopes that someone here was fluent in trigonometry AND
had a working idea of the galactocentric lat/long of the Sun
(all I know is the distance).

Here ya go!
http://vp.ispcal.com/sgc/sgc2.htm

And this:
http://www.astro.utu.fi/~cflynn/galdyn/lecture7.html

Includes gems like:
"Disk Star Orbits

Let us now examine some orbits for this range of kinematic properties.

In the last two lectures we developed a description of the Galactic
potential, and a numerical method for calculating the orbits of stars
given that we know their initial velocities and positions.

The Solar Orbit 

We can start with the Sun. The solar motion is (10, 5, 7) km/s
relative to the LSR. In lecture 4 we developed a model Galaxy fo which
we can compute the local LSR exactly. This turns out to be 228.5 km/s.

We therefore set the Sun's velocity components to V = (-10, 234, 7)
and place it at the spatial position (8000,0,5) pc, i.e. we have put
the Sun at

X0 = 8000 pc 
Y0 = 0 pc 
Z0 = 5 pc 

and 

U0 = -10 km/s  (i.e. moving inward) 
V0 =  234 km/s (i.e. faster than the LSR) 
W0 = 7 km/s (i.e moving upward) 

Note the Sun is slightly above the midplane at Z = 0. This is what we
will find later on when we look at the distribution of nearby stars in
Z.  Note also that Y0 = 0 by definition"

AND

"Galactic system of coordinates 
       An astronomical coordinate system using latitude measured north
and south from the galactic equator and longitude measured in the
sense of increasing right ascension from 0 to 360 degrees. Galactic
latitude is designated b, galactic longitude l. The reference points
for galactic coordinates were changed by action of the International
Astronomical Union in 1958. The new values are: the north galactic
pole lies in the direction right ascension = 12 hours 49 minutes,
declination = 27.4 degrees N (equinox 1950); the new zero of longitude
is the great semicircle originating at the new north galactic pole at
the position angle 0 = 123 degrees with respect to the equatorial pole
for 1950.

Source http://sulu.lerc.nasa.gov/dictionary/ 




The distance and position on the sky can be converted to the cartesian
system of Galactic coordinates (X,Y,Z) and, if the radial velocity is
known to the Galactic velocities (U,V,W). Note that in order to
recover the 6-D position vector and velocity vector of a star in space
one requires 6 measurements, distance, two coordinates on the sky, two
apparent motions on the sky, and the radial velocity.

X = Rsun - D cos(b)cos(l) 
Y = D cos(b)sin(l) 
Z = D sin(b) 

U = VR cos(l)cos(b)-Vlsin(l)-Vbcos(l)sin(b) 
V = VR sin(l)cos(b)+Vlcos(l)-Vbsin(l)sin(b) 
W = VR sin(b)+Vbcos(b) 

where Vl is the velocity of the star along the direction of increasing
Galactic latitude and  Vb  is the velocity along the direction of
increasing Galactic latitude.

The velocity of a star (in km/s) is related to the proper motion m (in
arcsec / year) via

V = 4.77 m / D. 
  

Hipparcos found about 8000 stars within 80 pc. We shall now examine
this sample in detail. "

I guess I should have looked first rather than try to work it out from
scratch, huh.

Thanks again!  I think this is enough for me to go on to close the
question.  Too bad no one could get paid for it.