Converting star coordinates to x,y,z

I'm working on a project of modeling the stars of the constellations in
3-D; not for their positions relative to Earth, but relative to any
given person being at 0,0,0.

Is there a simple formula for this? ...or better yet, do you know
anyone who has them on the net or in a book?

Thanks!

Rick.

Clarification:
I see that I too hastily posted the question. 0,0,0 would be Earth.
The
figure for 'x' would need to be the distance from Earth, with y and z
being
taken from RA and DEC...is that right?

If we use the stars of Orions belt as an example, can you help me put
them
to x,y,z?

Alnitak: 400ly, RA 05 40 45.5, DEC -01 56 34
Alnilam: 1340ly, RA 05 36 12.7, DEC -01 12 07
Mintaka: 600ly, RA 05 32 00.3, DEC -00 17 57

On or about 2004-12-09,
illuminated us with:
Clarification:
I see that I too hastily posted the question. 0,0,0 would be Earth.
The
figure for 'x' would need to be the distance from Earth, with y and z
being
taken from RA and DEC...is that right?

If we use the stars of Orions belt as an example, can you help me put
them
to x,y,z?

Alnitak: 400ly, RA 05 40 45.5, DEC -01 56 34
Alnilam: 1340ly, RA 05 36 12.7, DEC -01 12 07
Mintaka: 600ly, RA 05 32 00.3, DEC -00 17 57

Speaking as an astronomical newbie and a mathematical (very) oldie (i.e.
rusty):

RA represents "astronomical longitude" and is measured as time, i.e.
the numbers are hours, minutes and seconds on a 24 hour clock, 1 hour
therefore represents 15 degrees. DEC is a simple angle in degrees, minutes &
seconds, between -90 deg (astronomical south pole) and +90 deg which is
approximately where polaris is.
http://www.google.co.uk/search?as_q=...n%22& num=100

So, if we want x to be the distance along the RA origin axis, y to be the
"elevation" above the equatorial plane and z to be the distance from the RA
0 0 0 "plane":

You'll need to convert the angles into whatever form your functions need
(e.g. radians for most spreadsheets[1]), then:

x = DIST * COS(RA) * COS(DEC)
y = DIST * SIN(DEC)
z = DIST * SIN(RA)

I think. Or my O-Level maths teach will be along shortly to berate me ;-)

If you don't get there with that lot, I'll have anotehr go when I've got
more time.

[1] If you use Excel, then entering RA as hh:mm:ss.cc actually gives you a
number where 1.0 = 24 hours, so converting to radians is trivial by multiplying
by 2*PI(). DEC is a touch harder :-)
--
Mark
Real email address | One good thing about egotists is that they don't
is mark at | talk about other people.
ayliffe dot org |

On or about 2004-12-09,
Mark Ayliffe illuminated us with:
On or about 2004-12-09,
illuminated us with:
Clarification:
...
If we use the stars of Orions belt as an example, can you help me put
them
to x,y,z?

Alnitak: 400ly, RA 05 40 45.5, DEC -01 56 34
Alnilam: 1340ly, RA 05 36 12.7, DEC -01 12 07
Mintaka: 600ly, RA 05 32 00.3, DEC -00 17 57

...

So, if we want x to be the distance along the RA origin axis, y to be the
"elevation" above the equatorial plane and z to be the distance from the RA
0 0 0 "plane":

x = DIST * COS(RA) * COS(DEC)
y = DIST * SIN(DEC)
z = DIST * SIN(RA)

Arrgh! Sorry. z = DIST * COS(DEC) * SIN(RA)

I think. Or my O-Level maths teacher will be along shortly to berate me ;-)

(Sorry Mr. Shingles!)

I've checked this out with OpenOffice now (Excel seems not to like negative
time values which complicates DEC a bit) and get (you'll need fixed width
font to see this properly):

Item Dist RA Dec RA(RAD) DEC(RAD) X Y Z Check
Alnitak 400 05:40:45.50 -01:56:34 1.486839 -0.001413 33.54 -0.57 398.59 400
Alnilam 1340 05:36:12.70 -01:12:07 1.467000 -0.000874 138.84 -1.17 1332.79 1340
Mintaka 600 05:32:00.30 -00:17:57 1.448645 -0.000218 73.11 -0.13 595.53 600

The check is that SQRT(X^2+Y^2+Z^2) is the same as the original distance.
Sanity check that the numbers are about right:

At just under 6H of RA, these stars are about one quarter of the way around
the astronomical sphere, so values of X will be relatively small. At less
than 2 deg of DEC, they are very close to the equatorial plane, so values of
Y will be tiny. And therefore most of the distance will be in the z
direction. Seems to fit with my answers.

By the way, there is a freeware package for Windows called Celestia which
allows you to "fly" around the stars. Seems a bit of a gimmick for normal
astronomy, if fun, but it might be useful to you for this.

HTH

--
Mark
Real email address |
is mark at | Is "tired old cliche" one?
ayliffe dot org |

Rick,

Try getting your hands on "Practical Astronomy with your Calculator" by
Peter Duffett-Smith (Amazon UK Link http://tinyurl.com/4xym3). My
rule is, if the equation you need isn't in this book then it isn't
worth knowing ;-)

I'd be interested to know what package/language you are using to do
this simulation.
All the best,
John D. Tanner
http://physics.open.ac.uk/~jdtanner