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.

wrote:
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?

It's going to be very messy. Star positions practically BEG for
spherical coordinates, (which is what they are really), so converting to
Cartesian you are looking for trouble.

I don't know if anyone has been masochistic enough to try it, but you
could conceivably do it. Open any Calculus book and apply the conversion
formulas found there, although I suspect that you'd be better off if you
situated Earth's center at (0,0,0). That way, you avoid the problem of
those coordinates changing (even more slightly) everytime the season
changes and the observer's position changes on the surface of the Earth.

Since the celestial dome appears "spherical", be prepared to be
bombarded with thousands of sin's and cos's.

Good luck!

Thanks!

Rick.
--
I. N. G. --- http://users.forthnet.gr/ath/jgal/

wrote:
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?

Ioannis responded:

It's going to be very messy. ...
Since the celestial dome appears "spherical", be prepared to be
bombarded with thousands of sin's and cos's.

Yikes! You make this sound like a major undertaking; it's actually
a problem in elementary trigonometry that any kid in a high-school
pre-calculus class should be able to solve in 5 minutes.

Conversion from RA/Dec to rectangular coordinates requires two sines
and two cosines -- or two and three, depending how you count.

My only question is about the meaning of the phrase "relative not to
Earth but to a person at 0,0,0". I've been searching for 0,0,0 all
my life, and I still haven't found it!

- Tony Flanders

wrote:
[snip]
Ioannis responded:


It's going to be very messy. ...
Since the celestial dome appears "spherical", be prepared to be
bombarded with thousands of sin's and cos's.


Yikes! You make this sound like a major undertaking; it's actually
a problem in elementary trigonometry that any kid in a high-school
pre-calculus class should be able to solve in 5 minutes.

Conversion from RA/Dec to rectangular coordinates requires two sines
and two cosines -- or two and three, depending how you count.

My only question is about the meaning of the phrase "relative not to
Earth but to a person at 0,0,0". I've been searching for 0,0,0 all
my life, and I still haven't found it!

My impression was that he wanted a conversion for a person situated on
the surface of the Earth for EACH such position.

Since there are going to be so many such positions (taken as 0,0,0), I
thought he wanted a model that would simultaneously provide for every
position, as the Earth rotates.

For example, what are the rectangular coordinates of Alnitak with 0,0,0
taken at my position in Spring in Athens and at your position in Winter?

If this wasn't the case, I seem to have misunderstood him.

- Tony Flanders
--
I. N. G. --- http://users.forthnet.gr/ath/jgal/

Ioannis wrote:
For example, what are the rectangular coordinates of Alnitak with 0,0,0
taken at my position in Spring in Athens and at your position in Winter?

Position on the Earth is entirely negligible. The maximum difference
is, of course, twice the parallax angle, and that is at most about 1.5
arcseconds. The parallax induced by the Earth's rotation is smaller
than that by about a factor of 25,000--about 60 microarcseconds. We
don't have a snowball's chance in hell of detecting that.

I'm pretty sure the OP didn't care about that. I think he just wants
to be able to set (0, 0, 0) to whatever location in space he would like.

Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt

wrote in message
ups.com...
My only question is about the meaning of the phrase "relative not to
Earth but to a person at 0,0,0". I've been searching for 0,0,0 all
my life, and I still haven't found it!

In college I had a brick on my desk that I defined as 0,0,0. At times it was
quite therapeutic to just pick it up and shake it.

--
jeff

wrote in message
ups.com...
I've been searching for 0,0,0 all
my life, and I still haven't found it!

Didn't it get redefined as 127.0.0.1?

:-)

Craig in Tampa

wrote in message
ups.com...
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.

Search for the program "Celestia" on Google and download it. It's free. It
knows both the direction (RA and Dec) *and* distance (LY) of many stars and
allows you to position yourself anywhere in 3D space and see how things look
from there. If you go to the Navigation / StarBrowser menu item it brings up
a list of up to 500 stars giving, among other things, their distance in LY
from earth. Using these 3 polar coordinates (RA, Dec, LY) for any star you
can transform these polar coordinates into Cartesian coordinates (probably
using Z-axis is declination 90 degrees, X-axis is RA of zero hours, Y-axis
is RA of 6 hours). With these 3 computed X, Y, Z coordinates you would then
subtract a translation vector (delta-X, delta-Y, delta-Z) of a viewpoint
somewhere other than earth and you now have the X, Y, Z coordinates of your
chosen star *relative* to the new viewpoint.

Unfortunately, Celestia does not show you any of its own mathematics, but
this is essentially what it's doing.

Timothy R Oltrogge wrote:

Unfortunately, Celestia does not show you any of its own
mathematics, but this is essentially what it's doing.


Celestia is open source, so technically the mathematics are there
for the taking... if you can read C/C++, that is.

--Steve

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.

And WHERE is the location of 0,0,0 exactly???
Clear, Dark, Steady Skies!
(And considerate neighbors!!!)