How to find the position of a star in the sky

1. Jul 27, 2009

Imaginedvl

Hello,

Sorry if this is not the right place but I'm trying to find this for days now...

I'm trying to find a working formula to determine the position (r.asc/decl) of a star at any given time and for a specific location on earth.

So the parameter would be:

R. Asc. of the star
Decl. of the star
Long. on earth
Lat. on earth
Date
Hour

And the result would be

R. Asc. of the star
Decl. of the star

(EDIT: I forgot the Long/Lat parameters ;-))
Thanks!

Last edited: Jul 27, 2009
2. Jul 27, 2009

mgb_phys

The RA and DEC of a star are fixed (except for small real motions) and don't depend on your position on earth or the date/time.
The set of stars visible overhead do depend on the date/time and position.

A good place to start is probably "Practical astronomy with your calculator"

3. Jul 27, 2009

Imaginedvl

Hi mgb,

Yes I used the wrong term for the result. It is actually the position of the star overhead for a specific location/date time that I want to know. Not sure in what this should be expressed.

"Practical astronomy with your calculator", thanks for the reference. Actually I had a book like this and lost it few years ago. I guess it would be a good idea to buy a new one and this one sounds to be a very good start.

4. Jul 27, 2009

mgb_phys

Last edited by a moderator: May 4, 2017
5. Jul 27, 2009

ideasrule

Imagine a line starting from the center of the Earth and passing out through your head. When you look straight up, you're looking along that line. Due to the definition of declination, the declination of the star there is simply your latitude. The star's right ascension depends on where the vernal equinox is--and sidereal time tells you that, with 0:00 corresponding to it being at the prime meridian. Right ascension is the difference between your longitude and the sidereal time, converted to degrees. If your longitude is 70 degrees West, for example, and the sidereal time is 5:00 (75 degrees), that would mean the vernal equinox is 5 degrees to the west of the star. Right ascension would be 5 degrees.

Depending on the level of accuracy you want, you may have to account for precession. Since the Earth's axis is wobbling due to the torque of the Sun and Moon, the right ascension and declination of any star changes with time. Most star catalogues give coordinates valid for J2000.0: noon on January 1, 2000, so to get extremely accurate coordinates, you'd have to precess your calculated RA and Dec. to that epoch.

6. Jul 27, 2009

flatmaster

Are you interested in finding the formula, or deriving it? It would be a good exercise.

7. Jul 28, 2009

Imaginedvl

I never realized it even it is so obvious now that I'm reading it...

I'm trying to write a small program to draw the sky for specific location/position, so I'm actually I'm trying to find the formula :) And yes it is a good exercise.
I think I got enough information so far in the thread to be able to do it.

8. Jul 28, 2009

ideasrule

I've actually written such a program before, so if you have any questions, feel free to ask!

9. Jul 28, 2009

flatmaster

Well, if you want to derive it, how far do you want to take it? precession of earth? eliptical earth orbit?