# Need help understanding celestial coordinate system

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#### laroof

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Hello all,<br /> I just discovered a very deep interest for astronomy and the universe last summer and am only more and more fascinated every day. However, I want to understand how the coordinate system of the skies work. I understand there is more than one and the Alt. Az system seems fairly simple. I can't quite grasp the coordinate system of RA and Dec. Can anyone help me out or point me to a helpful website on this subject?<br />Thanks,<br />Brandon

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#### Saiph

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The actual coordinate system for RA and DEC is easy, it's figuring out how that interacts with the alt-az system that's a pain (i.e. figuring out where a star is in the sky, now, only from RA and DEC).<br /><br />Basics: Declination is the number of degrees away from the celestial equator (which is labled 0 degres), and towards the poles (at + or - 90). So the north star (polaris) is +90.<br /><br />This mimics (on purpose) the latitude coordinate for navigation on a globe. The latitude on earth is the angle made between the north star, the observer, and the horizon (so if the NS is 40 degrees above the northern horizon, you're at 40 degrees latitude)<br /><br />The Celestial equator is <i>not</i> parrallel to the plane of hte solar system (i.e. it's tilted with respect ot the sun and solar system).<br /><br />The Celestial equator is a mirror image of earth's equator. I.e. it is straight overhead when you are at the equator (nowhere else!).<br /><br />Now, for a little math: The altitude of the highest point on the celestial equator is 90-latitude. So if you're at 40 degrees latitude (i.e. north star is 40 degrees above the horizon), the celestial equator is 50 degrees (40+50=90, the angle of seperation between the pole and equator). <br /><br />The reason for this can be seen if youd raw out a diagram of the sky. Draw a circle, for the whole sky, cut it in half with a line (the horizon line). This line seperates the visible part (the top) from the ground blocked part (the bottom). From the center, draw a line out to somewhere on the top half. Label it NS for north star. The acute angle formed by this line, and the "horizon" line, is the altitude of the NS and the latitude of the observer. Now, 90 degrees from the NS line, draw another, starting from the center of the circle agian (so you should have a quarter circle wedge outlined between the NS line and the new one). Th is is the Celestial equator. The accute angle formed between it and the horizon line is the <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>

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