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Right Ascension and Declination

  1. Oct 8, 2005 #1
    Hi guys,

    This might be a stupid question and all, but I was wondering how would you know what Right Ascension and Declination coordinates are visible from a particular location? I mean I was having a look at an old almanac lying around here and couldn't really make sense of how you can see from those coordinate values that the object will be viewable from your location.
  2. jcsd
  3. Oct 8, 2005 #2
    In principle all the lines of RA pass through your location everyday. It's the declination that is limited by your latitude. If you were at the equator, you should be able to see all the declinations (again, that's in principle). Assuming that you're in the Northern Hemisphere, the north celestial pole, i.e. +90° declination, will be A° above your northern horizon, where A denotes your latitude. So the celestial sphere is 'tilted' and limits the southernmost declination visible to -(90-A)°.

    The lines of declination remain fixed to the Earth while the RA appears to go around. The situation changes if you were at any of the two poles. Both the lines of RA and Decliantion will appear to go around, but with the difference that all RA, at least a part of them, will always be above the horizon, meeting at the zenith.
    Last edited: Oct 9, 2005
  4. Oct 8, 2005 #3
    You might want to buy a Whitney star finder. It's an inexpensive and valuable tool that shows you what will be visible on a certain night from a certain loacation. It may seem overpriced, since it 's a bit of printed cardboard, plastic, etc, with a little rivet as a pivot. It's worth every penny and more.
  5. Oct 9, 2005 #4
    ahh k cheers guys.

    So neutrino you're saying that really the right ascension is the coordinate that determines what time you see the object?? And how does that work? I know the right ascension has to do with the distance from the first point of aries. So would you find out what time that is visible from your location or something?

    @ Turbot
    Yeah I think I've seen those before, so I will see about grabbing one. I don't think it was too expensive really. I think it was something like $5.
  6. Oct 9, 2005 #5


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    Let's see if we can visualize this a little more concretely. Take a look at the following picture:

    http://qonos.princeton.edu/nbond/sky_coord.gif" [Broken]

    Imagine that the sphere in the center is earth and the outer sphere is the celestial coordinate system (right ascension and declination). The coordinate system was basically defined so as to project out from the earth's coordinate system (longitude and latitude), so you don't need to worry about the earth's tilt.

    Ok, now find your latitude on the earth in that diagram and pick any point on that line of constant latitude. It'll probably be easier if you pick a point on the edge, but it doesn't really mater. Once you've chosen this, mentally (or graphically) consider a plane that's tangent to it; that is, a plane that's perpendicular to the line connecting the earth's center and your chosen location. It should roughly cut the bigger sphere in half. This plane delimits everything you can see at one moment in time. Since we haven't normalized the coordinate systems (that is, put numbers on the diagram), this could be any time, day or night, but it won't turn out to matter. Also note that, to do this exercise more correctly, the earth should be much, much smaller than the large sphere so that it the plane really does cut the bigger sphere in half.

    Anyway, as I was saying, above that plane is everything in the sky at one moment in time. To determine everything that you could see throughout the year, just rotate that plane 360 degrees around the earth's axis (that is, the line connecting the north and south pole). Every part of the outer sphere that is above that plane at any point in the rotation will be visible to you at some time throughout the year. To ensure that you understand what I mean, try the exercise for an observer at the north pole and at the equator. The person at the north pole will always see the same half of the sky throughout the year, while the person on the equator will see the entire sky as the year progresses.

    Given this conceptual explanation, maybe you can do a little geometry to determine that the range of declinations that you can't ever see is given by:

    [tex]\delta < l - 90^o~if~l>0[/tex]
    [tex]\delta > l + 90^o~if~l<0[/tex]

    where l is your latitude.
    Last edited by a moderator: May 2, 2017
  7. Oct 9, 2005 #6
    thanks again space tiger : )
  8. Oct 13, 2005 #7
    are you sure? this doesn't make sense to me... what makes sense yo me is just l > d for a star at that d to be just circumpolar
  9. Oct 14, 2005 #8


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    Pretty sure. Imagine the simple cases. If you live on the north pole (l = 90 degrees), you always see the same sky -- that is, the northern half of the celestial sphere. Plugging its latitude into my equation, it says that you can't see anything with declination below 0. In other words, it says you can't see the southern half of the sky, which is what we know to be true.

    The other extreme case is the equator (l = 0). My equation technically doesn't include that, but if you imagine being just slightly above or below it, you can plug into the first or second equation and get that your limiting declination is either 90 or -90 degrees; that is, you're not limited at all. This is consistent with the fact that folks at the equator can see the entire sky throughout the course of a year.

    That doesn't prove the equation right, but let's check that you're understanding these cases before I we verify the more general one.
  10. Oct 14, 2005 #9
    but what about a case where say L = 45 degress... then by your formula:

    d < l - 90*
    d < -45*

    so if your latitude it 45* you can always view a star of -45* dec... that doesn't make sense to me at all
  11. Oct 14, 2005 #10
    Think of it this way - as an approximation (given very flat terrain and no trees, buildings) you will be able to see a hemisphere (180 deg) of sky from your location (90 deg to the North and 90 deg to the south). If you are at the north pole, you can only see stars down to about 0 degrees. If you live at 45 deg North, you can see stars to the North that are 45 deg on the opposite side of the celestial pole, and you can see stars to the south that are 45 deg on the south side of the celestial equator (45 deg to the equator and 45 more southerly).
  12. Oct 14, 2005 #11


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    The formula only says which stars you can never see. If a star lies within your "visible range", that doesn't mean that it's always up, just that it's theoretically possible to see it from your location. At the very end of the visible range (like the case you mention above), the star will only just appear on the horizon. If you happen to live in a valley, then you'll never see it -- so the formula should be viewed as rough. In practice, there are a few other things that will limit a star's visibility (like clouds, local terrain, etc.), but they vary from place to place, so they can't be simply described in an equation like that.
  13. Oct 15, 2005 #12


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    From an observational perspective, stuff that never rises more than about 5 degrees above the horizon is hardly even worth the effort. Even 10 degrees above the horizon is pretty is tough to view most of the time.
  14. Oct 15, 2005 #13


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    To really convert the right ascension and declination of a celestial body into the azimuth and altitude of your local sky you have to solve the astronomical triangle. I suppose that is done by software these days.

    The triangle's three vertices are the pole, the body, and your zenith; its three sides are the altitude (arc from your zenith to the body), the body's co-declension, and your co-latitude. Its three angles are hour angle (converted from right ascension and your longitude) at the pole, azimuth, at you, and an unnamed angle at the body.
  15. Oct 15, 2005 #14
    Get a Weems star finder for a few dollars on eBay.
  16. Oct 15, 2005 #15
    RA and Declination

    Better than the Weems star finder, but more expensive, is a celestial globe which can be adjusted to show what stars might be available. I see them on eBay often, by searching under "star globe," "celestial globe," or "Russian globe." Most Russian ones label stars and constellations using the Cyrillic alphabet, but some use the Roman alphabet.

    I bought one for $120 once. They're usually more than that. Postage from Kaliningrad or Odessa, and a fee for transferring money can add to the cost.
  17. Oct 15, 2005 #16


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    Yeah, there are entire computer programs, kits, etc. devoted to helping amateur astronomers with this sort of calculation. If you're just looking for position information, I suggest using one of those. If you're trying to understand the celestial sphere, there are many problems (like the one here) that can be worked out simply without resorting to spherical trig. However, if you're really into astronomy, it wouldn't hurt to learn a little basic spherical trigonometry.
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