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Help with Right Ascension and Hour Angle

  1. Jun 28, 2011 #1
    I am seeking help calculating the Right Ascension.

    What I would like to know specifically, is if the Right Ascension can be calculated knowing the Zodiacal Longitude or Celestial Longitude of a plant, body or star.

    For example, for a body at 11° Pisces 57', which is 341°57' of Celestial Longitude, how can I determine the Right Ascension?

    Other variables that I know are the Sidereal Time and the Local Sidereal Time. I also know the terrestrial geographic Latitude.

    Variables that I can calculate (which perhaps might be helpful) are the Declination (the Arcsine[Sine(Obliquity) * Sine(Celestial Longitude)]).

    I can also calculate the Ascensional Difference (the Arcsine[Tan(Declination) * Tan(Latitude)]).

    Would it be possible to calculate the Oblique Ascension based on any the variables I know?

    If I could do that, then I could use the Oblique Ascension and Ascensional Difference to find Right Ascension.

    I am familiar with: Hour Angle = Local Sidereal Time - Right Ascension.

    However, I don't really understand what the Hour Angle is, other than I have seen it repeatedly defined as "the amount of time since an object has crossed the meridian." Which meridian? The location of the observer, or is the meridian a fixed point? Anyway, it appears that if the Right Ascension is 0° then the Hour Angle = Local Sidereal Time.

    For shats and gaggles, I set up a spread sheet and using the Local Sidereal Time, just calculated the Hour Angle using the Right Ascension in 15° increments to 360°.

    What I saw was the Hour Angle decreasing as the Right Ascension increases, but I don't understand the relationship between the two. Does the Hour Angle equate to Oblique Ascension or Celestial Longitude?

    Am I not approaching this right? Am I missing something here? Thanks in advance for anyone's help.
  2. jcsd
  3. Jun 29, 2011 #2


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    What's zodiacal longitude? What's celestial longitude? They're not standard astronomy terms, and a Google search doesn't give many clues. I suspect "zodiacal longitude" is the same thing as right ascension, just expressed in a different format.
  4. Jun 29, 2011 #3
    What you would do is to draw a spherical triangle between the north celestial pole, the north ecliptic pole, and the object. Then you'd you the trig identities to convert longitudes.

    None of those matter. The celestial longitude won't change as the earth rotates.

    The meridian is the line between the zenith and the celestial pole. The hour angle is zero when the star is on that line.

    No. The hour angle changes as the earth rotates.
  5. Jul 6, 2011 #4
    Thanks for your replies.

    I was afraid of that. I am 3-Dimensionally challenged and not at all good with trigonometry (I've heard people mention calculus on occasion).

    I am good at "plugging in" the variables and constants, if I know what the equation is.

    Would I be correct in thinking that the Hour Angle is the Oblique Ascension?

    I understand that certain Zodiac Signs cross the Meridian (or maybe it's the Horizon) faster than others.

    I have continued to search and did find a fairly good treatise on the subject called Modern Almagest. (It's in Adobe format here if anyone is interested http://farside.ph.utexas.edu/syntaxis/Almagest.pdf [Broken])

    I have attempted to read through it, and as best as I can tell, I think the Right Ascension is expressed as

    RA = Inverse Tangent (Cosine Obliquity * Tangent Longitude)

    I have no idea what that means, but my Excel 2002 spreadsheet has those functions, so I'll plug in some numbers and see what happens.

    I was kind of searching for something that would explain it in layman's terms. The Zodiacal Longitude is simply 7° Gemini 23' or 26° Sagittarius 56' and that can also (apparently) be expressed as 67°23' or 266°56' respectively and many seem to refer to that as Celestial Longitude or Absolute Longitude.
    Last edited by a moderator: May 5, 2017
  6. Jul 6, 2011 #5
    Right Ascension = arctan[(sin(Celestial or Ecliptical Longitude)*cos(Obliquity)-tan(Ecliptical Latitude)*sin(Obliquity))/sin(Ecliptical Longitude)]

    This formula among others can be found in Wikipedia under "Ecliptical Longitude".
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