Help with Right Ascension and Hour Angle

In summary, to calculate the Right Ascension of a celestial body, you can use its Celestial or Ecliptical Longitude and the Earth's Obliquity. This can be done using the arctan function in trigonometry. The Hour Angle is the amount of time since the object has crossed the meridian, which is an imaginary line between the observer's zenith and the celestial pole. This angle changes as the Earth rotates. The Zodiacal Longitude and Celestial Longitude are different ways of expressing the same thing, and can also be used in the calculation of Right Ascension. It is helpful to refer to resources such as the Modern Almagest for a better understanding of these concepts.
  • #1
GLee
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.
 
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  • #2
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.
 
  • #3
GLee said:
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.

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.

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

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

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?

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

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?

No. The hour angle changes as the Earth rotates.
 
  • #4
Thanks for your replies.

twofish-quant said:
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.

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.

twofish-quant said:
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.

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 )

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.

ideasrule said:
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.

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.
 
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  • #5
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

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".
 

What is Right Ascension?

Right Ascension is a celestial coordinate that measures the angular distance of a celestial object eastward along the celestial equator from the vernal equinox. It is often denoted by the symbol α (alpha) and is measured in hours, minutes, and seconds.

What is Hour Angle?

Hour Angle is a celestial coordinate that measures the angular distance of a celestial object westward from the observer's meridian to the hour circle passing through the object. It is often denoted by the symbol H and is measured in hours or degrees.

How do I calculate Right Ascension and Hour Angle?

To calculate Right Ascension, you will need the declination of the object and the local sidereal time. Use the formula α = arctan(sin(α) / (cos(δ) * tan(θ))) where δ is the declination and θ is the local sidereal time. To calculate Hour Angle, use the formula H = LST - α where LST is the local sidereal time and α is the Right Ascension of the object.

Why are Right Ascension and Hour Angle important in astronomy?

Right Ascension and Hour Angle are important in astronomy because they allow us to accurately locate and track celestial objects in the sky. They are also used to determine the position of the observer on Earth and to calculate the altitude and azimuth of an object.

What is the relationship between Right Ascension and Hour Angle?

The relationship between Right Ascension and Hour Angle is that they are complementary angles. This means that the sum of Right Ascension and Hour Angle is equal to 90 degrees. In other words, as the Right Ascension of an object increases, its Hour Angle decreases and vice versa.

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