Ascension Difference Equation

BlogAdvanced AstrologyMathematics of the Celestial SphereSpherical Geometry
Ascension Difference Equation

Feb. 28, 2023, 9:08 a.m. Mark Rusborn 1 min. to read

In the previous article on celestial coordinates, we mentioned that the ascension difference is the difference between the equatorial coordinate of the ascending point of the celestial sphere and the ascending degree of the equator itself. In this article, we will derive the equation for the right ascension difference and the oblique ascension of any celestial object.


Ascension Difference

Your geographic latitude $\phi$ is zero when you are at the Earth's equator. In this case, the axis of rotation of the celestial sphere is directed strictly to the North.

As you move to the northern latitudes, the axis of rotation of the celestial sphere will rise above the horizon. Moreover, your latitude coincides with the angle of elevation of the axis of rotation above the horizon.

Consider a time of the ascension of the interested planet at a given geographical latitude.

The task is to find that planet's ascension difference and oblique ascension.

ascension difference
Fig. 1 - Ascension Difference.

Let's denote equatorial coordinates of the planet by $RA$ for the right ascension and $D$ for declination.

As you can see from the figure above, we have the right spherical triangle with sides $AD$ (ascension difference), $D$ (declination), and angle $90° - \phi$

From the equation(3) of spherical triangles, we have:

$$ \tan(D) = \sin(AD)\tan(90° - \phi) $$

Equations (3) of the sum of two angles, we have

$$ \tan(90° - \phi) = \frac{1}{\tan\phi} $$

Eventually, we have the equation for the ascension difference:

$$ \sin AD = \tan\phi \tan D\tag{1} $$

where $\phi$ is geographical latitude of the observer, and $D$ is the declination fo the planet

Oblique Ascension

Now we can derive an oblique ascension from any equatorial coordinates $(RA, D)$ with a simple equation:

$$ \begin{aligned} OA & = RA - AD \\\ & = RA - \arcsin(\tan\phi \tan D) \end{aligned}\tag{2} $$

Mark Rusborn

Mark Rusborn

Search the Articles