# Ascension Difference Equation

BlogAdvanced AstrologyMathematics of the Celestial SphereSpherical Geometry Feb. 28, 2023, 9:08 a.m. Mark Rusborn 1 min. to read

In the previous article, we introduced the concept of ascension difference. This article will find an equation for the ascension difference and the oblique ascension of any celestial body.

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

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

I am a former Soviet physicist, now a professional astrologer. I have been lucky to get familiar with eminent scientists, including the Nobel laureate in physics. It helped me a lot in the proper structuring of thinking.