How to Convert Ecliptical coordinates to Equatorial And Vice Versa?

This article will derive equations for converting a planet’s ecliptic coordinates to equatorial coordinates. This conversion is necessary for calculating primary directions - an ancient technique of event prediction.

Notations

First, let’s define terms. We will denote

• Celestial longitude and latitude by $\lambda$ and $\delta$ letters
• Equatorial coordinates - right ascension and declination by $RA$ and $D$
• Ecliptic inclination by $ϵ$

Coordinate rotation

Let’s denote the vector (a pointer) to the observed planet with the letter $\vec{v}$. Then the spherical coordinates of this vector in the ecliptic coordinate system will be

$$\vec{v}=(\lambda ,\delta )$$

The Cartesian coordinates of the same vector, according to equation (2) will be equal to

$$\begin{array}{rl}\vec{v}& =\cos \delta \cos \lambda {\text{X}}_{\text{ecl}}\\ & +\cos \delta \sin \lambda {\text{Y}}_{\text{ecl}}\\ & +\sin \delta {\text{Z}}_{\text{ecl}}\end{array}$$

Here we set $R$ (the radius of the celestial sphere) to be equal to 1 for simplicity. The ${\text{X}}_{\text{ecl}}$-axis directs to 0° Aries, the ${\text{Y}}_{\text{elc}}$-axis to 0° Cancer, and the ${\text{Z}}_{\text{ecl}}$-axis to the northern celestial hemisphere.

The equatorial plane is inclined at an angle $-ϵ$ relative to the plane of the ecliptic in the ${\text{Y}}_{\text{ecl}}{\text{Z}}_{\text{ecl}}$ plane.

We can use the rotation matrix, which we introduced earlier, substituting ($-\sin ϵ$) instead of $\sin (-ϵ)$.

$${\mathbf{A}}_{YZ}=\left[\begin{array}{rrr}1& 0& 0\\ 0& \cos ϵ& -\sin ϵ\\ 0& \sin ϵ& \cos ϵ\end{array}\right]$$

It gives us cartesian coordinates of the vector $\vec{v}$ in an equatorial coordinate system:

$$\{\begin{array}{l}{x}_{eq}=\cos \delta \cos \lambda \\ y_{eq}=\cos ϵ\cos \delta \sin \lambda -\sin ϵ\sin \delta \\ z_{eq}=\sin ϵ\cos \delta \sin \lambda +\cos ϵ\sin \delta \end{array}$$

From the conversion equation (1) it follows that

$$\{\begin{array}{l}\tan (RA)=y_{eq}/x_{eq}\\ \sin (D)=z_{eq}\end{array}$$

It gives us the final equations for converting $(\lambda ,\delta )\to (RA,D)$:

$$\begin{array}{r}\tan (RA)=\frac{\cos ϵ\sin \lambda -\sin ϵ\tan \delta }{\cos \lambda }\\ \sin (D)=\sin ϵ\cos \delta \sin \lambda +\cos ϵ\sin \delta \end{array}\text{\hspace{1em}(1)}$$

Invert Conversion

For invert conversion we change $ϵ$ to $(-ϵ)$. It gives us the following:

$$\begin{array}{r}\tan (\lambda )=\frac{\cos ϵ\sin (RA)+\sin ϵ\tan (D)}{\cos RA}\\ \sin (\delta )=-\sin ϵ\cos (D)\sin (RA)+\cos ϵ\sin (D)\end{array}\text{\hspace{1em}(2)}$$