## 1. From EclipticalLongitude to Right Ascension

$$\tan α = \tan λ \cos ε$$

This is approximated by (to sixth order of $$ε$$)

\begin{align} α = & λ - \left( \frac{1}{4} ε^2 + \frac{1}{24} ε^4 + \frac{17}{2880} ε^6 \right) \sin(2 λ) \notag \\ & + \left( \frac{1}{32} ε^4 + \frac{1}{96} ε^6 \right) \sin(4 λ) \notag \\ & - \frac{1}{192} ε^6 \sin(6 λ) \end{align}

If $$ε$$ isn't close to 0° but to 180°, then we define $$ε = 180° + ε_1$$ and then we can approximate

\begin{align} α = & λ + \left( \frac{1}{4} ε_1^2 + \frac{1}{24} ε_1^4 + \frac{17}{2880} ε_1^6 \right) \sin(2 λ) \notag \\ & - \left( \frac{1}{32} ε_1^4 + \frac{1}{96} ε_1^6 \right) \sin(4 λ) \notag \\ & + \frac{1}{192} ε_1^6 \sin(6 λ) \end{align}

## 2. From Right Ascension to EclipticalLongitude

For this we find

\begin{align} λ = & α + \left( \frac{1}{4} ε^2 + \frac{1}{24} ε^4 + \frac{17}{2880} ε^6 \right) \sin(2 α) \notag \\ & + \left( \frac{1}{32} ε^4 + \frac{1}{96} ε^6 \right) \sin(4 α) \notag \\ & + \frac{1}{192} ε^6 \sin(6 α) \end{align}

and another approximation is

\begin{align} λ = & α - \left( \frac{1}{4} ε_1^2 + \frac{1}{24} ε_1^4 + \frac{17}{2880} ε_1^6 \right) \sin(2 α) \notag \\ & - \left( \frac{1}{32} ε_1^4 + \frac{1}{96} ε_1^6 \right) \sin(4 α) \notag \\ & - \frac{1}{192} ε_1^6 \sin(6 α) \end{align}

## 3. From EclipticalLongitude to Declination

$$\sin δ = \sin λ \sin ε$$

We can approximate this with

\begin{align} δ = & \left( ε - \frac{1}{6} ε^3 + \frac{1}{120} ε^5 \right) \sin(λ) \notag \\ & + \left( \frac{1}{6} ε^3 - \frac{1}{12} ε^5 \right) \sin(λ)^3 \notag \\ & + \frac{3}{40} ε^5 \sin(λ)^5 \end{align}

Another approximation follows if you change $$ε$$ into $$-ε_1$$ everywhere.

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Last updated: 2016-02-07