coordinates computation

Let's consider the Celestial Equatorial Frame:

$$ R_0 = (\vec{\gamma},\vec{y_0},\vec{N}) $$

where $\vec{\gamma}$ is the Vernal point, $\vec{N}$ the celestial North Pole and $\vec{y_0}$ completes the basis. We transfer to the intermediate basis:

$$ R_1 = (\vec{x_1},\vec{y_1},\vec{z_1}) $$

by a rotation of the angle $\alpha$ (the Right Ascension) around the North Pole $\vec{N}$. Consequently:

$$ \begin{matrix} \vec{x_1} = \cos\alpha \vec{\gamma} + \sin\alpha \vec{y_0} \cr \vec{y_1} = \cos\alpha \vec{y_0} - \sin\alpha \vec{\gamma} \cr \vec{z_1} = \vec{N} \end{matrix} $$

Finally we transfer to the last basis:

$$ R_2 = (\vec{x_2},\vec{y_2},\vec{z_2}) $$

by a rotation of the angle $\delta$ (the Declination) around the $\vec{y_1}$ axis. Consequently:

$$ \begin{matrix} \vec{x_2} = \cos\delta \vec{x_1} + \sin\delta \vec{z_1} = \cos\delta (\cos\alpha \vec{\gamma} + \sin\alpha \vec{y_0}) + \sin\delta \vec{N} \cr \vec{y_2} = \vec{y_1} = \cos\alpha \vec{y_0} - \sin\alpha \vec{\gamma} \cr \vec{z_2} = \cos\delta \vec{z_1} - \sin\delta \vec{x_1} = \cos\delta \vec{N} - \sin\delta (\cos\alpha \vec{\gamma} + \sin\alpha \vec{y_0}) \end{matrix} $$

The good news is that these equations of transfer from $R_0$ to $R_2$ can either apply from the Celestial Equatorial Frame to the star direction (aka the star RA-DEC coordinates), from the telescope frame to the star direction (aka the star position in the telescope frame), and from the Celestial Equatorial Frame to the telescope frame (aka the telescope position error). Let's define the index $T$ for the star coordinates in the telescope frame, and the index $E$ for the telescope coordinates in the Celestial Equatorial Frame. No index represents the true RA-DEC coordinates of the star.

The vector of the star in the Celestial Equatorial Frame is:

$$ \vec{x}_{star} = \cos\delta (\cos\alpha \vec{\gamma} + \sin\alpha \vec{y_0}) + \sin\delta \vec{N} $$

It can also be written in the telescope frame as:

$$ \vec{x}_{star} = \cos\delta_T (\cos\alpha_T \vec{x_T} + \sin\alpha_T \vec{y_T}) + \sin\delta_T \vec{z_T} $$

We must now write the telescope axis in the Celestial Equatorial Frame:

$$ \begin{matrix} \vec{x_T} = \cos\delta_E (\cos\alpha_E \vec{\gamma} + \sin\alpha_E \vec{y_0}) + \sin\delta_E \vec{N} \cr \vec{y_T} = \cos\alpha_E \vec{y_0} - \sin\alpha_E \vec{\gamma} \cr \vec{z_T} = \cos\delta_E \vec{N} - \sin\delta_E (\cos\alpha_E \vec{\gamma} + \sin\alpha_E \vec{y_0}) \end{matrix} $$

For sake of simplicity, we use a first order development in $\alpha_E$ and $\delta_E$:

$$ \begin{matrix} \vec{x_T} = \vec{\gamma} + \alpha_E \vec{y_0} + \delta_E \vec{N} \cr \vec{y_T} = \vec{y_0} - \alpha_E \vec{\gamma} \cr \vec{z_T} = \vec{N} - \delta_E \vec{\gamma} \end{matrix} $$

By replacing these coordinates in the star position, we find:

$$ \vec{x}_{star} = (\cos\delta_T\cos\alpha_T - \alpha_E\cos\delta_T\sin\alpha_T - \delta_E\sin\delta_T) \vec{\gamma} + (\cos\delta_T\sin\alpha_T + \alpha_E\cos\delta_T\cos\alpha_T) \vec{y_0} + (\sin\delta_T + \delta_E\cos\delta_T\cos\alpha_T) \vec{N} $$

Making use of the true star coordinates $\alpha$ and $\delta$ (cf some equations above), we find the system of equations:

$$ \begin{matrix} \cos\delta \cos\alpha = \cos\delta_T\cos\alpha_T - \alpha_E\cos\delta_T\sin\alpha_T - \delta_E\sin\delta_T \cr \cos\delta \sin\alpha = \cos\delta_T\sin\alpha_T + \alpha_E\cos\delta_T\cos\alpha_T \cr \sin\delta = \sin\delta_T + \delta_E\cos\delta_T\cos\alpha_T \end{matrix} $$

We write these equations as a matricial problem $Y=AX$ with:

$$ Y= \begin{pmatrix} \cos\delta \cos\alpha - \cos\delta_T\cos\alpha_T \cr \cos\delta \sin\alpha - \cos\delta_T\sin\alpha_T \cr \sin\delta - \sin\delta_T \end{pmatrix} $$

$$ A= \begin{pmatrix} -\cos\delta_T\sin\alpha_T & -\sin\delta_T \cr \cos\delta_T\cos\alpha_T & 0 \cr 0 & \cos\delta_T\cos\alpha_T \end{pmatrix} $$

$$ X= \begin{pmatrix} \alpha_E \cr \delta_E \end{pmatrix} $$

The solution is given by the pseudo-inverse $A^\dagger = (A^TA)^{-1}A^T$. We can compute:

$$ A^TA= \begin{pmatrix} \cos^2\delta_T & \cos\delta_T\sin\alpha_T\sin\delta_T \cr \cos\delta_T\sin\alpha_T\sin\delta_T & \sin^2\delta_T + \cos^2\delta_T\cos^2\alpha_T \end{pmatrix} $$

We directly find:

$$ \det A^TA = \cos^2\delta_T (\sin^2\delta_T + \cos^2\delta_T\cos^2\alpha_T - \sin^2\alpha_T\sin^2\delta_T) $$

Using the fun fact:

$$ \sin^2\delta_T = \sin^2\delta_T (\cos^2\alpha_T+\sin^2\alpha_T) $$

We simplify:

$$ \det A^TA = \cos^2\delta_T \cos^2\alpha_T $$

In order to solve the system, the star coordinates must satisfy both conditions $\alpha_T\neq\pi/2$ $[\pi]$ and $\delta_T\neq\pi/2$ $[\pi]$. It means that the star cannot be located at the North Pole, South Pole, nor RA=6h nor RA=18h.

Note: I was obviously expecting issues at North Pole and South Pole (since RA is not defined there), but I was not expecting a condition depending on the RA coordinates...