I have two 2D coordinate frames, one local and one regional. I have five points measured on both systems and I want to figure out the translation and rotation parameters between these two systems.
Two coordinate systems one local and one regional, both on a 2D plane
Picture above for reference, I know how to determine the translation parameters $tx$ and $ty$, but when I try to find the angle $a$ utilizing the inverse tangent I end up with many different answers for each set of x,y coordinates.
The relation between the two frames is
$ p = t + R q $
where $p$ is the coordinate vector in the regional frame (red), and $q$ is the coordinate vector in the local frame (green). $t = (t_x, t_y) $ is the offset of the local frame with respect to the regional frame, and $R$ is a rotation matrix.
Now, suppose you have two pairs $p_1, q_1$, and $ p_2, q_2 $, then
$ p_1 = t + R q_1 $
$ p_2 = t + R q_2 $
Therefore,
$ p_1 - p_2 = R (q_1 - q_2)$
Let $u = p_1 - p_2 = (u_x, u_y)$ and $v = q_1 - q_2 = (v_x, v_y) $
Then
$u_x = \cos(\alpha) v_x - \sin(\alpha) v_y $
$u_y = \sin(\alpha) v_x + \cos(\alpha) v_y $
Therefore,
$ u = A [\cos(\alpha) , \sin(\alpha) ] ^T$
where
$ A = \begin{bmatrix} v_x && - v_y \\ v_y && v_x \end{bmatrix} $
Therefore, it follows that
$\begin{bmatrix} \cos(\alpha) \\ \sin(\alpha) \end{bmatrix} = A^{-1} u $
Now that we have $\cos(\alpha)$ and $\sin(\alpha)$, we have $R$ given by
$ R = \begin{bmatrix} \cos(\alpha) && -\sin(\alpha) \\ \sin(\alpha) && \cos(\alpha) \end{bmatrix} $
And we have the angle $\alpha = \text{Atan2}(\cos(\alpha), \sin(\alpha)) $
Having found $R$, we go back to the first equation
$p_1 = t + R q_1 $
Now we can solve for the offset vector $t$
$ t = p_1 - R q_1 $