Camera calibration and absolute conic

• Let ${\cal C}$ and ${\cal R}$ be the optical center and the retinal plane of a camera.
• Consider a rigid motion ${\cal D}$ of the camera from configuration $({\cal C},{\cal R})$ to $({\cal D}({\cal C}),{\cal D}({\cal R}))$ .
• Let $\omega_1$ and $\omega_2$ be the two images of $\Omega$ corresponding to the two configurations.
• Clearly $\omega_1 = \omega_2$ because $\Omega = {\cal D}(\Omega)$ .
• Thus $\omega$ is determined by only the internal parameters.
• Consider the equation of $\Omega$
  $$X^2 + Y^2 + Z^2 = 0 = T = {\bf M}^T{\bf M}$$
  where ${\bf M}^T = \left[ X,Y,Z \right]$
• Images ${\bf x}$ of points ${\bf X}$ of $\Omega$ satisfy the equation
  $${\bf x} = \left[ {\bf P} \mid {\bf -Pt} \right] \left[ \begin{array}{c} X \\ Y \\ Z \\ 0 \end{array}\right] = {\bf P}{\bf M}$$

• Since ${\bf M}^T{\bf M} = 0$ , we have that the equation of $\omega$ is
  $${\bf x}^T{\bf P}^{-1T}{\bf P}^{-1}{\bf x} = 0$$

• In terms of pixel coordinates this can be re-written as:
  $$(\frac{u-u_0}{fk_u})^2 + (\frac{v-v_0}{fk_v})^2 + 1 = 0$$
  Thus, the image of the absolute conic are determined completely by the internal parameters.