Epipolar geometry: uncalibrated case

• Given the two cameras 1 and 2, we have the camera equations:
  $${\bf x_1} = \tilde{\bf P_1} {\bf X}$$    and $${\bf x_2} = \tilde{\bf P_2} {\bf X}$$

• The optical center projects as
  $$\tilde{\bf P_i}{\bf X} = {\bf0}$$

• Writing
  $$\tilde{\bf P_i} = \left[ {\bf P_i} \mid -{\bf P_i}{\bf t_i} \right]$$
  where ${\bf P_i}$ is $3 \times 3$ non-singular we have that ${\bf t_i}$ is the optical center.
  $$\left[ {\bf P_i} \mid -{\bf P_i}{\bf t_i} \right] \left[ \begin{array}{c} {\bf t_i} \\ 1 \end{array} \right] = {\bf0}$$

• The epipole ${\bf e_2}$ in the second image is the projection of the optical center of the first image:
  $${\bf e_2} = \tilde{\bf P_2} \left[ \begin{array}{c} {\bf t_1} \\ 1 \end{array} \right]$$

• The projection of point on infinity along the optical ray $<{\bf t_1},{\bf x_1}>$ on to the second image is given by:
  $${\bf x_2} = {\bf P_2}{\bf P_1}^{-1}{\bf x_1}$$

• The epipolar line $<{\bf e_2},{\bf x_2}>$ is given by the cross product ${\bf e_2} \times {\bf x_2}$ .
• If $\left[ {\bf e_2} \right]_{\times}$ is the $3 \times 3$ antisymmetric matrix representing cross product with ${\bf e_2}$ , then we have that the epipolar line is given by
  $$\left[ e_2 \right]_{\times} {\bf P_2}{\bf P_1}^{-1}{\bf x_1} = {\bf F}{\bf x_1}$$

• Any point ${\bf x_2}$ on this epipolar line satisfies
  $${\bf x_2}^{T}{\bf F}{\bf x_1} = 0$$

• ${\bf F}$ is called the fundamental matrix. It is of rank 2 and can be computed from 8 point correspondences.
• Clearly ${\bf Fe_1} = {\bf0}$ (degenerate epipolar line) and ${\bf e_2}^{T}{\bf F} = {\bf0}$ . The epipoles are obtained as the null spaces of ${\bf F}$ .