Planes and epipolar geometry

• The homography between two planes is represented by a $3 \times 3$ matrix
  $${\bf\tilde{x}'} = {\bf Tx}$$

• The epipoles are image projections of a 3D point lying on all planes, because the line joining the optical centers intersects any plane.
  $${\bf e'} = {\bf Te}$$

• Four coplanar points and two points off the plane determine the epipolar geometry.
  1. Compute the plane projective transformation ${\bf T}$ , such that ${\bf\tilde{x_i}'} = {\bf Tx_i},\;\;i=1 \ldots 4$ .
  2. Determine the epipole ${\bf e'}$ in the ${\bf x'}$ image as the intersection of the lines $({\bf Tx_5} \times {\bf x'_5})$ and $({\bf Tx_6} \times {\bf x'_6})$ .
     

  3. The epipolar line in the ${\bf x'}$ image of any other point ${\bf x}$ is given as $({\bf Tx} \times {\bf e'})$ .
  4. ${\bf F} = {\bf e'}_{\times}{\bf T}$