Projective structure using a five point basis

• Select the projection center of the first camera as the coordinate origin $(0,0,0,1)^T$ and that of the second camera as the unit point $(1,1,1,1)^T$ .
• Complete the 3D basis by choosing 3 other visible 3D points ${\bf A_1}$ , ${\bf A_2}$ and ${\bf A_3}$ such that no four of the five points are coplanar.
• Let ${\bf a_1}$ , ${\bf a_2}$ , ${\bf a_3}$ and ${\bf a'_1}$ , ${\bf a'_2}$ , ${\bf a'_3}$ be the image projections of ${\bf A_1}$ , ${\bf A_2}$ and ${\bf A_3}$ in the two images.
• Make a projective transformation of each image such that these 3 points and the epipoles form a standard basis
  $${\bf a_1} = {\bf a'_1} = \left[ \begin{array}{c}1 \\ 0 \\ 0 \end{... ...\; {\bf e} = {\bf e'} = \left[ \begin{array}{c}1 \\ 1 \\ 1 \end{array} \right]$$

• Also fix the 3D coordinates as
  $${\bf A_1} = \left[ \begin{array}{c}1 \\ 0 \\ 0 \\ 0 \end{array} \... ...\;\;\; {\bf A_3} = \left[ \begin{array}{c}0 \\ 0 \\ 1 \\ 0 \end{array} \right]$$

• It follows that
  $${\bf P} = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right]$$    and $${\bf P'} = \left[ \begin{array}{cccc} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{array}\right]$$

• Once ${\bf P}$ 's are known finding the projective structure of the other points is straightforward.

Localization errors in any of the points affect subsequent computation