Linear estimation of parameters

• Ignoring radial distortion (for the time being) and setting $d_x = d_y = 1$ (measuring $f$ in pixels), we have $x^d_i = x^u_i$ and $y^d_i = y^u_i$ .
• Then, combining equations we have
  $$\frac{x_i - u_0}{f} = s_x \frac{r_{11}X_w + r_{12}Y_w + r_{13}Z_w + t_x}{r_{31}X_w + r_{32}Y_w + r_{33}Z_w + t_z}$$
  and
  $$\frac{y_i - v_0}{f} = \frac{r_{21}X_w + r_{22}Y_w + r_{23}Z_w + t_y}{r_{31}X_w + r_{32}Y_w + r_{33}Z_w + t_z}$$

• Assuming $(u_0,v_0)$ to be known (at the center of the image) and setting $x'_i = x_i - u_0$ and $y'_i = y_i - v_0)$ , we have
  $$\frac{x'_i}{f} = s_x \frac{X_i}{Z_i}$$    and $$\frac{y'_i}{f} = \frac{Y_i}{Z_i}$$

• Eliminating $f$ we have
  $$\frac{x'_i}{y'_i} = s_x \frac{X_i}{Y_i} = s_x \frac{r_{11}X_w + r_{12}Y_w + r_{13}Z_w + t_x}{r_{21}X_w + r_{22}Y_w + r_{23}Z_w + t_y}$$

• Rearranging, we have
  $$s_x \left( r_{11}X_w + r_{12}Y_w + r_{13}Z_w + t_x \right)y'_i - \left( r_{21}X_w + r_{22}Y_w + r_{23}Z_w + t_y \right)x'_i = 0$$
  or
  $$(X_w y'_i) s_x r_{11} +(Y_w y'_i) s_x r_{12} + (Z_w y'_i) s_x r_{... ... t_x - (X_w x'_i) r_{21} +(Y_w x'_i) r_{22} + (Z_w x'_i) r_{23} + x'_i t_y = 0$$
  which is a linear homogeneous equation in the eight unknowns $s_x r_{11}$ , $s_x r_{12}$ , $s_x r_{13}$ , $r_{21}$ , $r_{22}$ , $r_{23}$ , $s_x t_x$ and $t_y$ .
• The unknown scale factor can be fixed by setting $t_y = 1$ . Image correspondences of seven points in general position are sufficient to solve for the remaining unknowns. Let the solution be $s_x r'_{11}$ , $s_x r'_{12}$ , $s_x r'_{13}$ , $r'_{21}$ , $r'_{22}$ , $r'_{23}$ , $s_x t'_x$ and $t'_y = 1$
• We can estimate the correct scale factor by noting that the two rows of the rotation matrix are supposed to be normal, i.e.,
  $$r_{11}^2 + r_{12}^2 + r_{13}^2 = r_{21}^2 + r_{22}^2 + r_{23}^2 = 1$$

• The scale factor $c$ for the solution can then be determined from
  $$c = 1/\sqrt{(r'_{21})^2 + (r'_{22})^2 + (r'_{23})^2}$$
  and
  $$c/s_x = 1/\sqrt{(s_x r'_{11})^2 + (s_x r'_{12})^2 + (s_x r'_{13})^2}$$
  This also allows recovery of $s_x$ .
• In the above procedure we didn't enforce orthogonality of the first two rows of $R$ . Given vectors ${\bf r'_1}$ and ${\bf r'_2}$ , we can find two orthogonal vectors ${\bf r_1}$ and ${\bf r_2}$ close to the originals as follows:
  $${\bf r_1} = {\bf r'_1} + \mu {\bf r'_2}$$    and $${\bf r_2} = {\bf r'_2} + \mu {\bf r'_1}$$
  which gives
  $${\bf r_1} \cdot {\bf r_2} = {\bf r'_1} \cdot {\bf r'_2} + \mu ({\... ...f r'_1} + {\bf r'_2} \cdot {\bf r'_2}) + \mu^2 {\bf r'_1} \cdot {\bf r'_2} = 0$$
  The solution of this quadratic in $\mu$ is numerically ill behaved because ${\bf r'_1} \cdot {\bf r'_2}$ will be quite small. We can use the approximate solution
  $$\mu \simeq -(1/2) {\bf r'_1} \cdot {\bf r'_2}$$
  since ${\bf r'_1} \cdot {\bf r'_1}$ and ${\bf r'_2} \cdot {\bf r'_2}$ are both near 1.
• ${\bf r_3}$ can then be recovered as ${\bf r_1} \times {\bf r_2}$ .
• Once we have $R$ we can estimate $f$ and $t_z$ from the basic equations above. This will require one more correspondence to be given.
• The above procedure may be problematic if $t_y$ is close to 0. In such a case the entire experimental data may have to be first translated by a fixed amount.