<aside> 💡 This is a personal review note. Most of the explanations come from textbooks or online resources, especially the classic vSLAM book: Basic Knowledge on Visual SLAM: From Theory to Practice. For further details, please refer to the References section at the end of this page. if you need more information, please leave a comment.

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Epipolar Geometry

<aside> 💡 The Epipolar constraint reduces the correspondence problem to a 1D search along an epipolar line.

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Basic Knowledge on Visual SLAM: From Theory to Practice

Two frames $I_1,I_2$. The motion from the $I_1$ to $I_2$ as $\bold{R}$,$\bold{t}$ and the centers of the two cameras as $O_1$,$O_2$. Consider we have a feature point $p_1$ in $I_1$ and it’s corresponds feature point $p_2$ in $I_2$ and the match is correct (no error). the $\overrightarrow{O_1p_1}$ and the $\overrightarrow{O_2p_2}$ will intersect at $P$ in the 3D space. we have Epipolar geometric relationship between them.

Epipolar geometry terms

• Baseline $\overline{O_1O_2}$: the line joining the two camera centers.
• Epipolar plane $Plane_{O_1O_2P}$: a given point contains the baseline and that point defines the plane.
• Epipolar line $l_1,l_2$ : the intersection of the epipolar plane with the image plane - comes in pairs. $l_1 = \overrightarrow{e_1p_1}$ , $l_2 = \overrightarrow{e_2p_2}$
  • if you have a point on one epipolar line, then its match point must be on the corresponding epipolar line in the other image.
• Epipole $e_1,e_2$: the inner intersection of the baseline ray with image planes.
  • the image plane has to be big enough.
  • all of the epipolar planes intersect the epipole, because it contains the baseline ray which defines the epipolar planes. so every epipolar line will also intersect the epipole.

Geometric relationship

• $\overrightarrow{O_1p_1}$ : All points on the ray will be projected to the same pixel ( $p_1$ ). (including the point $P$)
• $\overrightarrow{e_2p_2}$ : without point $P$ position the ray $\overrightarrow{O_1p_1}$ is projected to the epipolar line (ray) $\overrightarrow{e_2p_2}$ on the second frame $I_2$, because $P$ can be at any possible position on $\overrightarrow{e_2p_2}$ (we don’t know where is the $p_2$) .
• $\overline{e_2p_2}$ : give a point $P$ the line $\overline{O_1P}$ is projected to the epipolar line $\overline{e_2p_2}$ on the second frame $I_2$

Epipolar Constraint

Epipolar Constraint Derivation

Define the spatial position of $P$ in the first frame to be:

$$ \bold{P}=\begin{bmatrix} X &Y&Z\end{bmatrix}^T $$