---
title: Relative Pose
categories: session
---

**Reading** Ma 2004 Chapter 5

# The Epipolar Constraint

![Epipolar constraint due to Hamidur Rahman](Images/epipolar.jpg)

![Epipolar constraint due to Arne Nordmann (norro) - CC BY-SA 3.0](Images/epipolar2.svg)

+ Two cameras, two co-ordinate frames
+ Relative Pose: transformation $(R,T)$ from Camera Frame 1 to Camera Frame 2.
+ Consider a point $p$ in 3D, it has
    + co-ordinates $\mathbf{X}_i$ in Camera Frame $i$
+ The projection of $\mathbf{X}_i$ in homogeneous co-ordinates is
  called $\mathbf{x}_i$
    + $\mathbf{X}_i = \lambda_i\mathbf{x}_i$
    + Note $\mathbf{x}_i$ is 3D co-ordinates if the image plane
      is normalised to $z=1$
+ Combining this with the relation $\mathbf{X}_2=R\mathbf{X}_1+T$, we get
  $$\lambda_2\mathbf{x}_2 = R\lambda_1\mathbf{x}_1 + T$$
+ To simplify, we multiply by $\hat T$ to get
  $$\lambda_2\hat T\mathbf{x}_2 = \hat TR\lambda_1\mathbf{x}_1 + \hat TT$$
+ The last term is zero because $T\times T=0$
  $$\lambda_2\hat T\mathbf{x}_2 = \hat TR\lambda_1\mathbf{x}_1$$
+ Now $\mathbf{x}_2$ is perpendicular on $T\times\mathbf{X}_2$ so
  $$0=\mathbf{x}_2^T\lambda_2\hat T\mathbf{x}_2 = \mathbf{x}_2^T\hat TR\lambda_1\mathbf{x}_1$$
+ Since $\lambda_1$ is a scalar, we can simplify
  $$0= \mathbf{x}_2^T\hat TR\mathbf{x}_1$$
  This is the **epipolar constraint**
+ $E=\hat TR$ is called the **essential matrix**

## Some Jargon

+ For each point $p$, we have
    + a *epipolar plane* $\langle o_1,o_2,p\rangle$
    + *epipolar line* $\ell_i$ as the intersection of the epipolar plane
      and the image plane
+ The *epipoles* $\mathbf{e}_i$ is the projection of origo onto
  the image plane of the other camera.
+ Note that the epipoles are on the line $\langle o_1,o_2\rangle$, and
  hence in the epipolar plane

## Some properties

**Proposition 5.3(1)**

$$\mathbf{e}_2^TE = E\mathbf{e}_1=0$$

+ This is because
    1. $\mathbf{e}_2\sim T$ and $\mathbf{e}_1\sim R^TT$
    2. $E=\hat TR$
    3. $T\hat T = T\times T=0$

**Proposition 5.3(2)**

**Proposition 5.3(3)**

+ Both the image point and the epipole lie on the epipolar line

# Eight-Point Algorithm

$$\mathbf{x}_2^TE\mathbf{x}_1 = 0$$

Kronecker product: $\bigotimes$

Serialisation of a matrix: $(\cdot)^$

$$(\mathbf{x}_1\bigotimes\mathbf{x}_2)^TE^s = 0$$

$$\mathbf{a} = \mathbf{x}_1\bigotimes\mathbf{x}_2$$

$$\chi = [mathbf{a}_1, mathbf{a}_2, \ldots, mathbf{a}_n]$$

We can solve $\chiE^s = 0$ for $E^s$.
With eight points, we have unique solutions up to a scalar factor.

The solution is not necessarily a valid essential matrix, but we can
project onto the space of such matrices and correct the sign to 
get positive determinant.