---
title: Relative Pose
categories: lecture
---

**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**

# Epipolar Entities

+ 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

# Decomposition of the Essential Matix

+ Theorem 5.5

# Two Relative Poses

+ Theorem 5.7

**Repetition**

+ Rotation by angle $\theta$ around the vector $\omega$ is given
  by $R=e^{\hat\omega\theta}$ assuming $\omega$ has unit length.

**Rodrigues' formula** (2.16)

$$e^{\hat\omega} =
  I + \frac{\hat\omega}{||\omega||}\sin(||\omega||)
    + \frac{\hat\omega^2}{||\omega||^2}(1-\cos(||\omega||))$$

## Lemma 5.6

> If $\hat T$ and $\hat TR$ are both skew-symmetric for $R\in\mathrm{SO}(3)$,
  then $R$ is a rotation by angle $\pi$ around $T$.
  
+ Skew-symmetry gives $(\hat TR)^T=-\hat TR$
+ We also have $(\hat TR)^T=R^T\hat T^T=-R^T\hat T$
+ Hence $\hat TR = R^T\hat T$,
+ and since $R^T=R^{-1}$, we have 
  $$R\hat TR=\hat T$$
+ Write $R=e^{\hat\omega\theta}$ for some $\omega$ of unit length and some 
  $\theta$, to get
  $$e^{\hat\omega\theta}\hat Te^{\hat\omega\theta}=\hat T$$
+ multiply by $\omega$
  $$e^{\hat\omega\theta}\hat Te^{\hat\omega\theta}\omega=\hat T\omega$$
  This represents a stationary rotation of the vector $\hat T\omega$.

Note that $\omega$ is stationary under rotation by $R$, and hence
it is an eigenvector associated with eigenvalue 1.
Furthermore, it is the only such eigenvector, and $\hat T\omega$
cannot be such.  Hence $\hat T\omega=T\times\omega=0$.
This is only possible if $T\sim\omega$, and since $\omega$ has
unit length, we get 
$$\omega = \pm\frac{T}{||T||}$$

We now know that $R$ has to be a rotation around $T$, and therefore
$R$ and $T$ commute.  This can be checked in Rodrigues' formula
(Theorem 2.9). 

Hence $R^2\hat T = \hat T$.
This looks like two half-round rotations to get back to start.
If $\hat T$ had been a vector or a matrix of full rank, we would
have been done.  However, with the skew-symmetric $\hat T$ there
is a little more fiddling.