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---
title: Relative Pose
categories: lecture
---
**Reading** Ma 2004 Chapter 5
# The Epipolar Constraint
+ 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
$$E = U\mathsf{diag}\{\sigma,\sigma,0\}V^T,$$
where $U,V\in\mathsf{SO}(3)$
+ Tricky proof. Do not spend too much time on this.
$$
\begin{cases}
(\hat T_1,R_1) &=
(UR_Z(+\frac\pi2)\Sigma U^T, UR_Z(+\frac\pi2)V^T)
\\
(\hat T_2,R_2) &=
(UR_Z(-\frac\pi2)\Sigma U^T, UR_Z(-\frac\pi2)V^T)
\end{cases}
$$
where
$$R_Z(+\frac\pi2) =
\begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}
$$
is a rotation by $\pi/2$ radians around the $z$-axis.
+ Note that there are two solutions from one $U\Sigma V^T$ decomposition.
+ Are there more solutions?
# Two Relative Poses
+ Theorem 5.7
> There exist exactly two relative poses $(R,T)$ with $R\in\mathsf{SO}(3)$
and $T\in\mathbb{R}^3$ corresponding to a nonzero essential matrix
$E\in\mathcal{E}$
[Theorem 5.7]
**Repetition**
## 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)
$$e^{\hat\omega} =
I + \frac{\hat\omega}{||\omega||}\sin(||\omega||)
+ \frac{\hat\omega^2}{||\omega||^2}(1-\cos(||\omega||))$$
## Theorem 5.7
+ **Demo** read the proof (debrief?)
## 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 $T$.
+ **Demo** read the proof (debrief?)
+ 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.
