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title: Relative Pose
categories: session
---
**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
$$0=\mathbf{x}_2^T\lambda_2\hat T\mathbf{x}_2} = \mathbf{x}_2^T\hat TR\lambda_1\mathbf{x}_1$$
$$0=\mathbf{x}_2^T\lambda_2\hat T\mathbf{x}_2 = \mathbf{x}_2^T\hat TR\lambda_1\mathbf{x}_1$$
+ 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.
