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---
title: Lecture: Distorted Space
title: (Lecture) Distorted Space
categories: lecture
---
# Distorted Space
# Distorted Space (Ch. 6.1)
## What is a distorted space?
1. Consider a pixmap image with pixels $1\times2$ mm.
What is the distance from origo to the points $(0,10)$, $(10,0)$,
and $(\sqrt{50},\sqrt{50})\approx(7,7)$?
2. $\psi: \mathbb{R}^3 \to \mathbb{R}^3$,
$\psi: \mathbb{X}\mapsto \mathbb{X}' = K\mathbb{X}$
- recall that in camera calibration, we typically have
$$K =
\begin{bmatrix}
s_x & s_\theta & o_x \\
0 & s_y & o_y \\
0 & 0 & 1
\end{bmatrix}
$$
2. Redifining the Inner Product
- $\langle\psi^{-1}(u),\psi^{-1}(v)\rangle
= u^TK^{-T}K^{-1}v
=\langle u,v\rangle_{K^{-T}K^{-1}}
=\langle u,v\rangle_{S}$
- where $S=K^{-T}K^{-1}$
4. Norm $||u||_S=\sqrt{\langle u,u\rangle}$
4. This gives rise to a **distorted space**
- angles are different
- norms are different
## 3D Motion in Distorted Space
1. Movement in canonical space: $X = RX_0+T$
2. Co-ordinates in uncalibrated camera frame
- before: $X_0' = KX_0$
- after: $X' = KX = KRX_0 + KT = KRK^{-1}X_0' + T'$
- where $T'=KT$
3. Thus the movement in distorted (uncalibrated) space is
$(R',T') = (KRK^{-1},KT)$
## Conjugate Matrix Group
1. The set of all Euclidean motions:
$\mathsf{SE}(3)=\{(R,T)|R\in\mathsf{SO}(3), T\in\mathbb{R}^3\}$
2. Conjugate of $\mathsf{SE}(3)$
$$G' = \bigg\{ g' = \begin{bmatrix} KRK^{-1} & T'\\0&1\end{bmatrix}
\bigg|R\in\mathsf{SO}(3), T\in\mathbb{R}^3\bigg\}$$
3. *Note commutative diagram in Fig 6.3 in the textbook*
## Image Formation
1. Calibrated (5.1) $\lambda x = \Pi_0X$
1. Uncalibrated (6.1) $\lambda x' = K\Pi_0gX_0$
- $g$ is camera pose
- $K$ is camera calibration matrix
- $\Pi_0$ is the projection (as before)
2. $\lambda x' = KRX_0 + KT$
- **abuse of notation!** we switch between homogeneous
and non-homogeneous co-ordinates
4. $\lambda x' = KRK^{-1}KX_0 + KT$
5. Rewriting in uncalibrated co-ordinates:
2. Image transformation $g: X_0 \mapsto X = KRX_0 + KT$
- Uncalibrated: $X' = KRK^{-1}X'_0+T'$
- Projected:. $\lambda x' = KRK^{-1}X'_0 + T'$ (homogeneous co-ordinates)
5. Rewriting in uncalibrated, heterogeneous co-ordinates:
- $\lambda x'=KRK^{-1}X'_0 + T' = \Pi_0g'X_0'$
6. Note $\Pi_0$ translates from 3D/homogeneous to 2D.
# Uncalibrated Epipolar Geometry
# Uncalibrated Epipolar Geometry (Ch. 6.2)
Two views by the same camera.
This gives one and the same calibration matrix $K$ for both views.
+ **Recall** the calibrated case
$$x_2^TEx_1 = 0$$
where $E=\hat TR$
+ In the uncalibrated case, this becomes
$$x_2'^TK^{-T}\hat TRK^{-1}x_1' = 0$$
by substituting $x=K^{-1}x'$
+ We define the **fundamental matrix**
$$F = K^{-T}\hat TRK^{-1}$$
$$F = K^{-T}\hat TRK^{-1} \quad\text{(eq. 6.10)}$$
+ This gives the **epipolar constraint** for uncalibrated cameras
$$x_2^TFx_1 = 0$$
$$x_2^TFx_1 = 0 \quad\text{(eq. 6.8)}$$
+ This works essentially as in the calibrated case
+ In a perfect camera, $K=I$ and $F=E$
+ It can be shown that
$$F = \hat T' KRK^{-1}$$
$$F = \hat T' KRK^{-1} \quad\text{(eq. 6.14)}$$
by invoking Lemma 5.4, but we'll have to take this on trust.
+ $F$ has rank two because $\hat T'$ has rank two
- if $F$ has full rank, find the SVD $F=U\mathsf{diag}(\sigma_1,\sigma_2,\sigma_3)V^T$
- replace $F$ by $U\mathsf{diag}(\sigma_1,\sigma_2,0)V^T$
- more or less as in the calibrated case
+ Note that $F$ has eight degrees of freedom
- $\hat T'$ has two
- $K$ has five
- $R$ has three
Hence it is impossible to recover $\hat T'$ and $R$ from $F$,
without additional information.
+ Many sources of additional information
# Ambiguities and Constraints in Image Formation (Ch 6.3)
