---
title: (Lecture) Distorted Space
categories: lecture
---

# 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.  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 (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**
  $$F = K^{-T}\hat TRK^{-1} \quad\text{(eq. 6.10)}$$
+ This gives the **epipolar constraint** for uncalibrated cameras
  $$x_2^TFx_1 = 0 \quad\text{(eq. 6.8)}$$
+ This works essentially as in the calibrated case
+ In a perfect camera, $K=I$ and $F=E$
+ It can be shown that
  $$F = \hat T' KRK^{-1} \quad\text{(eq. 6.14)}$$
  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)