---
title: Lecture Planar Scenes
categories: lecture
---

**Reading** Ma 2004:Ch 5

# Review

Last weeks exercise

1.  Not designed as an exercise to check that you have learnt what 
    I know.
    + Rather, it is designed as an experiment, as I would use it to
      test my own understanding.
2.  It sets up a closed loop.
    + the final result can be checked against the original data
3.  It demonstrates the eight-point algorithm, but it also demonstrates
    image capture (projection).
    - but this requires that you take the time to comprehend each step ...

If you can complete and comprehend all the steps, you have understood the
core of 3D reconstruction ...

however, there is more

- the planar case
- uncalibrated cameras

# Degeneration

+ A plane $P$ is described by an equation
  $$N^T{X}=d$$
    + where $N=(n_1,n_2,n_3)$ is a vector orthogonal on $P$
+ Consider object points $X_1,X_2,\ldots,X_n\in P$.
    + they all satisfy $N^TX_i = d$
    + or $\frac1dN^TX=1$ (1)
+ Extra constraint compared to the case for the eight-point algorithm
+ Consider the transformation between camera frames
  $$X'=RX+T$$
+ inserting from (1), we have
  $$X'=RX+T\frac1dN^TX=(R+T\frac1dN^T)X=HX$$
    + where $H=R+T\frac1dN^T$
    + $H$ depends on $(R,T)$ as well as $(N,d)$.
+ Consider the image points $x'=X'/\lambda'$ and $x=X/\lambda$.
    + we get $x'\sim Hx$ *(planar) homography*
+ multiplying both sides by $\hat x'$, we get
+ **Planar epipolar constraint**
  $$\hat x'Hx=0$$
+ because $x'\sim Hx$, for any $u\in\mathbb{R}^3$, $u\times x'=\hat ux'\bot Hx$
+ hence $x'^T\hat u Hx=0$ for all $u\in\mathbb{R}^3$
+ thus the epipolar constraint is under-defined
+ it follows that the eight-point algorithm cannot work

# More theory

+ An image point $x$ corresponding to $p\in P$ uniquely determines $x'\sim Hx$
    + if $p\not\in P$, $x'$ only ends up on the epipolar line

# Four-Point Algorithm for Planar Scenes

# Homography versus Essential Matrix