---
title: Tracking Features
categories: session
---

- aperture problem [page 78]

# Briefing

## Corners

![Universitetsområdet i Ålesund](Images/ntnuaes1.jpg)

![Universitetsområdet i Ålesund (ny vinkel)](Images/ntnuaes2.jpg)

+ What are distinctive points in the image?
+ Distinctive points can (to some extent) be matched in two different images.

[More Images](https://www.flickr.com/photos/ntnu-trondheim/collections/72157632165205007/)

## Corner Correspondence

+ Two images of the same scene $I_1,I_2: \Omega\subset\mathbb{R}^3\to\R_+ ; \mathbf{x}\mapsto I_1(\mathbf{x}),I_2(\mathbf{x})$
+ Different in general 

*Why are they different?*

## Brightness Constancy Constraint

+  Suppose we photograph empty space except for a single point $p$
    - *Brightness Constancy Constraint*

$$I_1(\mathbf{x}_1) = I_2(\mathbf{x}_2) \sim \mathcal{R}(p)$$

+  Simple dislocation from $\mathbf{x}_1$ to $\mathbf{x}_2$ 
+  Problem: Globally, it is an infinite-dimentional transformation
+  Motion: $h: \mathbf{x}_1\mapsto\mathbf{x}_2$
    + so that $I_1(\mathbf{x}_1) = I_2(h(\mathbf{x}_1)) \forall
      \mathbf{x}_1\in\Omega\cap h^{-1}(\Omega)\subset\mathbb{R}^{2\times2}$
      


## Motion Models

+ Affine Motion Model: $h(\mathbf{x}_1) = A\mathbf{x}_1 + \mathbf{d}$
+ Projective Motion Model: $h(\mathbf{x}_1) = H\mathbf{x}_1$ where
  $H\in\mathbb{R}^{3\times3}$ is defined up to a scalar factor.
+ Need to accept changes to the intencity

## Aperture Problem


$$\hat h = \arg\min_h\sum_{\tilde\mathbf{x}\in W(\mathbf{x})} ||I_1(\tilde\mathbf x)-I_2(\tilde\mathbf x)||^2$$

+ The window, or aperture, $W(\vec{x})$
+ cannot distinguish points on a blank wall

+ Choose $h$ from a family of functions, parameterised by $\alpha$
    + translational: $\alpha=\Delta\mathbf{x}$
    + affine: $\alpha=\{A,\mathbf{d}\}$

## Feature Tracking

$$I_1(\textbf{x})= I_2(h(\textbf{x}))= I_2(\textbf{x}+\Delta\textbf{x})$$

+ Consider infitesimally small $\Delta\textbf x$

## Infinitesimal Model

+ Model on a time axis - two images taken infinitesimally close in time
    + ... under motion


$$I(\mathbf{x}(t),t) = I(\mathbf{x}(t)+t\mathbf{u},t+dt)$$

$$\nabla I(\mathbf{x}(t),t)^\mathrm{T}\mathrm{u} + I_t(\mathbf{x}(t),t) = 0$$

$$\nabla I(\mathbf{x},t) = \begin{bmatrix} I_x(\mathbf{x},t)\\ I_y(\mathbf{x},t) \end{bmatrix}
= \begin{bmatrix}\frac{\partial I}{\partial x}(\mathbf{x},t)\\ \frac{\partial I}{\partial y}(\mathbf{x},t) \end{bmatrix}
\in\mathbb{R}^2$$

$$I_t(\mathbf{x},t) = \frac{\partial I}{\partial t}(\mathbf{x},t)\in \mathbb{R}$$
   
*Brightness Constancy Constraint* for the simplest possible continuous model

+ Two applications
    - optical flow: fix a position $\mathbf x$ and consider particles passing through
    - feature tracking: fix a partical $x(t)$ an track it through space 

## Solving for $\textbf{u}$

$$\nabla I^\mathrm{T}\mathrm{u} + I_t = 0$$

+ There are infititly many solutions, due to the *aperture problem*
+ We can solve for the component in the direction of the gradient though

$$\frac{\nabla I^\mathrm{T}\mathrm{u}}{||\nabla I||} = - \frac{I_t}{||\nabla I||} $$

+ Left hand side is the scalar projection of $\mathbf u$ onto $\nabla I$.
+ Multiplying by $\nabla I/||\nabla I||$, we get the vector projection:

$$\mathbf u_n = \frac{\nabla I^\mathrm{T}\mathrm{u}}{||\nabla I||}\cdot\frac{\nabla I}{||\nabla I||} = - \frac{I_t}{||\nabla I||\cdot\frac{\nabla I}{||\nabla I||}} $$