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Tracking Features

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

**Date** 24 September 2021

**Briefing** [Tracking Features Lecture]()

# Exercises

+ Exercise 4.1 (Ma 2004)
    + Let $\mathbf{X}$ be a 3D point and $\mathbf{x}$ its projection 
    + Consider the image motion $h(\mathbf{x}) = \mathbf{x}+\Delta\mahtbf{x}$
    + What transformation $(R,T)$ must the scene undergo to relalise
      the given $h(x)$? 
    + *Hint*
        + We remember that $(R,T)$ has six degrees of freedom in general.
        + To support this given $h$, some of these six degrees have to be
          fixed.  Which ones?
+ Exercises 4.2, 4.3 (Ma 2004)
+ (Exercises 4.4 (Ma 2004))

# Debrief

## \PiR\mathbf{X}+Exercise 4.1 (Ma 2004)
## Exercise 4.1 (Ma 2004)

**Note** We discussed the solution in the debrief session,
and this discussion is far more instructive than the algebraic
solution outlined below.  Yet, this sketch may add some other 
insights.


**Sketch for a solution** 
Write 
$$\mathbf{x}=\Pi\mathbf{X},$$
where $\Pi$ is the projection matrix, including the camera parameters.

After the transformation, we have
$$\mathbf{x}+\Delta\mathbf{x}=\Pi(R\mathbf{X}+T)=\PiR\mathbf{X}+\Pi T$$
$$\mathbf{x}+\Delta\mathbf{x}=\Pi(R\mathbf{X}+T)=\Pi R\mathbf{X}+\Pi T$$

Inserting for $\mathbf{x}$ on the left hand side, we have
$$\Pi\mathbf{X}+\Delta\mathbf{x}=\PiR\mathbf{X}+\Pi T$$
$$\Pi\mathbf{X}+\Delta\mathbf{x}=\Pi R\mathbf{X}+\Pi T$$

To satisfy this for every $\mathbf{X}$, we require that
$$\Pi = \Pi R$$
and
$$\Delta\mathbf{x}=\Pi T$$

Recall that $\Pi$ is given as
$$\Pi=\frac1Z\cdot
\begin{bmatrix}
f & 0 & 0 \\
0 & f & 0 \\
0 & 0 & 1 
\end{bmatrix}
$$
Note that the projection matrix depends on the $Z$-coordinate of the
3D point.

Since $\Pi=\Pi R$, we must have $R=I$, i.e. there is no rotation.

Let us write $\Delta\mathbf{x} = (\Delta x, Delta y,0)$
Let us write $\Delta\mathbf{x} = (\Delta x, \Delta y,0)$
and $T^T=(x_t,y_t,z_t)$, to get
$$\Delta x = \frac{fx_t}{Z}\quad \Delta y = \frac{fy_t}{Z}$$
Furthermore, 
$$0=\frac{fz_t}{Z},$$
and hence $z_t=0$.

We conclude that the transformation $(R,T)$ has to be a translation 
parallel to the image plane.