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 \(h(\mathbf{x}) = \mathbf{x}+\Delta\mathbf{x}\)
  • 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

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 \[\mathbf{x}+\Delta\mathbf{x}=\Pi(R\mathbf{X}+T)=\Pi R\mathbf{X}+\Pi T\]

Inserting for \(\mathbf{x}\) on the left hand side, we have \[\Pi\mathbf{X}+\Delta\mathbf{x}=\Pi R\mathbf{X}+\Pi T\]

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 \(\Delta\mathbf{x} = (\Delta x, \Delta y,0)\) 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.