Graphics Programming
Cameras: Parallel Projection

• With parallel projection, objects with larger distance to the viewer or camera are not getting smaller
• There are no vanishing points

Perspective projection

Parallel projection

projection rays

image plane

$x$

$y$

$z$

$\tilde{\mathbf{P}}$

$\mathbf{P}$

• The projection rays are parallel, i.e., the projection center is located at infinity
• The formula for mapping a 3D point $\mathbf{P}=(p_x,p_y,p_z)^\top$ to a point $\tilde{\mathbf{P}}= (\tilde{p}_x,\tilde{p}_y,\tilde{p}_z)^\top$ located at the image plane of the camera is given by:

  $\tilde{\mathbf{P}}= \left(p_x, p_y, 0 \right)^\top$

• Using homogeneous coordinates the parallel projection can be written as a linear mapping using a $4 \times 4$ matrix:

  $\begin{align}\underline{\tilde{\mathbf{P}}} & = \begin{pmatrix}p_x \\ p_y \\ 0\\ 1\end{pmatrix} = \underbrace{\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}}_{\mathtt{A}} \begin{pmatrix}p_x \\p_y \\ p_z\\ 1\end{pmatrix}\\ \underline{\tilde{\mathbf{P}}} &=\mathtt{A}\, \underline{\mathbf{P}} \end{align}$

• The $x$-, $y$- and $z$- coordinates within viewing volume shall be mapped to the range $[-1;1]$
• To this end, for each coordinate two new parameters $\alpha$ and $\beta$ are added to the linear transformation

$\mathtt{A}$

$x$

$y$

$z$

$x$

$y$

$z$

$\mathtt{A} = \begin{bmatrix} \alpha_x & 0 & 0 & \beta_x \\ 0 & \alpha_y & 0 & \beta_y \\ 0 & 0 & -\alpha_z & \beta_z \\ 0 & 0 & 0 & 1 \end{bmatrix}$

• For example, for the $z$-coordinate of points on the near-plane it should hold:
  $p_z=-z_n \quad \mapsto \quad \tilde{p}_z=-1$
  and for points on the far-plane
  $p_z=-z_f \quad \mapsto \quad \tilde{p}_z=1$

$\underline{\tilde{\mathbf{P}}} = \begin{pmatrix}\alpha_x \, p_x + \beta_x \\ \alpha_y \, p_y + \beta_y \\ -\alpha_z \, p_z + \beta_z \\ 1\end{pmatrix} = \begin{bmatrix} \alpha_x & 0 & 0 & \beta_x \\ 0 & \alpha_y & 0 & \beta_y \\ 0 & 0 & -\alpha_z & \beta_z \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{pmatrix}p_x \\p_y \\ p_z\\ 1\end{pmatrix} \in \mathbb{H}^3$

• Now $\alpha_z$ and $\beta_z$ can be determined from the conditions for the mapping of the $z$-coordinates:
  $\begin{align}p_z=-z_n \,&\mapsto \, \tilde{p}_z=-1 \quad \Rightarrow \alpha_z \, z_n + \beta_z = -1 \\ p_z=-z_f \, &\mapsto \, \tilde{p}_z=\ 1 \,\,\,\,\quad \Rightarrow \alpha_z \, z_f + \beta_z = 1\end{align}$

$\mathtt{A}$

$x$

$y$

$z$

$x$

$y$

$z$

$x_r$

$x_l$

$y_t$

$y_b$

$z_n$

$z_f$

• Solving the equation system for $\alpha_z$ and $\beta_z$ provides:

  $\begin{align} \alpha_z &= \frac{2}{z_f-z_n}\\ \beta_z & = -\frac{z_f + z_n}{z_f-z_n}\end{align}$

• After analogous calculations for the $x$- and $y$-coordinate, the projection matrix is given by:

  $\mathtt{A} = \begin{bmatrix} \frac{2}{x_r-x_l} & 0 & 0 & -\frac{x_r + x_l}{x_r-x_l} \\ 0 & \frac{2}{y_t-y_b} & 0 & -\frac{y_t + y_b}{y_t-y_b} \\ 0 & 0 & \frac{-2}{z_f-z_n} & -\frac{z_f + z_n}{z_f-z_n} \\ 0 & 0 & 0 & 1 \end{bmatrix}$

$\begin{align} \mathtt{A} \mathtt{T}_{\mathrm{\small cam}}^{-1} & = \mathtt{A} \, \mathtt{T}_{r_x}(\beta) \, \mathtt{T}_{r_y}(\alpha)\\ & = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0& \cos \beta & -\sin \beta& 0 \\ 0 & \sin \beta & \cos \beta & 0 \\ 0 & 0 & 0 &1 \\ \end{bmatrix} \begin{bmatrix} \cos \alpha & 0& \sin \alpha& 0 \\ 0 & 1 & 0 & 0 \\ -\sin \alpha & 0 &\cos \alpha & 0 \\ 0 & 0 & 0 &1 \\ \end{bmatrix} \\ & = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos \alpha & 0 & \sin \alpha & 0 \\ \sin \beta \sin \alpha & \cos \beta & -\sin \beta \cos \alpha & 0\\ - \cos \beta \sin \alpha & \sin \beta & \cos \beta \cos \alpha & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\ & = \begin{bmatrix} \cos \alpha & 0 & \sin \alpha & 0 \\ \sin \beta \sin \alpha & \cos \beta & -\sin \beta \cos \alpha & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \end{align}$

• Thus, the basis vectors are mapped as follows:

  $ \underline{\tilde{\mathbf{e}}}_x = \begin{pmatrix}\cos \alpha\\ \sin \beta \sin \alpha \\ 0\\ 1\\ \end{pmatrix}, \quad \underline{\tilde{\mathbf{e}}}_y = \begin{pmatrix}0\\ \cos \beta \\ 0\\ 1\\ \end{pmatrix}, \quad \underline{\tilde{\mathbf{e}}}_z = \begin{pmatrix}\sin \alpha\\ -\sin \beta \cos \alpha \\ 0\\ 1\\ \end{pmatrix}$

• From this the lengths of the projected basis vectors may be calculated and the orthographic projection can be classified in isometric, dimetric, or trimetric:

  $\begin{align} |\tilde{\mathbf{e}}_x| &= \sqrt{\cos^2\alpha + (\sin \beta \sin \alpha)^2}\\ |\tilde{\mathbf{e}}_y| &= \sqrt{\cos^2\beta}\\ |\tilde{\mathbf{e}}_z| &= \sqrt{\sin^2\alpha + (\sin \beta \cos \alpha)^2}\\ \end{align}$

$\mathbf{e}_x$

$\mathbf{e}_y$

$\tilde{\mathbf{e}}_z$

$\mathbf{e}_z$

cavalier projection

cabinet projection

1

1

1

1

1

0.5

$\alpha$

$\beta$

• Oblique projections allow an additional shearing
• In classic oblique projections two coordinate axes are not changed by the mapping and a shearing is applied to the third
• The projection direction of the third axis is given by two angles $\alpha$ and $\beta$ (see figure)
• The angle $\alpha$ is preserved in the projection and corresponds to the angle between the projected $z$- and $x$-axis. The angle $\beta$ controls the ratio:
  • Cavalier projection $x:z=1:1$
  • Cabinet projection: $x:z=2:1$

• General oblique projection:

  $\mathtt{A} \mathtt{T}_{\mathrm{\small cam}}^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & \frac{-\cos \alpha}{\tan \beta} & 0 \\ 0 & 1 & \frac{\sin \alpha}{\tan \beta} & 0 \\ 0 & 0 & \frac{-1}{\sin \beta} & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 & \frac{-\cos \alpha}{\tan \beta} & 0 \\ 0 & 1 & \frac{\sin \alpha}{\tan \beta} & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$