2D translation

A translation is an affine transformation which is a linear transformation followed by some displacement

$$ \mathbf{v'} = \mathbf{Mv} + \mathbf{b} $$

Even though we can’t express 2D translation using a 2x2 matrix, we can express such a transformation as a shearing transformation in 3D projective geometry, to do so we have to imagine that the 2D Euclidean world exists as the plane $w = 1$ in a 3D space, under this geometry any point has the form $\begin{bmatrix} x & y & 1 \end{bmatrix}$

In Euclidean geometry a vector expressed as a linear combination of the standard basis has the form

$$ \mathbf{v} = v_x \unit{i} + v_y \unit{j} = \begin{bmatrix} v_x & v_y\end{bmatrix}^T $$

In Projective geometry a vector which exists in the plane $w = 1$ has the form

$$ \mathbf{v} = v_x \unit{i} + v_y \unit{j} + 1 \unit{w} = \begin{bmatrix} v_x & v_y & 1 \end{bmatrix}^T $$

This basis can be represented using the following transformation matrix

$$ \mathbf{M} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \mathbf{\cuv{i}} & \mathbf{\cuv{j}} & \mathbf{\cuv{w}} \end{bmatrix} $$

The translation transform then can be seen in Projective geometry as a simple shearing of the space by the coordinate $w$, using the shearing transform $\mathbf{H_{xy}}(\Delta{x}, \Delta{y})$ to transform a point $v$

$$ \mathbf{v'} = \mathbf{H_{xy}}(\Delta{x},\Delta{y}) \mathbf{v} = \begin{bmatrix} 1 & 0 & \Delta{x} \\ 0 & 1 & \Delta{y} \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} v_x \\ v_y \\ 1 \end{bmatrix} = \begin{bmatrix} v_x + \Delta{x} \\ v_y + \Delta{y} \\ 1 \end{bmatrix} $$

Now that we’re using perspective geometry to represent entities, let’s imagine a point $p = \begin{bmatrix} x & y & 0 \end{bmatrix}$ (a point that lies in the plane $w = 0$), whenever this point is transformed by a transformation matrix we can notice that the translation components of the matrix are cancelled because of $w = 0$, we can take advantage of this fact and represent vectors with this notation.

Let $v_{\infty}$ be a point located in the plane $w = 0$, applying the shearing operation $\mathbf{H_{xy}}(s, t)$ results in

$$ \mathbf{v_{\infty}'} = H_{xy}(\Delta{x},\Delta{y}) \mathbf{v_{\infty}} = \begin{bmatrix} 1 & 0 & \Delta{x} \\ 0 & 1 & \Delta{y} \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} v_x \\ v_y \\ 0 \end{bmatrix} = \begin{bmatrix} v_x \\ v_y \\ 0 \end{bmatrix} $$

It’s important to note that this matrix multiplication is still a linear transformation and that this trick of translating 2D points is actually a shearing of the 3D projective plane

3D translation

Similarly to 2D a 3D translation can be represented as a shearing of the 4D projective hyperplane which has the form

$$ \mathbf{T} = \mathbf{H_{xyz}}(\Delta{x},\Delta{y},\Delta{z}) = \begin{bmatrix} 1 & 0 & 0 & \Delta{x} \\ 0 & 1 & 0 & \Delta{y} \\ 0 & 0 & 1 & \Delta{z} \\ 0 & 0 & 0 & 1 \end{bmatrix} $$

When a 4D vector existing on the hyperplane $w = 1$ is transformed with this matrix the result is

$$ \mathbf{v'} = \mathbf{H_{xyz}}(\Delta{x},\Delta{y},\Delta{z})\mathbf{v} = \begin{bmatrix} 1 & 0 & 0 & \Delta{x} \\ 0 & 1 & 0 & \Delta{y} \\ 0 & 0 & 1 & \Delta{z} \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} v_x \\ v_y \\ v_z \\ 1 \end{bmatrix} = \begin{bmatrix} v_x + \Delta{x} \\ v_y + \Delta{y} \\ v_z + \Delta{z} \\ 1 \end{bmatrix} $$

The general 3D translation matrix is then denoted as

$$ \begin{equation} \label{general-translation-matrix} \mathbf{T} = \begin{bmatrix} 1 & 0 & 0 & T_x \\ 0 & 1 & 0 & T_y \\ 0 & 0 & 1 & T_z \\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} I_{3 \times 3} & T_{3 \times 1} \\ 0_{1 \times 3} & 1 \end{bmatrix} \end{equation} $$

References
  • Dunn, F. and Parberry, I. (2002). 3D math primer for graphics and game development. Plano, Tex.: Wordware Pub.