Orthographic projection (geometry)

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In Euclidean geometry, an orthographic projection is an orthogonal projection. In particular, in 3D it is an affine, parallel projection of an object onto a perpendicular plane.

A simple orthographic projection onto the plane z = 0 can be defined by the following matrix:

P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}

For each point v = (vx, vy, vz), the transformed point would be

Pv = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} v_x \\ v_y \\ v_z \end{bmatrix} = \begin{bmatrix} v_x \\ v_y \\ 0 \end{bmatrix}

Often, it is more useful to use homogeneous coordinates, since translation cannot be accomplished with a 3-by-3 matrix. The transformation above can be represented for homogeneous coordinates as

P = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}

For each homogeneous vector v = (vx, vy, vz, 1), the transformed vector would be

Pv = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} v_x \\ v_y \\ v_z \\ 1 \end{bmatrix} = \begin{bmatrix} v_x \\ v_y \\ 0 \\ 1 \end{bmatrix}

In computer graphics, one of the most common matrices used for orthographic projection can be defined by a 6-tuple, (left, right, bottom, top, near, far), which defines the clipping planes. These planes form a box with the minimum corner at (left, bottom, near) and the maximum corner at (right, top, far).

The box is translated so that its center is at the origin, then it is scaled to the unit cube which is defined by having a minimum corner at (-1,-1,-1) and a maximum corner at (1,1,1).

The orthographic transform can be given by the following matrix:

P = \begin{bmatrix} \frac{2}{right-left} & 0 & 0 & -\frac{right+left}{right-left} \\ 0 & \frac{2}{top-bottom} & 0 & -\frac{top+bottom}{top-bottom} \\ 0 & 0 & \frac{2}{far-near} & -\frac{far+near}{far-near} \\ 0 & 0 & 0 & 1 \end{bmatrix}

which can be given as a translation followed by a scaling of the form

P = ST = \begin{bmatrix} \frac{2}{right-left} & 0 & 0 & 0 \\ 0 & \frac{2}{top-bottom} & 0 & 0 \\ 0 & 0 & \frac{2}{far-near} & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & -\frac{left+right}{2} \\ 0 & 1 & 0 & -\frac{top+bottom}{2} \\ 0 & 0 & 1 & -\frac{far+near}{2} \\ 0 & 0 & 0 & 1 \end{bmatrix}
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