Vanishing points & the horizon, interactive demo

Where parallel lines meet

Take a 3D line through the point $\mathbf{X}$ with direction $\mathbf{d}$ . Its points $\mathbf{X} + s\,\mathbf{d}$ project to

w \begin{bmatrix} u \\ v \\ 1 \end{bmatrix} = K \left( R\,(\mathbf{X} + s\,\mathbf{d}) + t \right) = K (R\,\mathbf{X} + t) + s\, K R\, \mathbf{d}

As the point moves away ( $s \to \infty$ ), the second term dominates, and the image point approaches the vanishing point

\mathbf{v} = K R\, \mathbf{d}

Neither $\mathbf{X}$ nor $t$ appears in it, so all lines with the same direction share the same vanishing point, and moving the camera without turning it does not move it. In the 3D view, the dashed ray from the camera center $C$ parallel to the lines hits the image plane exactly at $\mathbf{v}$ : the vanishing point is the image of the point at infinity in direction $\mathbf{d}$ . As in homogeneous coordinates, it can be computed as the intersection $l_1 \times l_2$ of two image lines $l_i = p_i \times q_i$ .

If $\mathbf{d}$ is parallel to the image plane, the third coordinate of $K R\,\mathbf{d}$ is $w = 0$ . Then the vanishing point is itself at infinity, and the image lines stay parallel. This is why vertical walls stay vertical in a photo taken with a level camera.

Vanishing lines and the horizon

All directions within a plane have their vanishing points on one line, the vanishing line of the plane (and of all planes parallel to it). It passes through any two of these vanishing points. For the ground plane, spanned by $X$ and $Y$ , this is the horizon:

l = \mathbf{v}_X \times \mathbf{v}_Y, \qquad \mathbf{v} \cdot l = 0 \text{ for every horizontal direction}

The horizon is the image of the plane through the camera center parallel to the ground, so everything at the height of the camera lies on it. The ramp in the scene is tilted: its edges vanish above the horizon, on the vanishing line of the ramp's own surface (dashed purple).

One-, two- and three-point perspective

One-point: the camera looks straight along $Y$ . Only the depth lines converge, to the principal point; $X$ and $Z$ lines stay parallel.
Two-point: the camera turns (yaw) but stays level. $X$ and $Y$ lines converge to two points on the horizon, vertical lines stay parallel.
Three-point: the camera also tilts (pitch). Vertical lines converge as well, above the image when looking up, below it when looking down.

Conventions. The world frame is right-handed with

Z

up; the camera frame has

x

right,

y

down and

z

forward. The pose is set with yaw, pitch and roll as in the pinhole camera demo, with the camera at height

h

above the ground.

Try this

Set pitch to 0. The vertical lines become parallel: $\mathbf{v}_Z$ has $w = 0$ and moves to infinity.
Press “One-point perspective”. $\mathbf{v}_Y$ sits at the principal point, and $\mathbf{v}_X$ is at infinity as well.
Change the camera height. The horizon and all vanishing points stay where they are, since $K R\,\mathbf{d}$ does not depend on the position.
Set the height to 3 m: the top of the left building is at camera height and lies exactly on the horizon. Try 2 m for the right one.
Turn the yaw. $\mathbf{v}_X$ and $\mathbf{v}_Y$ slide along the horizon in opposite directions. Zoom out to follow them.
Roll the camera. The horizon tilts, but it still passes through $\mathbf{v}_X$ and $\mathbf{v}_Y$ .
Pitch the camera up and down. The horizon moves down and up, and $\mathbf{v}_Z$ jumps from far above the image to far below it when it passes through infinity.
Pick a family under “check family” and compare $l_1 \times l_2$ with $K R\,\mathbf{d}$ in the panel.

Vanishing points & horizon

Where parallel lines meet

Vanishing lines and the horizon

One-, two- and three-point perspective

Try this