Past exam of the computer science course of the University of Oxford / 2015 / Computer graphics / 2 / a / ii / Solution

Interpret the given focal point as the centre of perspective projection, $C=(0,6,-8)$. First translate it to the origin with $H(0,-6,8)$. The direction from $C$ to the world origin is then $(0,-6,8)$, of length 10. Choose the rotation matrix $G(\theta )$ so that this direction becomes $(0,0,10)$:Indeed, the transformed $y$ component is $-6(4/5)-8(-3/5)=0$, and the transformed $z$ component is $(-6)(-3/5)+8(4/5)=10$. Thus the required mapping, followed by the perspective divide, isFor a world point $(x,y,z)$, its camera coordinates and projected device coordinates areThe world origin projects to $(0,0)$ and lies at camera depth 10. A point five units along the unit viewing direction from $C$ has camera coordinates $(0,0,5)$, so it lies on the central image plane as expected. Here device coordinates are image-plane coordinates centred on the optical axis; an actual pixel viewport would require its own specified scale and offset.

Only rotations about the $x$-axis are available. Consequently the camera's horizontal axis stays parallel to the world $x$-axis; the viewing direction lies in the world $yz$ plane. The available orientation has pitch but no independent yaw or roll. Translations move the camera and the projected origin but cannot supply the missing orientation degrees of freedom.