Computer Graphics

Coordinate Frames

Let $A$ be the original basis and $B$ be the new basis

x y 1_{B} = a d 0 b e 0 c f 1 x y 1_{A}

Then:

$[a d]$ is $i_{A}$ , how to transform the $x$ coordinate
- $a$ is how much of $i_{B}$ we need to make one $i_{A}$
- $d$ is how much of $j_{B}$ we need to make one $i_{A}$
$[b e]$ is $j_{A}$ , how to transform the $y$ coordinate
- $b$ is how much of $i_{B}$ we need to make one $j_{A}$
- $e$ is how much of $j_{B}$ we need to make one $j_{A}$
$[c f]$ is $O_{A}$ , the translation of the entire frame
- $c$ is how much of $i_{B}$ we need to get from $O_{B}$ to $O_{A}$
- $f$ is how much of $j_{B}$ we need to get from $O_{B}$ to $O_{A}$

The translation from $P_{A}$ to $P_{B}$ can be represented as $P_{B} = O_{A} + x_{A} i_{A} + y_{A} j_{A}$

Translate(x,y,z)
- $x^{'} y^{'} z^{'} 1 = 111 a b c 1 x y z 1$
Rotate(z,theta)
- $x^{'} y^{'} z^{'} 1 = cos θ sin θ - sin θ cos θ 11 x y z 1$
Scale(x,y,z)
- $x^{'} y^{'} z^{'} 1 = a b c 1 x y z 1$

In a scene hierarchy, the Camera Coordinate Frame ( $F_{V CS}$ ) is generally the root.

Transformations in scene graphs are written right to left, starting with source frame and ending with target frame.

k = \frac{P _{eye} - P _{re f}}{∥ P _{eye} - P _{re f} ∥}

i = \frac{V _{u p} \times k}{∥ V _{u p} \times k ∥}

j = k \times i

M_{c am} = i_{1} i_{2} i_{3} 0 j_{1} j_{2} j_{3} 0 k_{1} k_{2} k_{3} 0 P_{eye 1} P_{eye 2} P_{eye 3} 1

M_{v i e w} = M_{c am}^{- 1}