Homogeneuous Coordinates

Besides rotation, there is also translation in the realm of transformations. But with translation, the origin of base frame and mobile frame is no longer shared, unlike rotations. Because of this limitation, a $3\times 3$ matrix is not able to express translation.

Homogeneuous Coordinate Frames

$$T=\left[\begin{array}{cc}R& p\\ 0& 1\end{array}\right]$$

• q is a point in $R^3$
• $F$ is the orthogonal coordinate frame for $R^3$
• homogeneuous coordinates of $q$ with repsect to $F$ are expressed as $[q{]}^F=[q_1,q_2,q_3,1{]}^T$
• the above $T$ is a $4\times 4$ homogeneuous tranformation matrix
• Note that the $R$ represents what we have learnt previously, which is a $3\times 3$ rotation matrix representing the orientation of mobile frame $M$ relative to the base frame $F$
• $p$ is the $3\times 1$ translation vector that represents the position of origin of mobile frame $M$ relative to origin at fixed frame $F$

Homogenuous Rotation

The $4\times 4$ homogeneuous transformation matrix can be expressed as such when there is only rotation involved:

$$\text{Rot}(\varphi ,k)=\left[\begin{array}{cc}{R}_k(\varphi )& \begin{array}{c}0\\ 0\\ 0\end{array}\\ \begin{array}{ccc}0& 0& 0\end{array}& 1\end{array}\right]$$

• this is essentially the same as $T$'s structure, but we just let $p=0$
• $\varphi$ represents the angle that $M$ is rotated about the $k$th unit vector of $F$

Worked Example

Homogeneuous Translation

$$\text{Tran}(p)=\left[\begin{array}{cccc}1& 0& 0& p_1\\ 0& 1& 0& p_2\\ 0& 0& 1& p_3\\ 0& 0& 0& 1\end{array}\right]$$

• this is the $4\times 4$ representation of homogeneous translation, without rotation
• $F$ and $M$ are initially coicident
• we translate the origin $M$ by an amount $p_k$ along the $k$th unit vector of $F$

Assuming a mobile frame $M$ moved translation $p$ from initial position at $F$. We can express homogeneuous coordinate of point $q$ relative to reference frame $F$ as such:

$$[q{]}^F=Tran(p)[q{]}^M$$