May 10, 2024

Homogeneous Transformation: Rotation and Translation


In our previous post, we have explained rotation matrices. In this post, we explain how to perform rotation and translation as a single matrix operation.

A YouTube video accompanying this post is given below.

Consider the figure below.

The coordinate system B is translated from the coordinate system A, and after that it has been rotated for the angle \theta. The location of the coordinate system B with respect to the coordinate system A is represented by the vector \leftindex^{A}{\mathbf{d}}. The notation \leftindex^{A}{\mathbf{d}} means that the vector \mathbf{d} is represented in the coordinate system A. That is, its components (projections) are represented in the coordinate system A. Similarly, the notation \leftindex^{B}{\mathbf{p}} means that the vector \mathbf{p} is represented in the coordinate system A. Let the coordinates of the vector \mathbf{p} expressed in the coordinate system B be given as follows:

(1)   \begin{align*}\leftindex^{B}{\mathbf{p}} =\begin{bmatrix} p_{x_{B}} \\   p_{y_{B}}   \end{bmatrix}\end{align*}

where p_{x_{B}} and p_{y_{B}} are the coordinates of the vector \mathbf{p} expressed in the coordinate system B.

Problem 1: Given the coordinates of the vector \leftindex^{B}{\mathbf{p}}, translation vector d, and the angle of rotation \theta, find the coordinates of the vector \leftindex^{A}{\mathbf{p}}.

Solution:

(2)   \begin{align*}  \leftindex^{A}{\mathbf{p}} =   \leftindex^{A}{\mathbf{d}} +   \leftindex^{A}_{B}{R}   \cdot \leftindex^{B}{\mathbf{p}} \end{align*}

where \leftindex^{A}_{B}{R} is the rotation matrix that transforms vectors from B to A coordinate systems.

That is

(3)   \begin{align*}\begin{bmatrix}  p_{x_{A}}  \\  p_{y_{A}}  \end{bmatrix} =   \begin{bmatrix}    d_{x_{A}}  \\  d_{y_{A}} \end{bmatrix} + \begin{bmatrix} cos(\theta) & -sin(\theta)  \\ sin(\theta)  & cos(\theta)    \end{bmatrix}  \begin{bmatrix}   p_{x_{B}}  \\  p_{y_{B}}    \end{bmatrix} \end{align*}

If you do not remember how the rotation matrix \leftindex^{A}_{B}{R}

(4)   \begin{align*} \leftindex^{A}_{B}{R} = \begin{bmatrix} cos(\theta) & -sin(\theta)  \\ sin(\theta)  & cos(\theta)    \end{bmatrix}\end{align*}

is constructed, see our previous post.

By multiplying vectors and matrices, and by adding the results, from (3), we have

(5)   \begin{align*} p_{x_{A}} = d_{x_{A}}  +  cos(\theta)   p_{x_{B}} - sin(\theta) p_{y_{B}}  \\  p_{y_{A}} = d_{y_{A}}  +  sin(\theta)   p_{x_{B}} + cos (\theta) p_{y_{B}}  \end{align*}

The tranformation (3), can be written as a single vector matrix multiplications. Namely, we can formally write

(6)   \begin{align*}\begin{bmatrix}  \leftindex^{A}{\mathbf{p}}  \\ 1   \end{bmatrix} =  \begin{bmatrix}   \leftindex^{A}_{B}{R} & \leftindex^{A}{\mathbf{d}} \\ \mathbf{0}_{2\times 1}  & 1 \end{bmatrix}       \begin{bmatrix}   \leftindex^{B}{\mathbf{p}}   \\ 1 \end{bmatrix} \end{align*}

where \mathbf{0}_{2\times 1} is 2 times 1 matrix of zeros. By expanding the last equation, we obtain

(7)   \begin{align*}\begin{bmatrix}    p_{x_{A}}  \\  p_{y_{A}}    \\ 1   \end{bmatrix} =  \begin{bmatrix}    cos(\theta) & -sin(\theta)    & d_{x_{A}} \\  sin(\theta) & cos(\theta) &  d_{y_{A}} \\ 0 & 0 & 1  \end{bmatrix}       \begin{bmatrix}   p_{x_{B}} \\  p_{y_{B}}   \\ 1 \end{bmatrix} \end{align*}

The matrix

(8)   \begin{align*}  \begin{bmatrix}    cos(\theta) & -sin(\theta)    & d_{x_{A}} \\  sin(\theta) & cos(\theta) &  d_{y_{A}} \\ 0 & 0 & 1  \end{bmatrix}  \end{align*}

is called a homogeneous transform.

In the next post, we we will see how to use this transform to solve the forward kinematics problem of robotic manipulators.