Denavit Hartenberg Analysis, Part 2: Homogeneous Matrices
Part 1/Part 2/Part 3/Part 4/Part 5/Part 6
The Problem
In Part 1 of this series, in which I attempt to explain the Denavit-Hartenberg convention for inverse kinematics I talked about coordinate frames and rotational transformations using matrices. This left open a pretty obvious question, though: turning coordinate frames is great and all, but how do we move them? If we want to describe the transformation from Frame 1 to Frame 0 here:
Then we need to express the rotation by the angle \(\theta\) (which we figured out last time) and the translation by the distance \(d\) (which we don’t know how to do yet).
To make this a little more clear, let’s look at an example of a translation by a vector \(\vec{t}\):
\[
\begin{align}
\vec{v}' &= \vec{v} + \vec{t} \\
\begin{bmatrix}
v_1' \\
v_2' \\
v_3'
\end{bmatrix} &=
\begin{bmatrix}
v_1 \\
v_2 \\
v_3
\end{bmatrix} + \begin{bmatrix}
t_1 \\
t_2 \\
t_3
\end{bmatrix}
\end{align}
\]
Given how we handled rotation matrices last time, we’d really like to find a matrix such that we can write down that translation as a matrix multiplication:
\[
\vec{v}' = \left[ T \right] \, \vec{v}
\]
For some magical unknown matrix \(\left[ T \right]\). But…we can’t. You can try. In fact, go for it! It’s a useful exercise, but it’s going to be futile, and we can show that pretty easily. As an example, let’s see what happens if we set \(\vec{v}\) to be all zeros:
\[
\vec{v}' = \left[ T \right] \, \begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}
\]
No matter what we set \(\left[ T \right]\) to be, \(\vec{v}'\) will always just be zeros. So clearly this method won’t work.
The Solution: Homogeneous Coordinates
It turns out that there’s a pretty simple fix for this, and it involves adding an extra term to each of our vectors. If we add a 1 to the end of each of our vectors \(\vec{v}\), then they end up looking like this:
\[
\vec{v} = \begin{bmatrix}
v_1 \\
v_2 \\
v_3 \\
1
\end{bmatrix}
\]
As it happens, we can now write out our translation matrix \(\left[ T \right]\) using this coordinate system:
\[
\begin{align}
\vec{v}' &= \vec{v} + \vec{t} \\
\begin{bmatrix}
v_1' \\
v_2' \\
v_3' \\
1
\end{bmatrix} &= \begin{bmatrix}
1 & 0 & 0 & t_1 \\
0 & 1 & 0 & t_2 \\
0 & 0 & 1 & t_3 \\
0 & 0 & 0 & 1
\end{bmatrix}\,
\begin{bmatrix}
v_1 \\
v_2 \\
v_3 \\
1
\end{bmatrix}
\end{align}
\]
If we write out the matrix multiplication, then we get the following equations:
\[
\begin{align}
v_1' &= v_1 + t_1 \\
v_2' &= v_2 + t_2 \\
v_3' &= v_3 + t_3
\end{align}
\]
which is exactly what we wanted! Rotations also work just fine in this coordinate system:
\[
\vec{v}' = \begin{bmatrix}
\cos{\theta_{1 1}} & \cos{\theta_{1 2}} & \cos{\theta_{1 3}} & 0 \\
\cos{\theta_{2 1}} & \cos{\theta_{2 2}} & \cos{\theta_{2 3}} & 0 \\
\cos{\theta_{3 1}} & \cos{\theta_{3 2}} & \cos{\theta_{3 3}} & 0 \\
0 & 0 & 0 & 1
\end{bmatrix} \, \begin{bmatrix}
v_1 \\
v_2 \\
v_3 \\
1
\end{bmatrix}
\]
The real kicker is that we can now combine translation and rotation just by multiplying our matrices together. For example, performing a rotation and then a translation might look like this:
\[
\vec{v}' = \begin{bmatrix}
1 & 0 & 0 & t_1 \\
0 & 1 & 0 & t_2 \\
0 & 0 & 1 & t_3 \\
0 & 0 & 0 & 1
\end{bmatrix} \,
\begin{bmatrix}
\cos{\theta_{1 1}} & \cos{\theta_{1 2}} & \cos{\theta_{1 3}} & 0 \\
\cos{\theta_{2 1}} & \cos{\theta_{2 2}} & \cos{\theta_{2 3}} & 0 \\
\cos{\theta_{3 1}} & \cos{\theta_{3 2}} & \cos{\theta_{3 3}} & 0 \\
0 & 0 & 0 & 1
\end{bmatrix} \,
\begin{bmatrix}
v_1 \\
v_2 \\
v_3 \\
1
\end{bmatrix}
\]
We can use the associative property of matrix multiplication to simplify this:
\[
\begin{align}
\vec{v}' &= \left(
\begin{bmatrix}
1 & 0 & 0 & t_1 \\
0 & 1 & 0 & t_2 \\
0 & 0 & 1 & t_3 \\
0 & 0 & 0 & 1
\end{bmatrix} \,
\begin{bmatrix}
\cos{\theta_{1 1}} & \cos{\theta_{1 2}} & \cos{\theta_{1 3}} & 0 \\
\cos{\theta_{2 1}} & \cos{\theta_{2 2}} & \cos{\theta_{2 3}} & 0 \\
\cos{\theta_{3 1}} & \cos{\theta_{3 2}} & \cos{\theta_{3 3}} & 0 \\
0 & 0 & 0 & 1
\end{bmatrix} \right) \,
\begin{bmatrix}
v_1 \\
v_2 \\
v_3 \\
1
\end{bmatrix} \\
&= \begin{bmatrix}
\cos{\theta_{1 1}} & \cos{\theta_{1 2}} & \cos{\theta_{1 3}} & t_1 \\
\cos{\theta_{2 1}} & \cos{\theta_{2 2}} & \cos{\theta_{2 3}} & t_2 \\
\cos{\theta_{3 1}} & \cos{\theta_{3 2}} & \cos{\theta_{3 3}} & t_3 \\
0 & 0 & 0 & 1
\end{bmatrix} \,
\begin{bmatrix}
v_1 \\
v_2 \\
v_3 \\
1
\end{bmatrix}
\end{align}
\]
Notice how the rotation and translation components of this resulting transformation stay in their own sections of the matrix. We’ll use that nice property to devise a shorthand for this kind of transformation matrix. Let:
\[
\textbf{H} \equiv \begin{bmatrix}
R & T \\
0 & 1
\end{bmatrix} \equiv \begin{bmatrix}
\cos{\theta_{1 1}} & \cos{\theta_{1 2}} & \cos{\theta_{1 3}} & t_1 \\
\cos{\theta_{2 1}} & \cos{\theta_{2 2}} & \cos{\theta_{2 3}} & t_2 \\
\cos{\theta_{3 1}} & \cos{\theta_{3 2}} & \cos{\theta_{3 3}} & t_3 \\
0 & 0 & 0 & 1
\end{bmatrix}
\]
Where R represents the rotation component and T represents the translation component.
Next time: putting it all together
In the next installment, we’ll use all of this coordinate transformation business to build a system that will let us figure out how to control our robot’s arm.
Robin Deits 09 June 2012