ME3120 · Dynamic Modelling

Theme 4 · Robot geometry

Frames & the rotation matrix

To write the energy of a real arm we first need to say exactly where each link is. It starts with one tool: a matrix that turns a vector by an angle.

Source: course notes, Week 3 (robot geometry); Spong, Lynch & Park.

Before you start

What you need first

Coordinates as a vector $\begin{bmatrix}x\\y\end{bmatrix}$, and the joint angle $\theta$ as a generalized coordinate.
$\sin,\cos$ and the angle-addition rules.

What you'll be able to do

Describe a point as a position vector and tell frames apart.
Build the rotation matrix $R(\theta)$ and use it to turn a vector.
Know its two handy properties.

Why geometry now?

The energy method (Theme 3) needs the position of every mass (for $V=mgh$) and its velocity (for $T=\tfrac12 mv^2$). For a robot, both depend on the joint angles $q$.

$$\text{joint angles } q \;\longrightarrow\; \boxed{\text{forward kinematics}} \;\longrightarrow\; \text{where each link \& tip is}$$

Forward kinematics = given the joint angles, compute where every part of the robot is. This theme builds it; today we start with rotation.

A point is a position vector

A point in the plane is two numbers — its coordinates — written as a column vector from the origin:

$$\mathbf{p}=\begin{bmatrix}x\\ y\end{bmatrix}$$

We treat every point as a position vector: an arrow from the origin out to the point. Rotating or shifting the point means acting on this vector.

A point $\mathbf{p}$ as a position vector from the origin.

Coordinate frames

A link frame $(x_1,y_1)$ rotated by $\theta$ from the fixed ground frame $(x_0,y_0)$.

A frame is a set of axes we measure from. A robot uses many: one fixed to the ground, and one riding on each link.

Ground frame — fixed to the base; the "world" we report answers in.
Link frame — glued to a moving link; simple for a point on that link.

The whole job: a point is easy to write in its link frame — we want it in the ground frame. For a revolute joint, the turn angle $\theta$ is the joint variable $q$.

The core move

Rotating a point by an angle $\theta$

Take a point $\mathbf{p}=(x,y)$ and spin it about the origin by $\theta$. It moves to $\mathbf{p}'=(x',y')$ at the same distance $r$, just turned. We want a formula for $(x',y')$.

Spin $\mathbf{p}$ by $\theta$ about the origin to get $\mathbf{p}'$ — same length $r$, new direction.

Derivation

Build the rotation formula

1Write $\mathbf{p}$ in polar form — distance $r$, angle $\phi$:

$$x=r\cos\phi, \qquad y=r\sin\phi$$

2Rotating by $\theta$ keeps $r$ and adds $\theta$ to the angle:

$$x'=r\cos(\phi+\theta), \qquad y'=r\sin(\phi+\theta)$$

3Expand with the angle-addition rules:

$$x'=r\cos\phi\cos\theta-r\sin\phi\sin\theta, \quad y'=r\sin\phi\cos\theta+r\cos\phi\sin\theta$$

4Substitute $r\cos\phi=x$ and $r\sin\phi=y$:

$$x'=x\cos\theta-y\sin\theta, \qquad y'=x\sin\theta+y\cos\theta$$

The rotation matrix $R(\theta)$

Write those two lines as one matrix multiply:

$$\begin{bmatrix}x'\\ y'\end{bmatrix}=\underbrace{\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}}_{R(\theta)}\begin{bmatrix}x\\ y\end{bmatrix}$$

where:
Symbol	Meaning	SI unit
$R(\theta)$	the rotation matrix (turns a vector by $\theta$)	—
$\theta$	rotation angle (positive = counter-clockwise)	rad
$(x,y)$	original point	m
$(x',y')$	rotated point	m

Multiplying any vector by $R(\theta)$ turns it by $\theta$. This one matrix is the workhorse of all robot geometry.

📐 Worked example

Rotate the point $(3,1)$ by $30^\circ$

Find the new coordinates after turning $(3,1)$ about the origin by $\theta=30^\circ$ ($\cos 30^\circ=0.866,\ \sin 30^\circ=0.5$).

1Build $R(30^\circ)$:

$$R(30^\circ)=\begin{bmatrix}0.866 & -0.5\\ 0.5 & 0.866\end{bmatrix}$$

2Multiply by $\begin{bmatrix}3\\1\end{bmatrix}$:

$$\begin{bmatrix}x'\\ y'\end{bmatrix}=\begin{bmatrix}0.866(3)-0.5(1)\\ 0.5(3)+0.866(1)\end{bmatrix}=\begin{bmatrix}2.10\\ 2.37\end{bmatrix}$$

The point swung counter-clockwise to $(2.10,\,2.37)$ — still the same distance $\sqrt{3^2+1^2}=\sqrt{10}$ from the origin (check: $\sqrt{2.10^2+2.37^2}\approx\sqrt{10}$ ✓).

Two handy facts about $R(\theta)$

Its columns are the new axes

Column 1 is where the old $x$-axis points after turning; column 2 is the new $y$-axis.

Undo with the transpose

$R(-\theta)=R(\theta)^{\mathsf T}$ — rotating back is just swapping rows and columns.

These keep the algebra clean. For this course you mainly need to apply $R(\theta)$ — but knowing the transpose undoes it is handy.

✏️ Try it yourself

Rotate the point $(4,0)$ by $\theta=60^\circ$ about the origin ($\cos 60^\circ=0.5,\ \sin 60^\circ=0.866$).

Step 1. $R(60^\circ)=\begin{bmatrix}0.5 & -0.866\\ 0.866 & 0.5\end{bmatrix}$ Step 2. $\begin{bmatrix}x'\\ y'\end{bmatrix}=\begin{bmatrix}0.5(4)-0.866(0)\\ 0.866(4)+0.5(0)\end{bmatrix}=\begin{bmatrix}2.00\\ 3.46\end{bmatrix}$ Check. Distance stays $4$: $\sqrt{2^2+3.46^2}=\sqrt{4+12}=4$ ✓

Common mistakes to avoid

Mistake	Fix
Sign slip in $R(\theta)$	The top-right entry is $-\sin\theta$; bottom-left is $+\sin\theta$.
Rotating with degrees in the matrix	Put numeric $\cos,\sin$ values in — they don't care about deg/rad, but your calculator must be in the right mode.
Swapping rows and columns of the point	The point is a column $\begin{bmatrix}x\\y\end{bmatrix}$; $R$ multiplies it from the left.
Expecting the length to change	A pure rotation keeps $r$ the same — a good sanity check.

Recap — the whole topic on one screen

$$\begin{bmatrix}x'\\ y'\end{bmatrix}=R(\theta)\begin{bmatrix}x\\ y\end{bmatrix}, \qquad R(\theta)=\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}$$

Idea	What you own now
Point as a vector	$\mathbf{p}=\begin{bmatrix}x\\y\end{bmatrix}$ from the origin
Frames	Ground (fixed) vs link (rides the link)
Rotation matrix	$R(\theta)$ turns any vector by $\theta$
Two facts	Columns = new axes; $R(-\theta)=R^{\mathsf T}$

Next topic

Homogeneous transforms & chaining

Rotation alone is not enough — a link is also shifted from the base. Next we fold rotation and shift into one matrix, the homogeneous transform $T$, and chain them down an arm — which is forward kinematics.

→ Homogeneous transforms & chaining