ME3120 · Dynamic Modelling

Theme 4 · Robot geometry

Homogeneous transforms & chaining

Rotation alone is not enough — a link is also shifted from the base. One clever matrix does rotation and shift at once, and chaining them is forward kinematics.

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

Before you start

What you need first

The rotation matrix $R(\theta)$ and how to apply it (Topic 16).
Matrix × vector multiplication.

What you'll be able to do

Combine rotation and shift into one operation.
Build and apply the homogeneous transform $T$.
Chain transforms down an arm.

Rotation and shift together

A link frame is turned by $\theta$ and shifted from the ground origin by a vector $\mathbf{d}$. A point $\mathbf{p}_{\text{link}}$ known in the link frame becomes, in the ground frame:

$$\mathbf{p}_{\text{ground}} = \mathbf{d} + R(\theta)\,\mathbf{p}_{\text{link}}$$

First rotate the point into ground directions with $R(\theta)$, then add the shift $\mathbf{d}$.

The link frame is shifted by $\mathbf{d}$ and rotated by $\theta$; the point $\mathbf{p}$ rides on it.

An annoyance worth fixing

That mapping mixes two different operations:

Rotation

= multiply by a matrix $R(\theta)$.

Translation

= add a vector $\mathbf{d}$.

When we chain many links, mixing "multiply" and "add" gets messy and error-prone. We want one single operation that does both.

The trick: add a 1

Write each point with an extra entry equal to $1$ — these are homogeneous coordinates:

$$\begin{bmatrix}x\\ y\end{bmatrix}\ \longrightarrow\ \begin{bmatrix}x\\ y\\ 1\end{bmatrix}$$

That harmless extra $1$ is what lets a single matrix do rotation and translation at once.

The homogeneous transform $T$

$$T=\begin{bmatrix} \cos\theta & -\sin\theta & d_x\\ \sin\theta & \cos\theta & d_y\\ 0 & 0 & 1 \end{bmatrix}$$

where:
Block	Meaning
top-left $2\times2$	the rotation $R(\theta)$
right column $(d_x,d_y)$	the shift $\mathbf{d}$
bottom row $[0\ 0\ 1]$	bookkeeping (keeps the extra $1$)

Rotation in the corner, shift up the side — one tidy package.

One matrix does both

Multiply the homogeneous point by $T$:

$$\begin{bmatrix}x'\\ y'\\ 1\end{bmatrix} = T\,\begin{bmatrix}x\\ y\\ 1\end{bmatrix}$$

Out comes $x' = \cos\theta\,x - \sin\theta\,y + d_x$ and $y' = \sin\theta\,x + \cos\theta\,y + d_y$: rotation and shift, in one multiply. The bottom row just carries the $1$ along.

📐 Worked example

Apply a transform to a point

A link frame is rotated $\theta = 90^\circ$ and shifted by $\mathbf{d}=(3,1)$. A point sits at $(2,0)$ in the link frame. Find it in the ground frame.

1Build $T$ (with $\cos 90^\circ=0,\ \sin 90^\circ=1$):

$$T=\begin{bmatrix}0 & -1 & 3\\ 1 & 0 & 1\\ 0 & 0 & 1\end{bmatrix}$$

2Multiply by the homogeneous point $(2,0,1)$:

$$\begin{bmatrix}x'\\ y'\\ 1\end{bmatrix}=\begin{bmatrix}0(2)-1(0)+3\\ 1(2)+0(0)+1\\ 1\end{bmatrix}=\begin{bmatrix}3\\ 3\\ 1\end{bmatrix}$$

The point lands at $(3,3)$. Check the two steps by hand: rotating $(2,0)$ by $90^\circ$ gives $(0,2)$; adding the shift $(3,1)$ gives $(3,3)$ ✓ — exactly what the one matrix produced.

Down the arm

Chaining transforms

If $T_0^1$ places link 1 in the ground frame, and $T_1^2$ places link 2 in link 1's frame, then link 2 in the ground frame is just the product:

$$T_0^2 = T_0^1 \; T_1^2$$

Walk from the base outward, multiplying one transform per joint. That product chain is forward kinematics.

Two joints, two transforms: $T_0^2 = T_0^1 T_1^2$ carries the base frame out to the tip.

Why chaining is the whole game

🔭 Looking ahead: a 6-joint industrial arm is just six transforms multiplied in a row, $T_0^6 = T_0^1 T_1^2 \cdots T_5^6$. The method does not change — only the number of links.

For 2–3 link arms we can skip the full matrices and read positions straight off the geometry — which is exactly what we do next, for the 2-link arm.

✏️ Try it yourself

A link frame is rotated $\theta=90^\circ$ and shifted by $\mathbf{d}=(1,2)$. A point sits at $(0,3)$ in the link frame. Use $T$ to find it in the ground frame.

Step 1. $T=\begin{bmatrix}0&-1&1\\ 1&0&2\\ 0&0&1\end{bmatrix}$ Step 2. $\begin{bmatrix}x'\\ y'\\ 1\end{bmatrix}=\begin{bmatrix}0(0)-1(3)+1\\ 1(0)+0(3)+2\\ 1\end{bmatrix}=\begin{bmatrix}-2\\ 2\\ 1\end{bmatrix}$ Check. Rotate $(0,3)$ by $90^\circ$ → $(-3,0)$; add $(1,2)$ → $(-2,2)$ ✓

Common mistakes to avoid

Mistake	Fix
Forgetting the shift $\mathbf{d}$	Rotation alone is not enough — the right column of $T$ carries the translation.
Dropping the extra $1$	The point must be $(x,y,1)$; without it the shift cannot enter.
Multiplying transforms in the wrong order	Chain base-outward: $T_0^2=T_0^1 T_1^2$, not $T_1^2 T_0^1$ (matrix order matters).
Putting $\mathbf{d}$ in the bottom row	The bottom row is always $[0\ 0\ 1]$; the shift goes in the right column.

Recap — the whole topic on one screen

$$T=\begin{bmatrix} \cos\theta & -\sin\theta & d_x\\ \sin\theta & \cos\theta & d_y\\ 0 & 0 & 1 \end{bmatrix}, \qquad T_0^2 = T_0^1\,T_1^2$$

Idea	What you own now
Rotation + shift	$\mathbf{p}_{\text{ground}}=\mathbf{d}+R(\theta)\mathbf{p}_{\text{link}}$
The trick	Add a $1$ → homogeneous coordinates
The transform	$T$ does rotation and shift in one multiply
Chaining	Multiply one transform per joint — that is forward kinematics

Next topic

Forward kinematics of the 2-link arm

Time for the centrepiece. We chain the geometry down a 2-link planar arm and read off the elbow, the tip, and each link's centre of mass — the positions next theme's energy needs.

→ Forward kinematics of the 2-link arm

Block	Meaning
top-left \(2\times2\)	the rotation \(R(\theta)\)
right column \((d_x,d_y)\)	the shift \(\mathbf{d}\)
bottom row \([0\ 0\ 1]\)	bookkeeping (keeps the extra \(1\))

Mistake	Fix
Forgetting the shift \(\mathbf{d}\)	Rotation alone is not enough — the right column of \(T\) carries the translation.
Dropping the extra \(1\)	The point must be \((x,y,1)\); without it the shift cannot enter.
Multiplying transforms in the wrong order	Chain base-outward: \(T_0^2=T_0^1 T_1^2\), not \(T_1^2 T_0^1\) (matrix order matters).
Putting \(\mathbf{d}\) in the bottom row	The bottom row is always \([0\ 0\ 1]\); the shift goes in the right column.

Idea	What you own now
Rotation + shift	\(\mathbf{p}_{\text{ground}}=\mathbf{d}+R(\theta)\mathbf{p}_{\text{link}}\)
The trick	Add a \(1\) → homogeneous coordinates
The transform	\(T\) does rotation and shift in one multiply
Chaining	Multiply one transform per joint — that is forward kinematics

Homogeneous transforms & chaining

Before you start

What you need first

What you'll be able to do

Rotation and shift together

An annoyance worth fixing

Rotation

Translation

The trick: add a 1

The homogeneous transform \(T\)

One matrix does both

Apply a transform to a point

Chaining transforms

Why chaining is the whole game

Common mistakes to avoid

Recap — the whole topic on one screen

Forward kinematics of the 2-link arm