---
title: Homogeneous Coordinates
categories: lectures 3D mathematics
geometry: margin=2cm
fontsize: 12pt
---


# Homogenous Co-ordinates

## Six degrees of Freedom

+ Translation - add $T=[y_1,y_2,y_3]$
+ Rotation - multiply by $\exp(\hat{[\omega_1,\omega_2,\omega_3]})$
+ $x\mapsto xR+T$ is affine, not linear

## Points in Homogenous Co-ordinates

+ Point $\textbf{X}=[X_1,X_2,X_3]^\mathrm{T}\in\mathbb{R}^3$
+ Embed in $\mathbb{R}^4$ as
    $\mathbf{\tilde X}=[X_1,X_2,X_3,1]^\mathrm{T}\in\mathbb{R}^4$
+ Vector $\vec{pq}$ is represented as
    $$\mathbf{\tilde X}(q)-\mathbf{\tilde X}(p) =
    \begin{bmatrix} \mathbf{ X}(q) \\ 1 \end{bmatrix}
    - \begin{bmatrix} \mathbf{ X}(p) \\ 1 \end{bmatrix}
    =
    \begin{bmatrix} \mathbf{X}(q) - \mathbf{X}(p) \\ 0 \end{bmatrix}$$
+ In homogenous co-ordinates,
    - points have 1 in last position
    - vectors have 0 in last position
+ Arithmetics
    + Point + Point is undefined
    + Vector + Vector is a Vector
    + Point + Vector is a Point

## Rotation 

Let $R$ be a $3\times3$ rotation matrix.

$$
   R\cdot\vec{x}=
   R
\cdot
\begin{bmatrix} x\\y\\z \end{bmatrix}
=
\begin{bmatrix} x'\\y'\\z' \end{bmatrix}
$$

$$
\begin{bmatrix}
   R & 0 \\
   0 & 1
\end{bmatrix}
\cdot
\begin{bmatrix} x\\y\\z\\1 \end{bmatrix}
=
\begin{bmatrix} x'\\y'\\z'\\1 \end{bmatrix}
$$

## Arbitrary motion

What happens if we change some of the zeroes?

$$
\begin{bmatrix}
   R & \vec{t} \\
   0 & 1
\end{bmatrix}
\cdot
\begin{bmatrix} x\\y\\z\\1 \end{bmatrix}
=
\begin{bmatrix} x'\\y'\\z'\\0 \end{bmatrix}
+
\begin{bmatrix} \vec{t}\\1 \end{bmatrix}
=R\vec{x}+\vec{t}
$$

We have rotated and translated!

## Motion as a function of time

$$
g =
\begin{bmatrix}
   R & T\\
   0 & 1
\end{bmatrix}
\in \mathrm{SE}(3)
$$

+ This is a group 

$$g_1\cdot g_2 =
\begin{bmatrix} R_1 & T_1\\ 0 & 1 \end{bmatrix}\cdot
\begin{bmatrix} R_2 & T_2\\ 0 & 1 \end{bmatrix}
=
\begin{bmatrix} R_1R_2 & R_1T_2+T_1\\ 0 & 1 \end{bmatrix}
\in \mathrm{SE}(3)
$$
    
$$g^{-1} =
\begin{bmatrix} R & T\\ 0 & 1 \end{bmatrix}^{-1}
=
\begin{bmatrix} R^T & -R^TT\\ 0 & 1 \end{bmatrix}
$$


# Canonical Exponential Co-ordinates


1.  A homogeneous motion matrix
    $$ g(t) = \begin{bmatrix} R(t) & T(t)\\ 0 & 1 \end{bmatrix} $$
2.  We can differentiate, invert, and multiply to get
    $$
    \dot g(t)\cdot g^{-1}(t) =
    \begin{bmatrix}
    \dot R(t) R^T(t) & \dot T(t)- \dot R(t)R^T(t)T(t) \\
    0 & 0 \end{bmatrix}
    \in\mathbb{R}^{4\times4}
    $$
3.  $\dot R(t) R^T(t)$ is skew-symmetric, hence
    $$
    \dot g(t)\cdot g^{-1}(t) =
    \begin{bmatrix}
    \hat\omega & \dot T(t)- \dot R(t)R^T(t)T(t) \\
    0 & 0 \end{bmatrix}
    $$
    for some $\omega\in\mathrm{so}(3)$
3.  Write 
    $$
    v(t)= \dot T(t)- \hat\omega(t)T(t) \\
    \dot g(t)\cdot g^{-1}(t) =
    \begin{bmatrix}
    \hat\omega & \dot v(t) \\
    0 & 0 \end{bmatrix}
    $$
4.  Call this matrix $\hat\xi(t)$
5.  This give the differential equation
    $$\dot g(t) = (\dot g(t) g^{-1})g(t)$$

+ $v(t)$ is linear velocity
+ $\omega(t)$ as angular velocity