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---
title: Representations of 3D Motion
categories: lectures 3D mathematics
---
# Representation of Rotations
Consider what happens when an object rotates continuously over time,
i.e. the rotational matrix is a function $R(t)$ of time.
## The derivative
1. Rotation is represented by an orthogonal matrix $R$
2. Consider rotation over time, i.e. the rotational matrix
is a function $R(t)$ of time.
$$R(t)\cdot R^T(t)=I$$
3. Implicit derivation
2. Implicit derivation
$$\dot R(t)\cdot R^T(t)+R(t)\cdot\dot R^T(t)=I$$
4. by transposing the product and moving one term across, we have
$$\dot R(t)\cdot R^T(t) = -(\dot R(t)\cdot R^T(t)^T$$
5. This is a skew-symmetric matrix, hence
3. by transposing the product and moving one term across, we have
$$\dot R(t)\cdot R^T(t) = -(\dot R(t)\cdot R^T(t))^T$$
4. This is a skew-symmetric matrix, hence
$$\exists \vec{\omega}\in\mathbb{R}^3, \text{s.t.}
\dot R^T(t)\cdot R^T(t) = \hat\omega(t)$$
5. Multiply by $R(t)$ to get
$$\dot R^T(t) = \hat\omega(t)\cdot R(t)$$
6. If $R(t_0)=I$ as an initial condition, then $\dot R(t)=\hat\omega(t)$
Note $so(3)$ is the space of all skew-symmetric matrices.
## The differential equation
Let $x(t)$ be a point rotated over time.
Assume that $\omega$ is constant.
1. **ODE:**
$$\dot x(t) = \hat\omega x(t), \quad x(t)\in\mathbb{R}^3$$
2. Solution:
$$x(t) = e^{\hat\omega t} x(0)$$
3. where
$$e^{\hat\omega t} = I + \sum_{i=1}^\infty \frac{(\hat\omega )^i}{i!}$$
4. The rotational matrix $$R(t)=e^{\hat\omega t}$$
signifies a rotation around the axis $\omega$ by $t$ radians.
$$\exp : \mathrm{so}(3)\to\mathrm{SO}(3); \hat\omega\mapsto e^{\hat\omega}$$
For any $R$, such an $\hat\omega$ can be found,
not necessarily unique.
Rotation is obviously periodic. A rotation by $2\pi$ is back to start.
# 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.
$$
R\cot\vec{x}=
R
\cdot
\begin{bmatrix} x\\y\\z \end{bmatrix}
=
\begin{bmatrix} x'\\y'\\z' \end{bmatrix}
$$
$$
\begin{bmatrix}
R & 0 \\
0 & 1
\end{bmatrix}
\cdot
\begin{bmatrix} x\\y\\z\\1 \end{bmatrix}
=
\begin{bmatrix} x'\\y'\\z'\\1 \end{bmatrix}
$$
## 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!