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---
title: Representations of 3D Motion
categories: lectures 3D mathematics
geometry: margin=2cm
fontsize: 12pt
---
# 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 $R$
$$R(t)\cdot R^T(t)=I$$
2. Implicit derivation
$$\dot R(t)\cdot R^T(t)+R(t)\cdot\dot R^T(t)=I$$
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.
$$\exp : \mathrm{so}(3)\to\mathrm{SO}(3); \hat\omega\mapsto e^{\hat\omega}$$
This is a map from a Lie algebra to a Lie group.
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.
**Note** Only three degrees of freedom.
## Logarithm
$$
R =
\begin{bmatrix}
r_{11} & r_{12} & r_{13} \\
r_{21} & r_{22} & r_{23} \\
r_{31} & r_{32} & r_{33}
\end{bmatrix}
= \exp(\hat\omega)
$$
where
$$
\DefineMathOperator{\tr}{trace}
||\omega|| = \cos^{-1}\big(\frac{\tr(R)-1}2\big)
$$
and
$$
\frac{\omega}{||\omega||} = \frac1{2\sin(||\omega||)}
\begin{bmatrix} r_{32}-r_{23}\\ r_{13}-r_{31}\\ r_{21}-r_{12} \end{bmatrix}
$$
# 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!
## 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
# Velocity
**Adjoint Map**
$\mathrm{ad}_g: \mathrm{se}(3)\to\mathrm{se}(3);\quad \hat\xi\mapsto g\hat\xi g^{-1}$