Representations of Rotation
A rotation, can be respresented as $so(3)$ (Lie Algebra of Special Orthogonal Group), or $SO(3)$, (Special Orthogonal Group) and rotation vector.
Representation 1 A rotation vector is $s = \theta [s_x, s_y, s_z] = [\omega_x, \omega_y, \omega_z]$, where:
\[\begin{gather*} \theta = \sqrt{\omega_x^2 + \omega_y^2 + \omega_z^2} \end{gather*}\]$[s_x, s_y, s_z]$ here is the axis of rotation, which is a unit vector.
Representation 2 Then we can write this rotation vector in the form of $so(3)$. It’s also called “skew symmetric matrix” of a rotation axis (notice how the matrix diagonal serve as the axis of symmetry and sign?)
\[\begin{gather*} \hat{\omega} = \begin{pmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \end{pmatrix} \end{gather*}\]Representation 3 SO(3)
Order is important
Different orders of Rotation about x, y, z axis could yield different final orientations. Try it yourself: first, rotate a random object you see now by z axis by +90 degrees followed by rotating about x axis by +90 degrees. Then, try rotating about x axis followed by about the z axis. A side note is that in robotics, people follow the right hand convention for frames. That is: x forward, y left, z up
So, it’s important to specify the order or rotation, if a single rotation can be decomposed into rotations about x,y,z axes.. Also, there are rotations about fixed axes, and Euler angles. Fixed axes refer to the axes of a fixed world frame, and euler angles refer to the X,Y,Z in the body frame. In either case, x,y,z are also known as “roll-pitch-yaw”.
In the robotics community:
-
Use world frame fixed axes. ROS uses the X-Y-Z order, so there is no ambiguity on order. There is no gimbal lock, either.
-
Euler angles can represent any orientation. There are common orders such as Z-X-Y, X-Y-Z, etc. There are 24 valid combinations. Below (from Wikipedia) is an illustration of Z-X-Z’
Multiple Rotations Leads To One Rotation
The final rotation of 3 rotations about fixed axes is \(\begin{gather*} R = R_z(\theta_z) R_y(\theta_y) R_x(\theta_x) \end{gather*}\)
Then we can get $\theta$ and rotation axis $u=[u_x, u_y, u_z]$
\[\begin{gather*} \theta = \cos^{-1} \left( \frac{\text{trace}(R) - 1}{2} \right) \\ \mathbf{u} = \frac{1}{2 \sin \theta} \begin{pmatrix} R_{32} - R_{23} \\ R_{13} - R_{31} \\ R_{21} - R_{12} \end{pmatrix} \end{gather*}\]Gimbal Lock, Singularities, and Quaternion
When using euler angles, certain axes in the body frame could align to each other. E.g., when a plane has a pitch of 90 degrees (as below), its z and x axes are aligned. Then, rotation about z and rotation about x are the same. From this configuration, the plane cannot rotate about the axis that are perpendicular to x,y,z axes, hence it loses 1 degree of freedom.
Mathematically, for a Z-X-Y system,
\[\begin{gather*} \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \gamma & -\sin \gamma \\ 0 & \sin \gamma & \cos \gamma \end{bmatrix} * \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} * \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{gather*}\]
See? The rotation about both the Z axis $\gamma$ and the X axis $\alpha$ will effectively create a combined rotation about the X axis, $\theta$. So, such rotations do not have unique angular values.
Implementations
- OpenCV: OpenCV provides rotation vector -> single rotation matrix. See here
Instantaneous Rotation
According to the Poisson Formula, $R’ = Rw^{\land}$, for a small time period $\Delta t$, the ODE can be solved:
\[\begin{gather*} R(t) = R(t_0)exp(w^{\land}(t - t_0)) = R(t_0) exp(w^{\land} \Delta t) \end{gather*}\]Rotation and Skew Matrices in 2D
Basic Stuff
Taylor Expansion
\[\begin{gather*} \begin{aligned} & Exp[\theta^{\land}] = I + \theta^{\land} + \frac{1}{2!} (\theta^{\land}) ^ 2 + \frac{1}{3!} (\theta^{\land}) ^ 3 \end{aligned} \end{gather*}\]Poisson Equation
\[\begin{gather*} \begin{aligned} & R^T R = I \Rightarrow \frac{d R^T R}{dt} = 0 \\ & = (R^T)' R + R' R^T = 0 \end{aligned} \end{gather*}\]If we define
\[\begin{gather*} \begin{aligned} & w^{\land} = R^T R' \end{aligned} \end{gather*}\]We get
\[\begin{gather*} \begin{aligned} & R' = R w^{\land} \end{aligned} \end{gather*}\]Instantaneous angular velocity is
\[\begin{gather*} \begin{aligned} & w = \vec{n} \theta \end{aligned} \end{gather*}\]Actually, we still need to establish the relationship between $w$ (instantaneous angular velocity) and rotation vector $\phi$:
\[\begin{gather*} \begin{aligned} & Rw^{\land} = R' = lim_{\Delta t \rightarrow 0}\frac{R(t + \Delta t) - R}{\Delta t} \\ & \approx lim_{\Delta t \rightarrow 0} \frac{R(t) Exp[J_r \Delta \phi] - R(t)}{\Delta t} \\ & = R(J_r \phi')^{\land} \end{aligned} \end{gather*}\]The above shows when an instantaneous rotation happens, its rotation vector change is not the instantaneous angular velocity, but:
\[\begin{gather*} \begin{aligned} & w = J_r \phi' \end{aligned} \end{gather*}\]An Important Accompanying Property for SLAM
In 2D, skew matrix is simply:
\[\begin{gather*} \begin{aligned} & w^{\land} = \begin{bmatrix} 0 & -a \\ a & 0 \end{bmatrix} \end{aligned} \end{gather*}\]Rotation matrix is:
\[\begin{gather*} \begin{aligned} & R = \begin{bmatrix} cos \theta & -sin \theta \\ sin \theta & cos \theta \end{bmatrix} \end{aligned} \end{gather*}\]- One Key property that doesn’t hold true in 2D is this:
- And equivalently,
But in 2D, one can easily find that: $\phi^{\land} R = R \phi^{\land}$
Commutative Property Of Cross Product
\[\begin{gather*} \begin{aligned} & \theta^{\land} v = - v^{\land} \theta \end{aligned} \end{gather*}\]Adjoint Action
For $R_0,R_1\in SO(3)$,
\[\begin{gather*} \begin{aligned} \operatorname{Ad}_{R_0}(R_1) \;\triangleq\; R_0\,R_1\,R_0^{-1}. \end{aligned} \end{gather*}\]Claim (equivariance of the exponential): For any $X\in \mathfrak{so}(3)$,
\[\begin{aligned} R_0 \, \operatorname{Exp}(X) \, R_0^{-1} &= \operatorname{Exp}\!\big(\operatorname{Ad}_{R_0} X\big) \quad \text{(definition of the adjoint action)} \\[6pt] &= \operatorname{Exp}\!\big(R_0 X R_0^{-1}\big) \quad \text{(adjoint action for $SO(3)$ is conjugation)}. \end{aligned}\]In particular with $X=\widehat{w}$ for $w\in\mathbb{R}^3$,
\[\begin{aligned} R_0\,\operatorname{Exp}(\widehat{w})\,R_0^{-1} & = \operatorname{Exp}\!\big(R_0\,\widehat{w}\,R_0^{-1}\big) \\ & = \operatorname{Exp}\!\big((R_0 w)^{\wedge}\big), \end{aligned}\]Proof
For matrices, since $R_0\,\widehat{w}\,R_0^{-1} = (R_0 w)^{\wedge}$, we get
\[\begin{aligned} R_0 \, \operatorname{Exp}(X) \, R_0^{-1} &= R_0 \left( \sum_{k=0}^{\infty} \frac{1}{k!} X^k \right) R_0^{-1} \quad \text{(series expansion of the exponential)} \\[6pt] &= \sum_{k=0}^{\infty} \frac{1}{k!} \, R_0 X^k R_0^{-1} \quad \text{(linearity of $R_0$ and $R_0^{-1}$)} \\[6pt] &= \sum_{k=0}^{\infty} \frac{1}{k!} \, (R_0 X R_0^{-1})^k \quad \text{(conjugation property)} \\[6pt] &= \operatorname{Exp}\!\big( R_0 X R_0^{-1} \big) \quad \text{(rebuild exponential from series)} \\[6pt] &= \operatorname{Exp}\!\big( R_0 (w)^{\wedge} R_0^{-1} \big) \quad \text{(substitute $X = (w)^{\wedge}$)} \\[6pt] &= \operatorname{Exp}\!\big( (R_0 w)^{\wedge} \big) \quad \text{(rotation acts on vector $w$)}. \end{aligned}\]QED.
First Order BCH Approximation For Small Angles
For a small angle $w$, its SO(3) is approximately
For the product of exp of k small angles, that’d be:
\[\begin{gather*} \begin{aligned} \prod_{k=1}^{n}\bigl(I + X_k + o(\varepsilon^2)\bigr) = I + \sum_{k=1}^{n} X_k + o(\varepsilon^2). \end{aligned} \end{gather*}\]But in the meantime,
\[\begin{gather*} \begin{aligned} & \operatorname{Exp} \!\Bigl(\sum_{k=1}^{n} X_k\Bigr) & = I + \sum_{k=1}^{n} X_k + o(\varepsilon^2). \end{aligned} \end{gather*}\]So
\[\begin{gather*} \begin{aligned} & \prod_{k=1}^{n} \operatorname{Exp}(\varepsilon^\wedge_k) = \operatorname{Exp} \!\Bigl(\sum_{k=1}^{n} \varepsilon^\wedge_k \Bigr) \end{aligned} \end{gather*}\]