Angular velocity

Rotation representations described the orientation of a body at one instant. This section describes how that orientation changes over time. It follows section 3.2.2 of Kevin Lynch and Frank Park, Modern Robotics: Mechanics, Planning, and Control (Kevin M. Lynch, 2017). You can download the chapter as a PDF. The section starts in the plane, where angular velocity is a single number. It then moves to space, where angular velocity is a vector whose direction can change with time.

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Angular velocities in the plane

A body turns about the origin at the rate $\omega = \dot\theta$. A point $p$ on it moves on a circle. The quarter-turn matrix $G$ turns $p$ by $90^\circ$, and scaling $Gp$ by $\omega$ gives the velocity $\dot p$ (red), which is tangent to the circle with length $|\omega|\,\|p\|$. A negative $\omega$ reverses it. The orientation as a whole obeys $\dot R = [\omega] R$ with $[\omega] = \omega G$. A body turns in the plane about the origin. Its orientation is the rotation matrix $R(\theta)$ from rotations in the plane, and the angle $\theta(t)$ changes with time. The angular velocity is the rate of that change, the scalar $$\omega = \dot\theta .$$ Differentiate $R(\theta)$ entry by entry. Each entry depends on time only through the angle $\theta(t)$, so the chain rule applies: the derivative with respect to time is the derivative with respect to $\theta$ times $\dot\theta$. For the top-left entry, $$\frac{d}{dt} \cos\theta(t) = \frac{d \cos\theta}{d\theta} \, \frac{d\theta}{dt} = -\sin\theta \; \dot\theta .$$ The other three entries work the same way, and every one of them picks up the same factor $\dot\theta$: $$\dot R = \dot\theta \begin{bmatrix} -\sin\theta & -\cos\theta \\ \cos\theta & -\sin\theta \end{bmatrix} .$$ The matrix on the right equals $G R$, where $G$ is the quarter-turn matrix $$G = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} ,$$ so $\dot R = \dot\theta\, G R$. Collect the rate and $G$ into one matrix: $$\dot R = [\omega]\, R , \qquad [\omega] = \omega G = \begin{bmatrix} 0 & -\omega \\ \omega & 0 \end{bmatrix} .$$ The bracket notation $[\omega]$ means “the skew-symmetric matrix of $\omega$”. The rest of the book uses it often. In the plane, $[\omega]$ belongs to $\mathfrak{so}(2)$, the tangent space at the identity in SO(2) as a manifold. A point fixed on the body has coordinates $p_b$ in the body frame. They do not change, because the point moves with the body. In the fixed frame its coordinates are $p_s = R\, p_b$, and these do change. Differentiate, using $\dot p_b = 0$: $$\dot p_s = \dot R\, p_b = [\omega] R\, p_b = [\omega]\, p_s = \omega\, G p_s .$$ Multiplying by $G$ turns a vector by $90^\circ$ counterclockwise. So the velocity is perpendicular to $p_s$, and its length is $|\omega| \, \|p_s\|$: the angular speed times the radius. This is the rule $v = \omega r$ of circular motion. Khan Academy covers it in angular velocity and speed and relating angular and regular motion variables. In space, two things change. The axis can point in any direction, not only out of the plane. And a point on the body need not lie in the plane of the circle it travels on. The cross product handles both, as the next part shows. From here on, $p$ without a subscript means the fixed-frame coordinates $p_s$. Lynch and Park make this argument for eq. 3.49 (page 83), and Kleppner and Kolenkow make it in section 7.2 (Kleppner & Kolenkow, 1973).

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Angular velocities in space

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A fixed rotating axis

The animation draws Figure 3.11 of Lynch and Park, and the derivation below goes through it step by step. A point $p$ on the body splits into $p_\parallel$ along the axis $\hat{\omega}$, which stays fixed, and $p_\perp$ across it, which sweeps a circle of radius $\|p\| \sin\phi$. A small turn $d\theta$ moves the point by $ds$ along the tangent $\hat{t}$. That gives $dp = (\hat{\omega} \times p)\, d\theta$, and dividing by $dt$ gives $\dot{p} = \omega \times p$ with $\omega = \hat{\omega}\dot\theta$. A rigid body turns about a fixed axis through the origin with unit direction $\hat{\omega}$, $\|\hat{\omega}\| = 1$. Let $p(t)$ be the position of a point on the body. Split it into a part along the axis and a part across it: $$p = p_\parallel + p_\perp , \qquad p_\parallel = (\hat{\omega} \cdot p)\, \hat{\omega} , \qquad p_\perp = p - (\hat{\omega} \cdot p)\, \hat{\omega} .$$ The axis does not move, so $p_\parallel$ stays constant. The part $p_\perp$ turns in the plane perpendicular to $\hat{\omega}$. The point therefore moves on a circle about the axis. Let $\phi$ be the angle between $p$ and $\hat{\omega}$. The radius of the circle is $\|p_\perp\| = \|p\| \sin\phi$. The projection $\hat{\omega} \cdot p$ is a dot product. Khan Academy’s dot product video reviews it. Assume the point is not on the axis, so $p_\perp \neq 0$ and the circle has a nonzero radius. When the body turns by a small angle $d\theta$, the point moves along the circle by the arc length $$ds = \|p\| \sin\phi \; d\theta .$$ This is the arc length $s = r\theta$ of a circle of radius $r$, written for a small angle. Khan Academy derives it in arc length from angular displacement. The point moves along the tangent to the circle. You can build the tangent direction from the two vectors you already have, $\hat{\omega}$ and $p_\perp$, using the cross product properties in the Note above:

• Direction. The vector $\hat{\omega} \times p_\perp$ is perpendicular to $\hat{\omega}$, so it lies in the plane of the circle. It is also perpendicular to $p_\perp$, the radius that points from the center to the point. On a circle, the direction perpendicular to the radius is the tangent.
• Which way. A tangent line can be followed in two opposite directions. The right-hand rule picks the one in which the point moves. With your thumb along $\hat{\omega}$, your fingers curl in the direction of a positive rotation, and $\hat{\omega} \times p_\perp$ points the same way.
• Length. $\hat{\omega}$ is a unit vector perpendicular to $p_\perp$, so $\|\hat{\omega} \times p_\perp\| = \|p_\perp\|$.

Any vector is its length times a unit vector in its direction. For the tangent vector $\hat{\omega} \times p_\perp$ this reads $$\hat{\omega} \times p_\perp = \|\hat{\omega} \times p_\perp\| \, \hat{t} = \|p_\perp\| \, \hat{t} .$$ Dividing by $\|p_\perp\|$ gives the unit tangent: $$\hat{t} = \frac{\hat{\omega} \times p_\perp}{\|p_\perp\|} .$$ For a small turn, the short arc is almost a straight segment along the tangent. So the displacement $dp$ has length $ds$ and direction $\hat{t}$: $$dp = ds \, \hat{t} .$$ Substitute the arc length, $ds = \|p\| \sin\phi \; d\theta = \|p_\perp\| \, d\theta$, and the unit tangent. The radius $\|p_\perp\|$ cancels: $$dp = \left( \|p_\perp\| \, d\theta \right) \frac{\hat{\omega} \times p_\perp}{\|p_\perp\|} = (\hat{\omega} \times p_\perp)\, d\theta .$$ The part $p_\parallel$ is parallel to $\hat{\omega}$, so $\hat{\omega} \times p_\parallel = 0$. Adding it back changes nothing: $$dp = (\hat{\omega} \times p)\, d\theta .$$ This holds for an infinitesimal turn only. A finite turn $\Delta\theta$ carries the point along the arc, not along the straight tangent, so the finite displacement is not $(\hat{\omega} \times p)\,\Delta\theta$. Now divide by $dt$. The cross product is linear, so the scalar $\dot\theta = d\theta/dt$ can move inside it: $$\dot{p} = (\hat{\omega} \times p)\, \dot\theta = (\hat{\omega} \dot\theta) \times p .$$ The vector in parentheses is the angular velocity (Lynch and Park eq. 3.25): $$\omega = \hat{\omega} \, \dot{\theta} , \qquad \dot{p} = \omega \times p .$$ Its direction is the rotation axis. Its magnitude $|\dot\theta|$ is the angular speed. Its sign follows the right-hand rule about $\hat{\omega}$. With $\dot\theta = 1$, this is Lynch and Park’s eq. 3.49, $\dot{p} = \hat{\omega} \times p$. The derivation assumed $p_\perp \neq 0$. A point on the axis has $p_\perp = 0$ and does not move. The result agrees: $p$ is then parallel to $\omega$, so $\omega \times p = 0$. The speed of the point is $\|\dot{p}\| = |\dot\theta|\, \|p\| \sin\phi$, the angular speed times the radius of its circle. This is the planar rule $v = \omega r$ again. When $\hat{\omega} = \hat{z}$ and $p$ lies in the plane, $\omega \times p$ is exactly $\omega\, G p$.

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A rotating axis that changes with time

In general the axis of rotation can change with time. Suppose a frame with unit axes $\{\hat{x}, \hat{y}, \hat{z}\}$ is attached to a rotating body. Consider the time derivatives of these axes. The length of $\hat{x}$ is fixed, so only its direction can change. The same is true of $\hat{y}$ and $\hat{z}$. Between times $t$ and $t + \Delta t$, the orientation changes by a rotation through an angle $\Delta \theta$ about a unit axis $\hat{\omega}$ through the origin. The axis $\hat{\omega}$ is independent of coordinates and has not yet been expressed in a reference frame. As $\Delta t$ approaches zero, the ratio $\Delta \theta / \Delta t$ approaches the rotation rate $\dot{\theta}$. The vector $\hat{\omega}$ becomes the instantaneous axis of rotation. Over that instant the body turns about a fixed axis, so the result above applies with $$\omega = \hat{\omega} \, \dot{\theta} .$$ The tip of each unit axis is a point on the body. With $p = \hat{x}$, it moves with velocity $\omega \times \hat{x}$, which is perpendicular to both $\omega$ and $\hat{x}$. A point on a rotating body has velocity $\dot{p} = \omega \times p$. Write out this cross product with the component formula from the Note above, for $\omega = (\omega_1, \omega_2, \omega_3)$ and $p = (p_1, p_2, p_3)$. Each component is a sum of terms in $p_1$, $p_2$ and $p_3$, so the whole product is a matrix times $p$: $$\omega \times p = \begin{bmatrix} \omega_2 p_3 - \omega_3 p_2 \\ \omega_3 p_1 - \omega_1 p_3 \\ \omega_1 p_2 - \omega_2 p_1 \end{bmatrix} = \underbrace{\begin{bmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{bmatrix}}_{[\omega]} \begin{bmatrix} p_1 \\ p_2 \\ p_3 \end{bmatrix} = [\omega]\, p .$$ So a cross product with $\omega$ is a product with the matrix $[\omega]$. Its entries satisfy $[\omega] = -[\omega]^T$. The same rule turns any vector $v \in \mathbb{R}^3$ into a matrix $[v]$. The set of all $3 \times 3$ real skew-symmetric matrices is the Lie algebra $\mathfrak{so}(3)$. The matrix $[\omega]$ is the hat of $\omega$, so an angular velocity is an element of this Lie algebra. Let $R(t)$ describe the orientation of the body frame relative to the fixed frame at time $t$. Its columns $r_1(t)$, $r_2(t)$ and $r_3(t)$ give $\hat{x}$, $\hat{y}$ and $\hat{z}$ in fixed-frame coordinates. Let $\omega_s$ be the angular velocity expressed in the fixed frame. Each column then moves according to $$\dot{r}_i = \omega_s \times r_i , \qquad i = 1, 2, 3 ,$$ Stack the three columns side by side to form $\dot R$: $$\dot R = \begin{bmatrix} \omega_s \times r_1 & \omega_s \times r_2 & \omega_s \times r_3 \end{bmatrix} .$$ Replace each cross product with the matrix product $[\omega_s] r_i$: $$\dot R = \begin{bmatrix} [\omega_s] r_1 & [\omega_s] r_2 & [\omega_s] r_3 \end{bmatrix} .$$ A matrix times $R$ multiplies each column of $R$ separately, so $[\omega_s]$ factors out: $$\dot R = [\omega_s] \begin{bmatrix} r_1 & r_2 & r_3 \end{bmatrix} = [\omega_s] R .$$ The axis of $\omega$ changes with time, and the yellow curve traces the tip of $\omega(t)$. At each instant the columns $r_1$, $r_2$ and $r_3$ of $R$ move with $\omega_s \times r_i$ (red arrows). Because the axis moves, the tip of $r_1$ no longer stays on one circle. The matrix $[\omega]$ turns each cross product into a matrix product, $\omega \times r_1 = [\omega] r_1$. Stacking the three velocities then gives $\dot R = [\omega_s] R$. Page 79 of Lynch and Park solves this equation for the angular velocity. In fixed-frame and body-frame coordinates, $$[\omega_s] = \dot{R} R^{-1} , \qquad [\omega_b] = R^{-1} \dot{R} ,$$ with $R^{-1} = R^\top$. The second form is the $\mathcal{X}^{-1}\dot{\mathcal{X}}$ of Lie I. The fixed-frame angular velocity $\omega_s$ is independent of the chosen body frame. Similarly, the body-frame angular velocity $\omega_b$ is independent of the chosen fixed frame. Both $R$ and $\dot{R}$ depend on $\{s\}$ and $\{b\}$. Their products remove one of these dependencies. The product $\dot{R} R^{-1}$ is independent of $\{b\}$, while $R^{-1} \dot{R}$ is independent of $\{s\}$. The two angular velocities are the same vector written in two frames, so $\omega_s = R\, \omega_b$. This holds for any pair of frames. If $R_{cd}$ is the orientation of $\{d\}$ relative to $\{c\}$, then $$\omega_c = R_{cd}\, \omega_d .$$ In the plane, $\omega_s = \omega_b = \dot\theta$, because every rotation is about the same axis and planar rotations commute. The velocity formulas $\dot p = \omega \times p$ and $\dot R = [\omega_s] R$ hold at every instant, whether or not the axis of $\omega$ moves. A moving axis changes only what happens over a finite time, as exponential coordinates shows.

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References

• Kevin M. Lynch, F. (2017). Modern Robotics.
• Kleppner, D., Kolenkow, R. (1973). An introduction to mechanics.

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