Instantaneous Axis Of Rotation

Question

What is the mathematical theory behind finding a Instaneous Axis/Centre of rotation for a rotation plus translation motion? Please explain in terms of the group of Rigid Motions(if possible) in $R^2$ & $R^3$ as why always such a centre will exist and why can a rigid motion can be described in terms of only a rotation about that point instead of a rotation plus translation.

http://en.wikipedia.org/wiki/Instant_centre_of_rotation

Answer 1

Let a Cartesian coordinate system be attached to a rigid body, say with its origin at the center of gravity of that body. Then, because the body is rigid, any movement of the coordinate system will perserve the length of the base vectors and the right angles between them. That is: any movement with respect to the bodies' origin is described by an orthogonal matrix (with positive determinant, because volumes are preserved as well). Such movements are commonly called rotations.
Rotations in 3-D space are represented by: $$ \vec{r_1}(t) = R(t)\,\vec{r} \qquad ; \qquad \left[ \begin{array}{c} x_1 \\ y_1 \\ z_1 \end{array} \right](t) = \left[ \begin{array}{ccc} R_{11}(t) & R_{12}(t) & R_{13}(t) \\ R_{21}(t) & R_{22}(t) & R_{23}(t) \\ R_{31}(t) & R_{32}(t) & R_{33}(t) \end{array} \right] \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] $$ Thus it is assumed that the vector $\vec{r}$ is a constant, $R(t)$ is a real valued orthogonal $3\times 3$ matrix and its coefficients are functions of time $t$ . Because $R$ is orthogonal, it is known that this matrix times its transpose is equal to the identity matrix (transpose = inverse). Differentiation according to the product rule and known rules for matrix transposition then gives: $$ R(t)R^T(t) = I \quad \Longrightarrow \quad \dot{R}(t)R^T(t) + R(t)\dot{R}(t)^T = 0 \quad \Longrightarrow \quad \dot{R}(t)R^T(t) = - \left[\dot{R}(t)R^T(t)\right]^T $$ Define: $$ \Omega(t) = \dot{R}(t)R^T(t) \qquad \Longrightarrow \qquad \Omega(t) = -\Omega^T(t) $$ It can be easily shown that the $\Omega$ matrix must have the following form: $$ \left[ \begin{array}{ccc} 0 & - \omega_z & +\omega_y \\ +\omega_z & 0 & -\omega_x \\ -\omega_y & +\omega_x & 0 \end{array} \right] $$ The vector $\;\vec{\omega}$ , not at all by coincidence, is defined in such a way that: $$ \vec{\omega} = \left[ \begin{array}{c} \omega_x \\ \omega_y \\ \omega_z \end{array} \right] \quad \Longrightarrow \quad \Omega\,\vec{r} = \left[ \begin{array}{ccc} 0 & - \omega_z & +\omega_y \\ +\omega_z & 0 & -\omega_x \\ -\omega_y & +\omega_x & 0 \end{array} \right] \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{c} \omega_y z - \omega_z y \\ \omega_z x - \omega_x z \\ \omega_x y - \omega_y x \end{array} \right] = \vec{\omega} \times \vec{r} $$ Where $\times$ denotes the usual cross product of two vectors. Consequently: $$ \dot{\vec{r}}_1(t) = \dot{R}(t)\vec{r} = \dot{R}(t)R^T(t)R(t)\vec{r}=\vec{\omega}(t)\times\vec{r}_1(t) $$ Thus the velocity of a vector rotating in time is a rotation of the same vector around an axis $\vec{\omega}(t)$ that itself is moving in time. Especially the fact that in the general case everything is "moving in time" makes it not very easy to comprehend conceptually.

Answer 2

Suppose $T(t)$ is a rigid motion, that is: $$ T(t)x = R(t)x + y(t) $$ where $R(t)$ is a rotation matrix. We assume that $T(0)x = x$. Note $R(t)^TR(t) = I$ because $R(t)$ is an orthogonal matrix.

We would like to see if $T(0)$ is a rotation about a fixed point or line. Note that in the 3D case, the line will be fixed, but the rigid motion might be a corkscrew motion around the axis, so it need not necessarily be a purely a rotation about the line.

Suppose $T(0)$ is not a translation.

In the 2D case, we want to show that there is a point $x_0$ such that $$ T(\delta t)x = x + R'(0)(x-x_0)\delta t +o (\delta t). \quad (1) $$ That is, $x_0$ is a fixed point. $T$ performs an infinitesimal rotation $x-x_0 \mapsto R(\delta t)(x-x_0)$.

In the 3D case, we want to show that there is a point $x_0$, a unit vector $v$ such that $R'(0) v = 0$, and a scalar $\mu$ such that $$ T(\delta t)x = x + (R'(0)(x-x_0) + \mu v)\delta t + o(\delta t). \quad (2) $$ That is, $x_0 + \alpha v, (\alpha \in \mathbb R)$ is a fixed line. $T$ performs an infinitesimal rotation $x-x_0 \mapsto R(\delta t)(x-x_0)$, and then add an infinitesimal translation along the axis of rotation $v$, adding $\mu v \delta t$.

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Proof: Differentiate $R(t)^TR(t) = I$ with respect to $t$: $$ R'(t)^T R(t) + R(t)^T R'(t) = 0 = (R(t)^T R'(t))^T + (R(t) R'(t)^T) $$ that is $$ R(t)^T R'(t) = S(t) $$ is anti-symmetric. Therefore rewriting as $R'(t) = R(t) S(t)$, we have $$ T'(t)x = R(t) S(t) x + y'(t) = 0 $$ or $$ S(t) x = -R(t)^T y'(t) .$$ From now on we will assume $t = 0$, and write $T$ for $T(0)$, $T'$ for $T'(0)$, etc. Note that $R = I$ and $y = 0$, so $R' = S$.

Since $S$ is skew symmetric, its kernel and range are orthogonal spaces. This is because $\text{ker}(S) = \text{range}(S^T)^\perp = \text{range}(S)^\perp$.

Supose all of the eigenvalues of $S$ are zero. Since $S$ is normal, i.e., $SS^T = S^TS$, it is diagonalizable. Hence then $S = 0$. In that case $T'x = x+y'$. If $y'=0$, then $T$ is instantaneously motionless. If $y'\ne0$ then $T$ is a pure translation. From now on we will assume that $S \ne 0$.

Since $S$ is skew symmetric, this means that $\lambda$ is an eigenvalue if and only if $-\lambda$ is an eigenvalue, because $$ \det(S-\lambda I) = \det((S-\lambda I)^T) = -\det(S + \lambda I) .$$ Therefore if $S \ne 0$, the range of $S$ must be two-dimensional. (Note also that the non-zero eigenvalues are imaginary, and come in conjugate pairs.)

Now let's look at the two different cases.

2D Case: Then $S$ is invertible. In that case there is exactly one solution $x_0$ to the equation $Sx_0 = -y'$, and so $$ T'x = R'x + y' = Sx - Sx_0 = R'(x - x_0),$$ which proves (1).

3D Case: One eigenvalue of $S$ must be zero, so the kernel is one dimensional. Let $v$ be a unit vector such that $\{v\}$ is a basis of $\text{ker}(S)$, so that $\text{range}(S) = \{v\}^\perp$.

Therefore there is a unique $\mu \in \mathbb R$ and $x_0$ such that $$ Sx_0 = - y' + \mu v.$$ Then $$ T'x = R'x + y' = Sx - Sx_0 + \mu v = R'(x-x_0) + \mu v,$$ which proves (2).