Kinematic Loop Formula

Here is an application of Theorem 6.3 (Kinematic Loop Formula).

Two referentials ${\cal A} (A,{\bf \hat{a}} _1,{\bf \hat{a}} _2,{\bf \hat{e}} _3)$ and ${\cal B} (B,{\bf \hat{b}} _1,{\bf \hat{b}} _2,{\bf \hat{e}} _3)$ are in rotational motion about axis $(A,{\bf \hat{e}} _3)$ and $(B, {\bf \hat{e}} _3)$, respectively, relative to a referential ${\cal E} (A,{\bf \hat{e}} _1,{\bf \hat{e}} _2,{\bf \hat{e}} _3)$. Points $A$ and $B$ are located on axis $(O,{\bf \hat{e}} _1)$, with $O$ midpoint of $AB$. ${\cal A}$ rotates at constant angular velocity $\omega$ in the counterclockwise direction. ${\cal B}$ rotates at constant angular velocity $2\omega$ in the clockwise direction. The axes $(A,{\bf \hat{a}} _1)$ and $(B,{\bf\hat{ b}} _1)$ both coincide with axis $(O,{\bf \hat{e}} _1)$ at $t=0$. Define $2l$ the distance $|AB|$.

a. Find the kinematic screw $\{ {\cal V}_{\cal B/A} \}$ resolved at point $A$, then at point $B$.

b. Deduce that $\cal B$ is in instantaneous rotation relative to $\cal A$. Determine the corresponding instantaneous axis of rotation $(I,{\bf \hat{e}} _3)$. Describe the loci of $I$ relative to $\cal A$ and $\cal B$. Then characterize the motions of $\cal A$ and $\cal B$ relative to $\cal E$.

Set of Screws as a Lie Algebra

A Lie algebra is an vector space $L$ equipped with a bilinear map called Lie bracket \[ (x, y) \in L\times L \mapsto [x,y]\in L \] which satisfies the following properties: \[ [x, x] = 0 \text{ for all } x\in L \qquad\qquad{(1)} \] \[ [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 \qquad {(2)} \] for all $x$, $y$, and $z \in L$.

Consider the set of screws over space $\cal E$ denoted $S({\cal E})$ and equipped with the bilinear map \[ (\{{\cal U} \}, \{{\cal V} \} ) \in S({\cal E}) \times S({\cal E}) \mapsto \{{\cal U} \} \times \{{\cal V} \} \] corresponding to the vector field \[ P \in {\cal E} \mapsto {\bf U}\times {\bf v}_P - {\bf V}\times {\bf u}_P \] We can show that $\{{\cal U} \} \times \{{\cal V} \}$ has the properties of a Lie bracket and that the set $S(\cal E)$ is a Lie algebra.

First we must show that $\{{\cal U} \} \times \{{\cal V} \}$ is a screw: \[ {\bf U}\times {\bf v}_Q - {\bf V}\times {\bf u}_Q = {\bf U}\times ( {\bf v}_P+ {\bf V}\times {\bf r}_{PQ}) - {\bf V}\times ({\bf u}_P+ {\bf U}\times {\bf r}_{PQ}) ={\bf U}\times {\bf v}_P - {\bf V}\times {\bf u}_P +({\bf U}\times{\bf V}) \times {\bf r}_{PQ} \] This shows that $\{{\cal U} \} \times \{{\cal V} \}$ is a screw of resultant ${\bf U}\times{\bf V}$. Property (1) is obvious. Property (2) must be written as \[ \{{\cal U} \} \times (\{{\cal V} \} \times \{\cal W \}) + \{{\cal V} \} \times (\{\cal W \} \times \{{\cal U} \})+\{\cal W \} \times (\{{\cal U} \} \times \{{\cal V} \}) = \{ 0 \} \] It follows from Jacobi's identity: \[ ({\bf U} \times {\bf V}) \times {\bf W} + ({\bf V} \times {\bf W}) \times {\bf U} + ({\bf W} \times {\bf U}) \times {\bf V} = {\bf 0} \]

SUM of N SLIDERS

Theorem 4.3 cannot be generalized to $N$ sliders $(N\geq 3)$: in other words, the sum of 3 or more sliders can yield a slider without their axes intersecting or being parallel. Here is an example: in $(O, {\bf \hat{e}}_1 , {\bf \hat{e}}_2, {\bf \hat{e}}_3)$ consider the three sliders \[ \{ {\cal S}_1 \} = \begin{Bmatrix}{\bf \hat{e}}_1 \\ {\bf 0} \end{Bmatrix}_{O}, \quad \{ {\cal S}_2 \} = \begin{Bmatrix}{\bf \hat{e}}_2 \\ {\bf 0} \end{Bmatrix}_{A}, \quad \{ {\cal S}_3 \} = \begin{Bmatrix}{\bf \hat{e}}_3 \\ {\bf 0} \end{Bmatrix}_{B} \] with ${\bf r}_{OA}= {\bf \hat{e}}_3$ and ${\bf r}_{OB}= {\bf \hat{e}}_1+ {\bf \hat{e}}_2$. Consider the screw \[ \{ {\cal V} \} = \alpha \{ {\cal S}_1 \}+ \beta \{ {\cal S}_2 \}+\gamma \{ {\cal S}_3 \} \] with $\alpha$, $\beta$, and $\gamma$ non-zero scalars. It is easy to see that the axes of these sliders are neither parallel nor intersecting. Yet it is possible to find a condition on the scalars $(\alpha,\beta,\gamma)$ for which $\{ {\cal V} \}$ is a slider: we impose the pitch of $\{ {\cal V} \}$ to be zero: \[ (\alpha {\bf \hat{e}}_1 + \beta {\bf \hat{e}}_2 + \gamma {\bf \hat{e}}_3 ) \cdot ( \beta {\bf \hat{e}}_2 \times {\bf \hat{e}}_3 + \gamma {\bf \hat{e}}_3 \times ({\bf \hat{e}}_1+ {\bf \hat{e}}_2) ) =0 \] giving the equation \[ \alpha \gamma = \alpha\beta + \beta\gamma \]

SUM OF TWO SLIDERS

Consider two sliders $\{ {\cal S}_1 \}$ and $\{ {\cal S}_2 \}$ defined by \[ \{ {\cal S}_1 \} = \begin{Bmatrix}{\bf S}_1 \\ {\bf 0} \end{Bmatrix}_{A_1}, \quad \{ {\cal S}_2 \} = \begin{Bmatrix} {\bf S}_2 \\ {\bf 0} \end{Bmatrix}_{A_2}, \quad \] Assuming ${\bf S}_1 + {\bf S}_2\neq {\bf 0}$, let's find the conditions for which the sum $\{ {\cal S}_1 \}+ \{ {\cal S}_2 \}$ is a slider. A necessary and sufficient condition is that $\{ {\cal S}_1 \}\cdot \{ {\cal S}_2 \} =0$ which can be stated as \[ {\bf S}_1 \cdot ( {\bf S}_2 \times {\bf r}_{A_1 A_2} ) = 0 \] Case 1: if ${\bf r}_{A_1 A_2} ={\bf 0}$, then the sliders' axes are intersecting.

Case 2: if ${\bf r}_{A_1 A_2} \neq {\bf 0}$, then the three vectors ${\bf S}_1$, ${\bf S}_2$, and ${\bf r}_{A_1 A_2}$ are coplanar, or two of these vectors are collinear. In the first case, the sliders' axes are intersecting or parallel. In the second case, either ${\bf S}_1$ and ${\bf S}_2$ are collinear or ${\bf r}_{A_1 A_2}$ and ${\bf S}_1$ (or ${\bf S}_2$) are collinear: again, this implies that the sliders' axes are either intersecting or parallel.

In conclusion, if the sum of two sliders is a slider, then their axes are either intersecting or parallel. The converse is true if ${\bf S}_1 + {\bf S}_2\neq {\bf 0}$.