SCREWS AS SUMS OF SLIDERS

We know that the set of sliders do not form a vector space: the sum of two sliders is, in general, not a slider. In fact, we can show that a screw can be written in infinitely many ways as the sum of two sliders.

Consider a screw $\{ {\cal V} \}$ of non-zero resultant $\bf V$. Let $\{ {\cal S}_1 \}$ be a slider whose axis is not parallel to $\bf R$ and which satisfies $\{ {\cal S}_1 \} \cdot \{ {\cal V} \} \neq 0$. Then there exist a unique slider $\{ {\cal S}_2 \}$ and a unique scalar $\lambda$ such that \[ \{ {\cal V} \} = \lambda \{ {\cal S}_1 \}+ \{ {\cal S}_2 \} \]

Given $\{ {\cal S}_1 \}$ and a scalar $\lambda$, define the screw $\{ {\cal W} \} = \{ {\cal V} \} -\lambda \{ {\cal S}_1 \}$. Can a value of $\lambda$ be found such that $\{ {\cal W} \}$ defines a slider? The resultant of $\{ {\cal W} \}$ is non-zero since it is the sum of two non-collinear vectors. Hence a necessary and sufficient condition for $\{ {\cal W} \}$ to be a slider can be stated by imposing $\{ {\cal W} \}\cdot \{ {\cal W} \} =0$: \[ \{ {\cal V} \}\cdot \{ {\cal V} \} -2 \lambda \{ {\cal V} \}\cdot \{ {\cal S}_1 \}+ \lambda^2 \{ {\cal S}_1 \} \cdot \{ {\cal S}_1 \} =0 \] However, we have $\{ {\cal S}_1 \} \cdot \{ {\cal S}_1 \} =0$ since $\{ {\cal S}_1 \}$ is a slider. This gives a unique value of $\lambda$: \[ \lambda = \frac{1}{2} \frac{ \{ {\cal V} \}\cdot \{ {\cal V} \} } { \{ {\cal V} \}\cdot \{ {\cal S}_1 \} } \] This value of $\lambda$ corresponds to unique slider $\{ {\cal W} \}= \{ {\cal S}_2 \}$.

PLANAR SCREWS

The most general screw is neither a couple nor a slider. There is an exception which is important in the context of planar kinematics.

Consider a screw $\{ {\cal V} \}$ (of resultant ${\bf V}$) in referential $\cal E$ which satisfies the following property: \[ {\bf \hat{e}}_3 \cdot {\bf v}_P = 0 \] where ${\bf \hat{e}}_3$ is a unit vector of $\cal E$. Then $\{ {\cal V} \}$ is either a couple or a slider.

For any two points $P$ and $Q$, we have ${\bf \hat{e}}_3 \cdot ({\bf v}_Q - {\bf v}_P)= 0$. Then we can write ${\bf \hat{e}}_3 \cdot ( {\bf V} \times {\bf r}_{PQ} ) =0$. This gives the triple scalar product $({\bf \hat{e}}_3 , {\bf V} , {\bf r}_{PQ} ) =0$. Since this is true for all $P$ and $Q$, this implies that ${\bf \hat{e}}_3 \times {\bf V} = {\bf 0}$. We then deduce that the screw satisfies \[ {\bf V} \cdot {\bf v}_P = 0 \] A screw whose scalar invariant ${\bf V} \cdot {\bf v}_P$ vanishes is either a couple or a slider.

From this result, we conclude that the kinematic screw of a body in planar motion is either a couple or a slider at any given instant: the body is in instantaneous translation or in instantaneous rotation.