Section 4.7 Product of Two Screws
The definition of the product of two screws is critical to many concepts of rigid body mechanics.
Definition 4.7.1. Product of Two Screws.
The product (or comoment) of two screws \(\{ \cU \}\) and \(\{ \cV \}\) is the scalar defined by:
\begin{equation}
\{ \cU \} \cdot \{ \cV \} =
{\bf U} \cdot \bvv_P +
{\bf V} \cdot \buu_P
\label{}\tag{4.7.1}
\end{equation}
Example 4.7.5.
Consider two sliders \(\{ \cS_1 \}\) and \(\{ \cS_2 \}\text{,}\) show that \(\{ \cS_1 \} \cdot \{ \cS_2 \} = 0\) if and only if their axes are coplanar.
Solution.
The product of two sliders can be expressed as
\begin{equation*}
\{ \cS_1 \} \cdot \{ \cS_2 \} =
\left\{
\begin{array}{c}
\bS_1 \\
\bze
\end{array}
\right\}_{Q_1} \cdot
\left\{
\begin{array}{c}
\bS_2 \\
\bze
\end{array}
\right\}_{Q_2}=
\left\{
\begin{array}{c}
\bS_1 \\
\bze
\end{array}
\right\}_{Q_1} \cdot
\left\{
\begin{array}{c}
\bS_2 \\
\bS_2 \times \br_{Q_2 Q_1}
\end{array}
\right\}_{Q_1}
=\bS_1 \cdot (\bS_2 \times \br_{Q_2 Q_1})
\end{equation*}
where
\(Q_i\) is an arbitrary point of the axis of
\(\{ \cS_i \}\text{.}\) It then suffices to use the fact that the triple scalar product
\((\bS_1, \bS_2 ,\br_{Q_2 Q_1})\) is zero if and only if the 3 vectors
\(\bS_1\text{,}\) \(\bS_2\) and
\(\br_{Q_2 Q_1}\) are coplanar (see
Subsection A.1.4).
Example 4.7.6.
Consider two screws \(\{ \cU \}\) and \(\{ \cV \}\) of non-zero and orthogonal resultants (\(\bU \cdot \bV =0\)). Assume that their product \(\{ \cU \} \cdot \{ \cV \} = 0\) is zero.
Show that their axes must necessarily intersect. Is the converse true?
Solution.
Axis \(\Delta_U\) of screw \(\{ \cU \}\) is described as the set of points \(P\) such that \(\br_{AP} = \br_{AI} + \lambda \bU\) where \(\lambda\) is an arbitrary real scalar and where point \(I\) is given by \(\br_{AI} = (\bU \times \buu_A)/\bU^2\) (\(A\) arbitrarily chosen). Similarly axis \(\Delta_V\) of screw \(\{ \cV \}\) is the set of points \(Q\) given by \(\br_{AQ} = \br_{AJ}+ \mu \bV\) with \(\br_{AJ} = (\bV \times \bvv_A)/\bV^2 \mu\) is scalar).
For these two axes to intersect there must exist two scalars \(\lambda\) and \(\mu\) such that
\begin{equation*}
\br_{IJ}= \lambda \bU -\mu \bV
\end{equation*}
Hence a necessary condition to obtain a solution can be stated as
\begin{equation*}
\br_{IJ} \cdot (\bU \times \bV ) = 0
\end{equation*}
To show that this condition is satisfied, we use the expressions of \(\br_{AI}\) and \(\br_{AJ}\text{:}\)
\begin{equation*}
\br_{IJ} \cdot (\bU \times \bV ) =
\frac{(\bV \times \bvv_A) \cdot (\bU \times \bV )}{\bV^2}
-
\frac{(\bU \times \buu_A) \cdot (\bU \times \bV )}{\bU^2}
\end{equation*}
We can use the properties of the scalar and vector triple products to find that
\begin{align*}
(\bV \times \bvv_A) \cdot (\bU \times \bV ) \amp =
\bvv_A \cdot \left( \bV \times (\bU \times \bV ) \right) =
\bvv_A \cdot \left( \bV^2 \bU - (\bU\cdot \bV) \bV ) \right)\\
\amp =
\bV^2 (\bvv_A \cdot \bU)
\end{align*}
after using \(\bU \cdot \bV =0\text{.}\) Likewise we find
\begin{equation*}
(\bU \times \buu_A) \cdot (\bU \times \bV )=
- \bU^2 (\buu_A \cdot \bV)
\end{equation*}
This leads to
\begin{equation*}
\br_{IJ} \cdot (\bU \times \bV ) =
\bvv_A \cdot \bU + \buu_A \cdot \bV =
\{ \cU \} \cdot \{ \cV \} = 0
\end{equation*}
Hence the necessary condition to find an intersection point is equivalent to having \(\{ \cU \} \cdot \{ \cV \} = 0\text{.}\) If this condition is satisfied then the scalars \(\lambda\) and \(\mu\) are found to be
\begin{equation*}
\lambda = \frac{\br_{IJ}\cdot\bU}{\bU^2}, \qquad \mu =- \frac{\br_{IJ}\cdot\bV}{\bV^2}
\end{equation*}
