Theorem 4.3.1. Scalar Invariant.
The quantity \(\bV \cdot \bvv_P\) is independent of point \(P\text{.}\) It is referred to as the scalar invariant (or automoment) of screw \(\{ \cV \}\text{.}\)
Section 4.3 Properties of Screws
Screws have a few interesting basic properties.
Proof.
To show that \(\bV \cdot \bvv_P\) is independent of point \(P\text{,}\) we take the scalar product of both sides of (4.1.1) with \(\bV\text{:}\) this leads to \(\bV \cdot \bvv_Q = \bV \cdot \bvv_P\) for any two points \(P\) and \(Q\) since \(\bV \times \br_{PQ}\) is orthogonal to \(\bV\text{.}\) \(\square\)
By proceeding in a similar manner, we can state the following theorem.
Theorem 4.3.2. Equiprojectivity.
The vector field \(\bvv_P\) associated with screw \(\{ \cV \}\) is equiprojective, that is, it satisfies the property:
\begin{equation}
\bvv_P \cdot \br_{PQ} = \bvv_Q \cdot \br_{PQ}\tag{4.3.1}
\end{equation}
for any two points \(P\) and \(Q\) of \(\cE\text{.}\)
Hence, the projections of the moments about \(P\) and \(Q\) on the line joining \(P\) and \(Q\) are identical. Equiprojectivity is quite fundamental to screws: it can be shown (see Exercise 4.10.1) that, to every equiprojective vector field \(\bvv: P\in \cE\mapsto \bvv_P\text{,}\) that is, satisfying \(\bvv_P \cdot \br_{PQ} = \bvv_Q \cdot \br_{PQ}\text{,}\) corresponds a unique screw. The study of screws and the study of equiprojective vector fields are one and the same thing.
We end this section with a useful theorem to characterize a screw from the knowledge of a vector field at three given points.
Theorem 4.3.3. Three-Point Theorem.
Let \(A\text{,}\) \(B\text{,}\) and \(C\) be three non-collinear points of \(\cE\text{,}\) and \(\bV_A\text{,}\) \(\bV_B\text{,}\) and \(\bV_C\) be three vectors which satisfy the following equiprojection conditions:
\begin{equation*}
(\bV_B -\bV_A)\cdot \br_{AB} =0, \quad
(\bV_C -\bV_A)\cdot \br_{AC}=0,\quad
(\bV_C -\bV_B)\cdot \br_{BC}=0
\end{equation*}
Then, the knowledge of vectors \(\bV_A\text{,}\) \(\bV_B\text{,}\) and \(\bV_C\) defines a unique screw \(\{ \cV \}\) associated with a vector field \(P\in \cE \mapsto \bvv_P\) such that
\begin{equation*}
\bvv_A =\bV_A, \quad \bvv_B =\bV_B, \quad\bvv_C =\bV_C
\end{equation*}
Proof.
The existence of a screw \(\{ \cV \}\) is examined by solving for the vector \(\bV\) solution of the following equations
\begin{equation*}
\bvv_B - \bvv_A = \bV \times \br_{AB}, \qquad
\bvv_C - \bvv_A = \bV \times \br_{AC}
\end{equation*}
Thanks to equiprojection, the solutions of each of these equations is guaranteed. Therefore, there exist two real numbers \(\la\) and \(\mu\) such that
\begin{equation*}
\bV = \frac{\br_{AB}}{|AB|^2} \times (\bvv_B - \bvv_A) + \lambda \br_{AB}
\end{equation*}
and
\begin{equation*}
\bV = \frac{\br_{AC}}{|AC|^2} \times (\bvv_C - \bvv_A) + \mu \br_{AC}
\end{equation*}
For both equations to be satisfied, there must exist at least one solution \((\la, \mu) \in \mathbb{R}^2\) satisfying
\begin{equation*}
\frac{\br_{AB}}{|AB|^2} \times (\bvv_B - \bvv_A) + \lambda \br_{AB}
=
\frac{\br_{AC}}{|AC|^2} \times (\bvv_C - \bvv_A) + \mu \br_{AC}
\end{equation*}
Two independent equations solving for \((\la, \mu)\) are found by projection on \(\br_{AB}\)
\begin{equation*}
\lambda |AB|^2 = \br_{AB}\cdot (\frac{\br_{AC}}{|AC|^2}\times(\bvv_C - \bvv_A)) + \mu \br_{AC}\cdot \br_{AB}
\end{equation*}
then on \(\br_{AC}\)
\begin{equation*}
\br_{AC}\cdot (\frac{\br_{AB}}{|AB|^2}\times(\bvv_B - \bvv_A)) + \lambda \br_{AB}\cdot \br_{AC} = \mu |AC|^2
\end{equation*}
The last equation obtained by projection on \(\br_{AB}\times \br_{AC}\) can be shown to be identically satisfied. Therefore, a screw of resultant \(\bV\) can be found. If two screws \(\{ \cV_1 \}\) and \(\{ \cV_2 \}\) are assumed to exist, then the resultant \(\bV\) of screw \(\{ \cV \}= \{ \cV_1 \}-\{ \cV_2 \}\) would satisfy
\begin{equation*}
\bV \times\br_{AB} = \bV\times \br_{AC} = \bze
\end{equation*}
Since \(\br_{AB}\) and \(\br_{AC}\) are linearly independent, we must have \(\bV=\bze\text{,}\) and thus \(\{ \cV \}= \{ 0 \}\) and unicity is proved. \(\square\)
