# Screws (Part 3): Application to Mechanisms

### 1. Kinematic loop formula

Consider 3 rigid bodies which may or may not be physically interconnected. To simplify notations, we denote them by the numerals 1, 2, and 3. Hence, we denote \(\{{\cal V}_{i/j}\}\) the kinematic screw of body \(i\) relative to body \(j\). Recall that it defines the vector field \(P\in i\mapsto {\bf v}_{P\in i/j}\). If \(\boldsymbol{\omega}_{i/j}\) is the angular velocity of body \(j\) relative to body \(i\), then we can write \[ \{ {\cal V}_{i/j} \} = \begin{Bmatrix} \boldsymbol{\omega}_{i/j} \\ {\bf v}_{A\in i/j} \end{Bmatrix} \] The notation \({\bf v}_{A\in i/j}\) emphasizes that point \(A\) is attached to body \(i\), even it is so only instantaneously.

We can show that the pairwise relationships formed between bodies 1, 2, and 3 are not
independent from the following identities:
\[
\boldsymbol{\omega}_{1/2}+ \boldsymbol{\omega}_{2/3}+ \boldsymbol{\omega}_{3/1} = {\bf 0}, \qquad {\bf v}_{P\in 1/2}+ {\bf v}_{P\in 2/3}+ {\bf v}_{P\in 3/1} = {\bf 0}
\]
These two equations can be assembled into a single formula, known as the *kinematic loop formula*
\[
\boxed{
\{ {\cal V}_{1/2} \}+\{ {\cal V}_{2/3} \}+\{ {\cal V}_{3/1} \} = \{0\}
}
\]

As a special case, we obtain \(\{ {\cal V}_{i/j} \} = - \{ {\cal V}_{j/i} \}\).

This formula can be generalized to \(n\) bodies.

The velocity loop formula can be shown by using the change of referential formula for a particle \(P\) in motion relative to 2 referentials \({\cal E}\) and \({\cal F}\): \({\bf v}_{P/{\cal E}}= {\bf v}_{P/{\cal F}} + {\bf v}_{P\in{\cal F}/{\cal E}}\).

This formula is very useful for the kinematic analysis of

*closed-loop mechanisms*, that is, mechanims constituted of rigid bodies such that each body is connected with 2 or more bodies.

### 2. Example of a closed loop mechanism

In the planar mechanism shown below, 4 rigid bodies form a single closed loop:

The joint 1/0 is slider along axis \((C,\boldsymbol{\hat{\jmath}})\): \(\{ {\cal V}_{1/0} \} = \begin{Bmatrix} {\bf 0}\\ \dot{y}_B \boldsymbol{\hat{\jmath}}\end{Bmatrix}\).

The joint 2/1 is a no-slip point-contact at \(I\): \(\{ {\cal V}_{2/1} \} = \begin{Bmatrix} \omega_{2/1}\boldsymbol{\hat{k}}\\ {\bf 0}\end{Bmatrix}_I\).

The joint 3/2 is a pivot of center \(A\): \(\{ {\cal V}_{3/2} \} = \begin{Bmatrix} \omega_{3/2}\boldsymbol{\hat{k}}\\ {\bf 0}\end{Bmatrix}_A\).

The joint 3/0 is a pivot of center \(O\): \(\{ {\cal V}_{3/0} \} = \begin{Bmatrix} \dot{\theta}_3 \boldsymbol{\hat{k}}\\ {\bf 0}\end{Bmatrix}_O\).

The loop equation \(\{ {\cal V}_{1/0} \}+\{ {\cal V}_{2/1} \}+\{ {\cal V}_{3/2} \} +\{ {\cal V}_{0/3} \} = \{0\}\) gives \[ \begin{Bmatrix} {\bf 0}\\ \dot{y}_B \boldsymbol{\hat{\jmath}}\end{Bmatrix} +\begin{Bmatrix} \omega_{2/1}\boldsymbol{\hat{k}}\\ {\bf 0}\end{Bmatrix}_I +\begin{Bmatrix} \omega_{3/2}\boldsymbol{\hat{k}}\\ {\bf 0}\end{Bmatrix}_A -\begin{Bmatrix} \dot{\theta}_3 \boldsymbol{\hat{k}}\\ {\bf 0}\end{Bmatrix}_O =\begin{Bmatrix} {\bf 0}\\ {\bf 0}\end{Bmatrix} \] To properly guarantee this equality, we resolve all screws about the same point (we choose \(A\)): \[ \begin{Bmatrix} {\bf 0}\\ \dot{y}_B \boldsymbol{\hat{\jmath}}\end{Bmatrix}_A +\begin{Bmatrix} \omega_{2/1}\boldsymbol{\hat{k}}\\ \omega_{2/1}\boldsymbol{\hat{k}}\times {\bf r}_{IA} \end{Bmatrix}_A +\begin{Bmatrix} \omega_{3/2}\boldsymbol{\hat{k}}\\ {\bf 0}\end{Bmatrix}_A -\begin{Bmatrix} \dot{\theta}_3 \boldsymbol{\hat{k}}\\ \dot{\theta}_3\boldsymbol{\hat{k}}\times {\bf r}_{OA} \end{Bmatrix}_A =\begin{Bmatrix} {\bf 0}\\ {\bf 0}\end{Bmatrix} \] This gives the following equations: \[ \omega_{2/1}+\omega_{3/2}=\dot{\theta}_3, \qquad \dot{y}_B\boldsymbol{\hat{\jmath}}+ \omega_{2/1}\boldsymbol{\hat{k}}\times r\boldsymbol{\hat{\jmath}}=\dot{\theta}_3 \boldsymbol{\hat{k}}\times l(\cos\theta_3 \boldsymbol{\hat{\imath}}+\sin\theta_3 \boldsymbol{\hat{\jmath}}) \] These equations allow for the determination of the kinematic outputs \((\dot{\theta}_3,\omega_{2/1},\omega_{3/2})\) in terms of input \(\dot{y}_B\): \[ \dot{\theta}_3 =\frac{\dot{y}_B}{l\cos\theta_3} , \qquad \omega_{2/1}= \frac{\dot{y}_B}{r}\tan\theta_3, \qquad \omega_{3/2} = \frac{\dot{y}_B}{l\cos\theta_3} (1- \frac{l}{r} \sin\theta_3) \]