# 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 $$\dot{y}_B$$: $\dot{\theta}_3 =\frac{\dot{y}_B}{l\cos\theta_3} , \qquad \omega_{2/1}= \frac{\dot{y}_B}{r}\tan\theta_3, \qquad \omega_{3/2} = \frac{\dot{y}_B}{l\cos\theta_3} (1- \frac{l}{r} \sin\theta_3)$

##### Reference: Advanced Engineering Dynamics, R. Valéry Roy, Hyperbolic Press, (2015). 