Hoeken Linkage [gnuplot]

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Formulation

Let $$\begin{eqnarray*} OA:OB:AC:BC:CD=1:2:\frac{5}{2}:\frac{5}{2}:\frac{5}{2} \end{eqnarray*}$$ $$\begin{cases} O(0, 0)\\ A(r\cos\theta, r\sin\theta)\\ B(2r, 0)\\ \overrightarrow{AC}=(p,q)^T \end{cases}$$ Then $$\begin{eqnarray*} AC^2 = CB^2 \end{eqnarray*}$$ $$\begin{eqnarray} \labeq{condition} p^2+q^2 &= \left[(r\cos\theta+p)-2r\right]^2 + \left[(r\sin\theta+q)-0\right]^2 \end{eqnarray}$$ By finding $p$ versus $q$ from eq. ($\ref{eq:condition}$), we obtain $$\begin{equation} \labeq{pvsq} p = \frac{(2\sin\theta)q+(5-4\cos\theta)r}{2(2-\cos\theta)} \end{equation}$$ Considering the Euclidean norm of $\overrightarrow{AC}$, we have the relational expression between $p$ and $q$. $$\begin{equation} \labeq{norm} \|\overrightarrow{AC}\|^2=p^2+q^2=\frac{25}{4}r^2 \end{equation}$$ Inserting eq. ($\ref{eq:pvsq}$) into eqs. ($\ref{eq:norm}$) and eliminating $p$, we get $$\begin{equation} \labeq{q} q = -\frac{r\sin\theta}{2} + r(2-\cos\theta)\sqrt{\frac{5+\cos\theta}{5-4\cos\theta}} \end{equation}$$ Substitution of eq. ($\ref{eq:q}$) in eq. ($\ref{eq:pvsq}$) leads to eq. ($\ref{eq:p}$). $$\begin{equation} \labeq{p} p = \frac{r(2-\cos\theta)}{2}+r\sin\theta\sqrt{\frac{5+\cos\theta}{5-4\cos\theta}} \end{equation}$$ On the other hand, the position of joint $C$ and $D$ are $$\begin{eqnarray} \labeq{posC} \overrightarrow{OC} &=& \overrightarrow{OA} + \overrightarrow{AC} &=& (r\cos\theta+p, r\sin\theta+q)^T\\ \labeq{posD} \overrightarrow{OD} &=& \overrightarrow{OA} + 2\overrightarrow{AC} &=& (r\cos\theta+2p, r\sin\theta+2q)^T \end{eqnarray}$$ Inserting eq. ($\ref{eq:q}$), ($\ref{eq:p}$) into eqs. ($\ref{eq:posC}$), we get $$\begin{eqnarray} \labeq{posC2} \left\{ \begin{array}{l} C_x=\frac{r\left(2+\cos\theta\right)}{2}+r\sin\theta\sqrt{\frac{5+\cos\theta}{5-4\cos\theta}} \\ C_y=\frac{r\sin\theta}{2}+r(2-\cos\theta)\sqrt{\frac{5+\cos\theta}{5-4\cos\theta}} \end{array} \right. \end{eqnarray}$$ Likewise, substitution of eq. ($\ref{eq:q}$), ($\ref{eq:p}$) in eqs. ($\ref{eq:posD}$) leads to eqs. ($\ref{eq:posD2}$). $$\begin{eqnarray} \labeq{posD2} \left\{ \begin{array}{l} D_x=2r+2r\sin\theta\sqrt{\frac{5+\cos\theta}{5-4\cos\theta}} \\ D_y=2r\left(2-\cos\theta\right)\sqrt{\frac{5+\cos\theta}{5-4\cos\theta}} \end{array} \right. \end{eqnarray}$$

Simulation [gnuplot]

Source code (PLT file)

Output (GIF file)

case1: $r=1$

case2: $r=2$