Justin
(1)
\begin{align} \vec{r_1} + \vec{r_2} + \vec{r_3} + \vec{r_4} = 0 \end{align}

where,

(2)
\begin{split} & \vec{r_1} - Ground Vector && \vec{r_2} - Vector for Link2 && \vec{r_3} - Vector for Link3 &&\vec{r_4} - Vector for Link4 \end{split}

For a fourbar, the link 2 is considered as the input i.e. the crank. Depending on the variation of the crank angle, the coupler angle and the follower angle can be found from the loop equation adn hence their respective positions.

(3)
\begin{align} \theta_4 = 2 \arctan \left[ \frac {-B \pm \sqrt {B^2 - 4AC}}{2A} \right] \end{align}
(4)
\begin{align} \theta_3 = 2 \arctan \left[ \frac {-E \pm \sqrt {E^2 - 4DF}}{2D} \right] \end{align}

where,

(5)
\begin{align} A = \cos\theta_2 - k1-k2\cos\theta_2 + k3 \end{align}
(6)
\begin{align} B = -2\sin\theta_2 \end{align}
(7)
\begin{align} C = k1 -(k2+1)\cos\theta_2+k3 \end{align}
(8)
\begin{align} D = \cos\theta_2-k1-k4\cos\theta_2+k5 \end{align}
(9)
\begin{align} E = -2\sin\theta_2 \end{align}
(10)
\begin{align} F = -k1+(k4-1)\cos\theta_2+k5 \end{align}
(11)
\begin{align} k1= \frac {r_4}{r_1} \end{align}
(12)
\begin{align} k2= \frac{r_4}{r_3} \end{align}
(13)
\begin{align} k3= \frac{\left(r_1^2-r_2^2+r_3^2+r_4^2)\right}{2r_1r_3} \end{align}
(14)
\begin{align} k4= \frac{r_4}{r_2} \end{align}
(15)
\begin{align} k5= \frac{\left(r_3^2-r_4^2+r_1^2+r_2^2)\right}{2r_1r_2} \end{align}

Velocity Analysis

The Link1 is the ground link which is static. Therefore velocity of Link1 is zero

For Link2

(16)
\begin{align} \vec{r_2} = r_2e^{i\theta_2} \end{align}
(17)
\begin{align} v_2 = r_2 \dot{\theta_2} i e^{i\theta_2} \end{align}
(18)
\begin{align} v_{2x} = -r_2 \dot{\theta_2} \sin{\theta_2} \end{align}
(19)
\begin{align} v_{2y} = r_2 \dot{\theta_2} \cos{\theta_2} \end{align}
(20)
\begin{align} a_{2x} = -r_2 \ddot{\theta_2} \sin\theta_2 - r_2 \dot{\theta_2}^2 \cos\theta_2 \end{align}
(21)
\begin{align} a_{2y} = r_2 \ddot{\theta_2} \cos\theta_2 - r_2 \dot{\theta_2}^2 \sin\theta_2 \end{align}

For Link 3

(22)
\begin{align} \vec{r_3} = r_3e^{i\theta_3} \end{align}
(23)
\begin{align} v_3 = r_3 \dot{\theta_3} i e^{i\theta_3} \end{align}
(24)
\begin{align} v_{3x} = -r_3 \dot{\theta_3} \sin{\theta_3} \end{align}
(25)
\begin{align} v_{3y} = r_3 \dot{\theta_3} \cos{\theta_3} \end{align}
(26)
\begin{align} a_{3x} = -r_3 \ddot{\theta_3} \sin\theta_3 - r_3 \dot{\theta_3}^2 \cos\theta_3 \end{align}
(27)
\begin{align} a_{3y} = r_3 \ddot{\theta_3} \cos\theta_3 - r_3 \dot{\theta_3}^2 \sin\theta_3 \end{align}

For Link4

(28)
\begin{align} \vec{r_4} = r_4e^{i\theta_4} \end{align}
(29)
\begin{align} v_4 = r_4 \dot{\theta_4} i e^{i\theta_4} \end{align}
(30)
\begin{align} v_{4x} = -r_4 \dot{\theta_4} \sin\theta_4 \end{align}
(31)
\begin{align} v_{4y} = r_4 \dot{\theta_4} \cos{\theta_4} \end{align}
(32)
\begin{align} a_{4x} = -r_4 \ddot{\theta_4} \sin\theta_4 - r_4 \dot{\theta_4}^2 \cos\theta_4 \end{align}
(33)
\begin{align} a_{4y} = r_4 \ddot{\theta_4} \cos\theta_4 - r_4 \dot{\theta_4}^2 \sin\theta_4 \end{align}

For finding velocities

(34)
\begin{align} L_1 = \vec{r_1} + \vec{r_2} + \vec{r_3} + \vec{r_4} = 0 \end{align}
(35)
\begin{align} \frac {d L_1}{dt} = r_2 \dot{\theta_2} i e^{i\theta_2} + r_3 \dot{\theta_3} i e^{i\theta_3} + r_4 \dot{\theta_4} i e^{i\theta_4} = 0 \end{align}
(36)
\begin{align} \frac {d L_{1x}}{dt} = -r_2 \dot{\theta_2} \sin\theta_2 - r_3 \dot{\theta_3} \sin{\theta_3} - r_4 \dot{\theta_4} \sin{\theta_4} = 0 \end{align}
(37)
\begin{align} \frac {d L_{1y}}{dt} = r_2 \dot{\theta_2} \cos\theta_2 + r_3 \dot{\theta_3} \cos{\theta_3} + r_4 \dot{\theta_4} \cos{\theta_4} = 0 \end{align}

writing Eqns (36) and (37) in matrix form, we get

(38)
\begin{align} \begin{bmatrix} \--r_3\sin\theta_3& \--r_4\sin\theta_4\\ \-r_3\cos\theta_3& \-r_4\cos\theta_4 \end{bmatrix} \begin{bmatrix} \-\dot{\theta_3}\\ \-\dot{\theta_4} \end{bmatrix} = \begin{bmatrix} \-r_2\sin\theta_2\\ \--r_2\cos\theta_2 \end{bmatrix} \dot{\theta_2} \end{align}
(39)
\begin{align} \begin{bmatrix} \-\dot{\theta_3}\\ \-\dot{\theta_4} \end{bmatrix} = \begin{bmatrix} \--r_3\sin\theta_3& \--r_4\sin\theta_4\\ \-r_3\cos\theta_3& \-r_4\cos\theta_4 \end{bmatrix}^{-1} \begin{bmatrix} \-r_2\sin\theta_2\\ \--r_2\cos\theta_2 \end{bmatrix} \dot{\theta_2} \end{align}

Thus the velocities are found.

Acceleration analysis

For finding accelerations

we differentiate the loop equation twice with respect to time.

(40)
\begin{align} \frac {d^2 L_{1x}}{dt} = -r_2 \ddot{\theta_2} \sin\theta_2 - r_2 \dot{\theta_2}^2\cos\theta_2 - r_3 \ddot{\theta_3} \sin{\theta_3} - r_3 \dot{\theta_3}^2\cos\theta_3 - r_4 \ddot{\theta_4} \sin{\theta_4} - r_4 \dot{\theta_4}^2\cos\theta_4= 0 \end{align}
(41)
\begin{align} \frac {d^2 L_{1y}}{dt} = r_2 \ddot{\theta_2} \cos\theta_2 - r_2 \dot{\theta_2}^2\sin\theta_2 + r_3 \ddot{\theta_3} \cos{\theta_3} - r_3 \dot{\theta_3}^2\sin\theta_3 + r_4 \ddot{\theta_4} \cos{\theta_4} - r_4 \dot{\theta_4}^2\sin\theta_4= 0 \end{align}

writing Eqns (40) & (41) in matrix form we get

(42)
\begin{align} \begin{bmatrix} \--r_3\sin\theta_3& \--r_4\sin\theta_4\\ \-r_3\cos\theta_3& \-r_4\cos\theta_4 \end{bmatrix} \begin{bmatrix} \-\ddot{\theta_3}\\ \-\ddot{\theta_4} \end{bmatrix} = \begin{bmatrix} \-r_2\sin\theta_2\\ \--r_2\cos\theta_2 \end{bmatrix} \ddot{\theta_2} + \begin{bmatrix} \-r_2\cos\theta_2\\ \-r_2\sin\theta_2 \end{bmatrix} \dot{\theta_2}^2 + \begin{bmatrix} \-r_3\cos\theta_3\\ \-r_3\sin\theta_3 \end{bmatrix} \dot{\theta_3}^2 + \begin{bmatrix} \-r_4\cos\theta_4\\ \-r_4\sin\theta_4 \end{bmatrix} \dot{\theta_4}^2 \end{align}
(43)
\begin{align} \begin{bmatrix} \-M\\ \end{bmatrix} \left(- \begin{bmatrix} \-V(q,\.q)\\ \end{bmatrix}- \begin{bmatrix} \-G(q)\\ \end{bmatrix}- \begin{bmatrix} \-F(q)\\ \end{bmatrix} \right) \end{align}

Force Analysis

After finding the respective velocities and accelerations, the modelling of the components was done using Pro-E to obtain the masses and the mass moment of inertia of the respective components. These are required to compute the forces in the joints whilst using the matrix method of force analysis. The component properties are shown below.

(44)
\begin{split} &m_2=71.83 g &&I_{g2}=1.1454*10^{-4} kgm^{2} &&m_3=2 kg &&I_{g3}=716.41*10^{-4} kgm^{2} &&m_4=92.69 g &&I_{g4}=2.3878*10^{-4} kgm^{2} \end{split}

Matrix method

'C' is defined by the matrix shown below.

(45)
\begin{align} \setcounter{MaxMatrixCols}{20} C = \begin{bmatrix} \-1&0&1&0&0&0&0&0&0&0&0&0&0\\ \-0& 1& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ \--r_{21y}&r_{21x}&-r_{22y}&r_{22x}& 0& 0& 0& 0& 0& 0& 0& 0& 1\\ \-0& 0& 0& 0& 1& 0& 1& 0& 0& 0& 0& 0& 0\\ \-0& 0& 0& 0& 0& 1& 0& 1& 0& 0& 0& 0& 0\\ \-0& 1& 0& 1& 0& -r_{32y}& r_{32x}& -r_{33y}& r_{33x}& 0& 0& 0& 0\\ \-0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 1& 0& 0\\ \-0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 1& 0\\ \-0& 0& 0& 0& 0& 0& 0& 0& -r_{43y}& r_{43x}& -r_{44y}& r_{44x}& 0\\ \-0& 0& 1& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ \-0& 0& 0& 1& 0& 1& 0& 0& 0& 0& 0& 0& 0\\ \-0& 0& 0& 0& 1& 0& 1& 0& 0& 0& 0& 0& 0\\ \-0& 0& 0& 0& 0& 1& 0& 1& 0& 0& 0& 0& 0\\ \end{bmatrix} \end{align}

and 'm' is defined by the matrix

(46)
\begin{align} m = \begin{bmatrix} \-m_2a_{g2x}\\ \-m_2a_{g2y} + m_2g\\ \-I_{g2}\alpha_2\\ \-m_3a_{g3x}\\ \-m_3a_{g3y}+m_3g\\ \-I_{g3}\alpha_3\\ \-m_4a_{g4x}\\ \-m_4a_{g4y}+m_4g\\ \-I_{g4}\alpha_4\\ \-0\\ \-0\\ \-0\\ \-0\\ \end{bmatrix} \end{align}

'X_{s}' is the state matrix that is defined by

(47)
\begin{align} X_{s} = \begin{bmatrix} \-l \\ \-i \\ \-w\\ \-ctwd\\ \-x\\ \-y\\ \-z\\ \-\theta_{W}\\ \-v_{x}\\ \-v_{y}\\ \-v_{z}\\ \-\omega_{W}\\ \end{bmatrix} = \begin{bmatrix} \-x_{1} \\ \-x_{2} \\ \-x_{3}\\ \-x_{4}\\ \-x_{5}\\ \-x_{6}\\ \-x_{7}\\ \-x_{8}\\ \-x_{9}\\ \-x_{10}\\ \-x_{11}\\ \-x_{12}\\ \end{bmatrix} \end{align}

'X_{s}' is the state matrix that is defined by

(48)
\begin{align} \.X_{s} = \begin{bmatrix} \-\frac{d X_{s}}{dt}\\ \end{bmatrix} = \begin{bmatrix} \-\.l \\ \-\.i \\ \-\.w\\ \-ct\.wd\\ \-\.x\\ \-\.y\\ \-\.z\\ \-\omega_{W}\\ \-\.v_{x}\\ \-\.v_{y}\\ \-\.v_{z}\\ \-\alpha_{W}\\ \end{bmatrix} = \begin{bmatrix} \-\.x_{1} \\ \-\.x_{2} \\ \-\.x_{3}\\ \-\.x_{4}\\ \-x_{9}\\ \-x_{10}\\ \-x_{11}\\ \-x_{12}\\ \-\.x_{9}\\ \-\.x_{10}\\ \-\.x_{11}\\ \-\.x_{12}\\ \end{bmatrix} \end{align}
(49)
\begin{align} \.x_{1}=u_{1}-(\frac{C_{2} \rho}{\pi r_{w}^{2}}) x_{1} x_{2}^{2} +(\frac{C_{1} \rho}{\pi r_{w}^{2}}) x_{2} \end{align}
(50)
\begin{align} \.x_{2}=u_{2}-(R_{a}+R_{s}+ \rho x_{1}) x_{2} - V_{0} - E_{a}(x_{4}-x_{1}) \end{align}
(51)
\begin{align} \.x_{3}=\frac{-x_{3} - \frac{\kappa_{1}}{\vec{V_{torch}}} + \kappa_{2} X_{2} - c}{\tau}} \end{align}
(52)
\begin{align} \.x_{4}=x_{0}-S(y) \end{align}
(53)
\begin{bmatrix} \-\.x_{5} \\ \-\.x_{6} \\ \-\.x_{7}\\ \-\.x_{8}\\ \end{bmatrix} = \begin{bmatrix} \-x_{9} \\ \-x_{10} \\ \-x_{11}\\ \-x_{12}\\ \end{bmatrix} = \begin{bmatrix} \-v_{x} \\ \-v_{y} \\ \-v_{z}\\ \-\omega_{W}\\ \end{bmatrix} = K_{eq} \begin{bmatrix} \-\omega_{l}\\ \-\omega_{r}\\ \-d_{4}\\ \-d_{1} + \omega_{4}\\ \end{bmatrix}
(54)
\begin{align} \begin{bmatrix} \-\.x_{9} \\ \-\.x_{10} \\ \-\.x_{11}\\ \-\.x_{12}\\ \end{bmatrix} = \begin{bmatrix} \-\"v_{x} \\ \-\"v_{y} \\ \-\"v_{z}\\ \-\alpha_{W}\\ \end{bmatrix} = \begin{bmatrix} \-M\\ \end{bmatrix} \left(- \begin{bmatrix} \-V(q,\.q)\\ \end{bmatrix}- \begin{bmatrix} \-G(q)\\ \end{bmatrix}+ \begin{bmatrix} \-F\\ \end{bmatrix} \right) \end{align}
(55)
\begin{align} \-MIN(f(q)) = \vec{e}^T \begin{bmatrix} \-W \end{bmatrix} \-\vec{e} \end{align}