Acceleration Analysis (Vector Loop Method)
Fourbar Linkage Vector Loop Equations
The derivation of the vector loop equation for a fourbar linkage mechanism involves defining the position vectors of the links and the angles between them. The fourbar linkage consists of four links connected by four joints, forming a closed loop as shown in the figure below.
Vector Loop Equation
The vector loop equation is derived by summing the position vectors around the closed loop of the fourbar linkage and setting their sum to zero:
\[\mathbf{R}_{A} + \mathbf{R}_{BA} - \mathbf{R}_{BO4} - \mathbf{R}_{O4} = 0 \nonumber\]To obtain an alternative form of the equations, often used in analysis, complex number notation can be used for the position vectors. Using complex notation, the following is obtained
\[a e^{j\theta_2} + b e^{j\theta_3} - c e^{j\theta_4} - d e^{j\theta_1} = 0 \nonumber\]where $a$, $b$, $c$ and $d$ represent the scalar length of each of the links.
Euler’s equation, $e^{j\theta} = (\cos \theta + j \sin \theta)$ may also be substituted into the equations to obtain this form
\[\begin{align*} &a\left(\cos \theta_2 + j \sin \theta_2\right) + b\left(\cos \theta_3 + j \sin \theta_3\right) - \\ & c\left(\cos \theta_4 + j \sin \theta_4\right) - d\left(\cos \theta_1+j \sin \theta_1\right) = 0 \end{align*}\]Solving for Accelerations
Determining accelerations from the vector loop equation follows the same structure as the position and velocity analyses.
Previously, in Velocity Analysis we differentiated the position loop
\[a e^{j\theta_2} + b e^{j\theta_3} - c e^{j\theta_4} - d e^{j\theta_1} = 0.\]Because the ground link is fixed, $\theta_1$ is constant and its time derivatives are zero. The differentiated velocity loop is therefore
\[j a \omega_2 e^{j\theta_2}+j b \omega_3 e^{j\theta_3}-j c \omega_4 e^{j\theta_4}=0.\]Differentiating once more with respect to time gives
\[\left(j a \alpha_2 e^{j\theta_2}-a \omega_2^2 e^{j\theta_2}\right) +\left(j b \alpha_3 e^{j\theta_3}-b \omega_3^2 e^{j\theta_3}\right) -\left(j c \alpha_4 e^{j\theta_4}-c \omega_4^2 e^{j\theta_4}\right)=0.\]This equation contains both tangential terms proportional to $\alpha$ and normal terms proportional to $\omega^2$. In vector form, it is the acceleration loop
\[\mathbf{A}_A+\mathbf{A}_{BA}-\mathbf{A}_B=0.\]where
\(\mathbf{A}_A=\left(\mathbf{A}_A^t+\mathbf{A}_A^n\right)=\left(j a \alpha_2 e^{j\theta_2}-a\omega_2^2 e^{j\theta_2}\right),\) \(\mathbf{A}_{BA}=\left(\mathbf{A}_{BA}^t+\mathbf{A}_{BA}^n\right)=\left(j b \alpha_3 e^{j\theta_3}-b\omega_3^2 e^{j\theta_3}\right),\) \(\mathbf{A}_B=\left(\mathbf{A}_B^t+\mathbf{A}_B^n\right)=\left(j c \alpha_4 e^{j\theta_4}-c\omega_4^2 e^{j\theta_4}\right).\)
Similarly to the process in Velocity Analysis, substitute $e^{j\theta}=\cos\theta+j\sin\theta$ to give
\[\begin{aligned} &\left[a \alpha_2 j\left(\cos\theta_2+j\sin\theta_2\right)-a \omega_2^2\left(\cos\theta_2+j\sin\theta_2\right)\right] \\ &\quad+\left[b \alpha_3 j\left(\cos\theta_3+j\sin\theta_3\right)-b \omega_3^2\left(\cos\theta_3+j\sin\theta_3\right)\right] \\ &\quad-\left[c \alpha_4 j\left(\cos\theta_4+j\sin\theta_4\right)-c \omega_4^2\left(\cos\theta_4+j\sin\theta_4\right)\right]=0 \end{aligned}\]which simplifies to
\[\begin{aligned} &\left[a \alpha_2\left(-\sin\theta_2+j\cos\theta_2\right)-a \omega_2^2\left(\cos\theta_2+j\sin\theta_2\right)\right] \\ &\quad+\left[b \alpha_3\left(-\sin\theta_3+j\cos\theta_3\right)-b \omega_3^2\left(\cos\theta_3+j\sin\theta_3\right)\right] \\ &\quad-\left[c \alpha_4\left(-\sin\theta_4+j\cos\theta_4\right)-c \omega_4^2\left(\cos\theta_4+j\sin\theta_4\right)\right]=0 \end{aligned}\]Collecting the terms into real and imaginary components gives, first, the real part:
\[-a \alpha_2 \sin \theta_2-a \omega_2^2 \cos \theta_2-b \alpha_3 \sin \theta_3-b \omega_3^2 \cos \theta_3+c \alpha_4 \sin \theta_4+c \omega_4^2 \cos \theta_4=0,\]and then the imaginary part:
\[a \alpha_2 \cos \theta_2-a \omega_2^2 \sin \theta_2+b \alpha_3 \cos \theta_3-b \omega_3^2 \sin \theta_3-c \alpha_4 \cos \theta_4+c \omega_4^2 \sin \theta_4=0.\]These two equations are solved simultaneously for $\alpha_3$ and $\alpha_4$:
\[\begin{aligned} \alpha_3 & =\frac{(a \alpha_2 \sin \theta_2 + a \omega_2^2 \cos \theta_2 + b \omega_3^2 \cos \theta_3 - c \omega_4^2 \cos \theta_4) (c \cos \theta_4)}{(c \sin \theta_4)(b \cos \theta_3)-(b \sin \theta_3)(c \cos \theta_4)} \\ & \quad - \frac{(c \sin \theta_4)(a \alpha_2 \cos \theta_2 - a \omega_2^2 \sin \theta_2 - b \omega_3^2 \sin \theta_3 + c \omega_4^2 \sin \theta_4)}{(c \sin \theta_4)(b \cos \theta_3)-(b \sin \theta_3)(c \cos \theta_4)} \\ \alpha_4 & =\frac{(a \alpha_2 \sin \theta_2 + a \omega_2^2 \cos \theta_2 + b \omega_3^2 \cos \theta_3 - c \omega_4^2 \cos \theta_4)(b \cos \theta_3)}{(c \sin \theta_4)(b \cos \theta_3)-(b \sin \theta_3)(c \cos \theta_4)} \\ & \quad - \frac{(b \sin \theta_3)(a \alpha_2 \cos \theta_2-a \omega_2^2 \sin \theta_2-b \omega_3^2 \sin \theta_3+c \omega_4^2 \sin \theta_4)}{(c \sin \theta_4)(b \cos \theta_3)-(b \sin \theta_3)(c \cos \theta_4)} \end{aligned}\]Having calculated $\alpha_3$ and $\alpha_4$, the linear accelerations $\mathbf{A}A$, $\mathbf{A}{BA}$, and $\mathbf{A}_B$ can be determined directly from the complex rigid-body acceleration expressions.
\[\begin{aligned} \mathbf{A}_A & =a \alpha_2\left(-\sin \theta_2+j \cos \theta_2\right)-a \omega_2^2\left(\cos \theta_2+j \sin \theta_2\right) \\ \mathbf{A}_{BA} & =b \alpha_3\left(-\sin \theta_3+j \cos \theta_3\right)-b \omega_3^2\left(\cos \theta_3+j \sin \theta_3\right) \\ \mathbf{A}_B & =c \alpha_4\left(-\sin \theta_4+j \cos \theta_4\right)-c \omega_4^2\left(\cos \theta_4+j \sin \theta_4\right) \end{aligned}\]