Velocity 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 Velocities
Determining velocities from the vector loop equation is a fairly simple process.
Differentiating the loop equation from above gives
\[a e^{j\theta_2}+b e^{j\theta_3}-c e^{j\theta_4}-d e^{j\theta_1}=0\]with respect to time:
\[j a e^{j\theta_2}\frac{d\theta_2}{dt} +j b e^{j\theta_3}\frac{d\theta_3}{dt} -j c e^{j\theta_4}\frac{d\theta_4}{dt} -j d e^{j\theta_1}\frac{d\theta_1}{dt}=0\]Since the ground link is fixed, $\theta_1$ is constant and the last term is zero. The resulting velocity loop is
\[\mathbf{V}_A+\mathbf{V}_{BA}-\mathbf{V}_B=0\]where
\[\mathbf{V}_A=j a \omega_2 e^{j\theta_2}, \qquad \mathbf{V}_{BA}=j b \omega_3 e^{j\theta_3}, \qquad \mathbf{V}_B=j c \omega_4 e^{j\theta_4}\]using the substitution $\frac{d\theta}{dt}=\omega$. The relative velocity equation may therefore be written as
\[j a \omega_2 e^{j\theta_2}+j b \omega_3 e^{j\theta_3}-j c \omega_4 e^{j\theta_4}=0\]The above can be solved to find $\omega_3$ and $\omega_4$. First substitute Euler’s identity, $e^{j\theta}=\cos\theta+j\sin\theta$, and then multiply through by $j$ to give
\[a\omega_2(-\sin\theta_2+j\cos\theta_2)+b\omega_3(-\sin\theta_3+j\cos\theta_3)-c\omega_4(-\sin\theta_4+j\cos\theta_4)=0\]Separating into real and imaginary components gives, first, the real part:
\[-a\omega_2\sin\theta_2-b\omega_3\sin\theta_3+c\omega_4\sin\theta_4=0\]and second, the imaginary part:
\[a\omega_2\cos\theta_2+b\omega_3\cos\theta_3-c\omega_4\cos\theta_4=0\]These equations can be solved simultaneously to give
\(\omega_3=\omega_2\frac{a\sin(\theta_4-\theta_2)}{b\sin(\theta_3-\theta_4)}\) \(\omega_4=\omega_2\frac{a\sin(\theta_2-\theta_3)}{c\sin(\theta_4-\theta_3)}\)
Once $\omega_3$ and $\omega_4$ are known, it is possible to find $\mathbf{V}A$, $\mathbf{V}{BA}$, and $\mathbf{V}_B$.
\[\begin{align*} \mathbf{V}_A &= j a \omega_2(\cos\theta_2+j\sin\theta_2) \\ &= a \omega_2(-\sin\theta_2+j\cos\theta_2) \\[1em] \mathbf{V}_{BA} &= j b \omega_3(\cos\theta_3+j\sin\theta_3) \\ &= b \omega_3(-\sin\theta_3+j\cos\theta_3) \\[1em] \mathbf{V}_B &= j c \omega_4(\cos\theta_4+j\sin\theta_4) \\ &= c \omega_4(-\sin\theta_4+j\cos\theta_4) \end{align*}\]To determine these velocities, the current values of $\theta_3$ and $\theta_4$ must already be known from the position analysis. Given $\theta_2$ and the link lengths $a$, $b$, $c$, and $d$, those angles can be found using the method discussed in Position Analysis.