Velocity Analysis

Some notation before we get started

$R_P=p{e}^{j\theta }$
$V_P=pj{e}^{j\theta }\frac{d\theta }{dt}$ which can be represented in the following form: $V_p=p\omega j\left(\cos \theta +jsin\theta \right)$

$V_{PA}$ is the linear velocity of point P, and is always perpendicular to $R_P$ and tangent to the circular path of travel.

There are different types of velocity:

• Absolute Velocity: Velocity with respect to some point on a global coordinate system
• Velocity Difference: Velocity with respect to a point on the same body that isn’t fixed

Vector Loop Analysis for Four Bar Linkages

![[Pasted image 20240718130025.png|650]]

Notes

• Note that $V_A$ is always perpendicular to $R_2$
• The same thing applied with $V_B$ and $R_4$ Steps

1. Check the reference frame $R_1$ should have no angle w.r.t the X-axis, otherwise change the coordinate system
2. Check to see if you have all the angles you need, you may need to do position analysis to solve ${\theta }_3$ and ${\theta }_4$
3. Calculator angular velocities ${\omega }_3$ and ${\omega }_4$
   ${\omega }_3={\frac{a\omega 2}{b}}_{}\frac{\sin \left({\theta }_4-{\theta }_2\right)}{\sin \left({\theta }_3-{\theta }_4\right)}$
   ${\omega }_4={\frac{a\omega 2}{c}}_{}\frac{\sin \left({\theta }_2-{\theta }_3\right)}{\sin \left({\theta }_4-{\theta }_3\right)}$
4. Solve for the linear velocities, note that if you need to solve for open and crossed positions, you need to repeat steps 3 and 4 twice, once with ${\theta }_3$ and ${\theta }_4$ for the open position, and once for the closed position. Here $V_A$ is of point A on link 2 and $V_B$ is of point B on link 4

$$\begin{array}{rl}{\mathbf{V}}_A& =ja{\omega }_2\left(\cos {\theta }_2+j\sin {\theta }_2\right)=a{\omega }_2\left(-\sin {\theta }_2+j\cos {\theta }_2\right)\\ {\mathbf{V}}_{BA}& =jb{\omega }_3\left(\cos {\theta }_3+j\sin {\theta }_3\right)=b{\omega }_3\left(-\sin {\theta }_3+j\cos {\theta }_3\right)\\ {\mathbf{V}}_B& =jc{\omega }_4\left(\cos {\theta }_4+j\sin {\theta }_4\right)=c{\omega }_4\left(-\sin {\theta }_4+j\cos {\theta }_4\right)\end{array}$$

Vector Loop Analysis For Crank-Sliders

Here, the crank is the input and the slider is the output

1. Check the reference frame, $R_1$ should have no angle similar to step 1 for four bar linkages
2. Check for missing angles (same as step 2 for four bar linkages)
3. Calculate angular velocity ${\omega }_3$ and linear velocity of the slider block

$$\begin{array}{rl}\dot{d}& =-a{\omega }_2\sin {\theta }_2+b{\omega }_3\sin {\theta }_3\\ {\omega }_3& =\frac{a\cos {\theta }_2}{b\cos {\theta }_3}{\omega }_2\end{array}$$

4. Solve for the linear velocities. Similar to step 4 in the four bar linkage process, you may need to solve steps 3 and 4 twice for both the open and crossed configurations

$$\begin{array}{rl}{\mathbf{V}}_A& =a{\omega }_2\left(-\sin {\theta }_2+j\cos {\theta }_2\right)\\ {\mathbf{V}}_{AB}& =b{\omega }_3\left(-\sin {\theta }_3+j\cos {\theta }_3\right)\\ {\mathbf{V}}_{BA}& =-{\mathbf{V}}_{AB}\\ {\mathbf{V}}_B& =-a{\omega }_2\sin ({\theta }_2)+b{\omega }_3\sin ({\theta }_3)\end{array}$$

Tips from solving questions

For four bar linkages:

• Just like position analysis but now you also apply the V and $\omega$ equations
• ${\theta }_2$ should be given as a part of the problem, if not, try to get it using trig (you should not have to do this)??? For slider cranks:
• Given offsets are actually negative ($-c$)
• Calculate ${\theta }_3$ and $d$
• Calculate ${\omega }_3$
• Calculate $V_a$ and angle
• Calculate $V_b$ Generally pretty straight forward