PVA Intro

Position, Velocity, Acceleration (PVA) analysis is used to analytically determine the motion of a mechanism. Position determines the configuration of the mechanism. Velocity is used to understand the kinetic energy and momentum of the system. Finally, acceleration is used to understand the forces in the system. PVA takes the known geometry, initial configurations, and PVA data of one link (the inputs) and determines the PVA data for the other links (the outputs).

Vector Calculus

In PVA analysis, links are represented using vectors. Using complex vectors is the easiest way to represent the link data as it utilizes the link length \( |R| \) and starting link angle \( \theta \) to create a vector. There are two forms of complex vectors that we will be focusing on: complex number form and polar form. Complex number form takes a Cartesian approach to complex numbers. Meanwhile polar form takes a polar approach (as the name suggests).

The \( j \) in each vector represents the imaginary number \( sqrt{-1 \). The Euler formula can be used to break the polar form into the complex form.

Let's look at the polar vector form. Angle \( \theta \) depends on time because the system is not static. This means that taking the derivative of the angle with respect to time will give the angular velocity \( \omega \).

Also note that when you rotate a vector \( \overrightarrow R_{AB} \) by 180 degrees, it becomes \( -\overrightarrow R_{AB} \). Upon inspection, we can see that \( -\overrightarrow R_{AB} \) is also equal to \( \overrightarrow R_{BA} \). This can be a useful transformation to use when creating vector loops.

Vector Loops

Vector loops are used to create equations that can be used to solve complex linkage problems. Vector loops work by creating a "loop" of vectors that start and end in the same location. This also means that the sum of these vectors is equal to zero. For example, take a look at a four bar linkage. If you choose a joint and trace around the links in one direction (making position vectors along the way) you end up creating a vector loop.

You can start the vector loop at any joint and go in any direction to finish the loop.

Alternatively, you can start at one point and go two different directions to end at a different point. As long as the two paths start and end at the same point, a vector loop can be created.

Since position, velocity, and acceleration are all path-independent vectors, we can usually write an equation in the form of \( \Sigma_0^n \overrightarrow R_n=0 \). This gives us an equation we are able to solve.

Position Analysis

1. Identify known and unknown variables. In the four-bar linkage example, the linkage lengths (\( a \), \( b \), \( c \), and \( d \)) and the initial angle of the crank \( (\theta_2) \) is known. The angle of the grounded link \( (\theta_1) \) is known to be zero. The initial angle of the coupler \( (\theta_3) \) and rocker \( (\theta_4) \) are unknown.

2. Establish a vector loop and write out the vectors in polar form. Recall that the polar form of a vector is in the form of \( \overrightarrow R=|R| e^{j\theta} \). Since the vector loop we established uses two different paths to get from the same starting point to the same ending point, we have to set the sum of the two paths equal to each other.

Position Vector Loop Equations
$$ \begin{gather} \overrightarrow R_2+\overrightarrow R_3=\overrightarrow R_4+\overrightarrow R_1 \\ \overrightarrow R_2+\overrightarrow R_3-\overrightarrow R_4-\overrightarrow R_1 = 0 \\ \overrightarrow R_{AO_2}+\overrightarrow R_{BA}-\overrightarrow R_{BO_4}-\overrightarrow R_{O_4O_2} = 0 \\ ae^{j\theta _2}+be^{j\theta _3}-ce^{j\theta _4}-de^{j\theta _1} = 0 \end{gather} $$
3. Break vector loop equation into real and imaginary parts. This can be done by using the Euler formula above.
   Position Vector Loop Equations in Cartesian Form

   $$ \begin{gather} ae^{j\theta_2}+be^{j\theta_3}-ce^{j\theta _4}-de^{j\theta _1} = 0 \\ a(cos\theta _2+jsin\theta _2)+b(cos\theta _3+jsin\theta _3)-c(cos\theta _4+jsin\theta _4)-d(cos\theta _1+jsin\theta _1) = 0 \end{gather} $$
   Remember that \( \theta_1 \) is equal to zero. This means that \( d(cos\theta_1+jsin\theta_1) \) simplifies to just \( d+0j \). Now we can easily break up the vector loop equation into its real and imaginary components.
   Real Component

   $$ a(cos\theta _2)+b(cos\theta _3)-c(cos\theta _4) - d = 0 $$
   Imaginary Component

   $$ a(sin\theta _2)+b(sin\theta _3)-c(sin\theta _4) +0= 0 $$
4. Solve for unknown positions and/or angles. Since lengths \( a \), \( b \), \( c \), and \( d \) are given along with angles \( (\theta_1) \) and \( (\theta_2) \), the unknowns are \( (\theta_3) \) and \( (\theta_4) \). Since there are two equations and two unknowns, this system can be solved using an equation solver.

Velocity Analysis

1. Identify known and unknown variables. All of the variables found in position analysis are known (including \( (\theta_3) \) and \( (\theta_4) \) as those were solved for in the last step of position analysis). In addition to these variables, \( \omega_2 \) is given. This leaves us with \( \omega_3 \) and \( \omega_4 \) as unknowns. Also, recall the position vector loop.
   Position Vector Loop

   $$ ae^{j\theta _2}+be^{j\theta _3}-ce^{j\theta _4}-d = 0 $$
   
2. Take the time derivative of the position vector loop to get the velocity vector loop.
   Velocity Vector Loop Equations

   $$ \begin{gather} \frac{d}{dt}(ae^{j\theta _2}+be^{j\theta _3}-ce^{j\theta _4}-de^{j\theta _1})=0 \\ ja\frac{d\theta _2}{dt}e^{j\theta _2}+jb\frac{d\theta _3}{dt}e^{j\theta _3}-jc\frac{d\theta _4}{dt}e^{j\theta _4}-jd\frac{d\theta _1}{dt}e^{j\theta _1} = 0 \end{gather} $$

   Because linkage 1 is grounded, we also know that \( \frac{d\theta _1}{dt}=\omega _1 = 0 \)

   Simplified Vector Loop Equation

   $$ j\omega _2ae^{j\theta _2} + j\omega _3be^{j\theta _3}-j\omega _3ce^{j\theta _4}=0 $$

   Or more specifically,

   Reduced Vector Loop Equation

   $$ \overrightarrow V_A + \overrightarrow V_{BA} - \overrightarrow V_B = 0 $$

   In this velocity vector loop \( \overrightarrow V_{AB} \) represents the velocity of joint \( B \) with respect to \( A \).

   
3. Break vector loop equation into real and imaginary parts.
   Velocity Vector Loop Equations in Cartesian Form

   $$ \begin{gather} j\omega _2ae^{j\theta _2} + j\omega _3be^{j\theta _3}-j\omega _3ce^{j\theta _4}=0 \\ j\omega _2a(cos\theta _2+jsin\theta _2) + j\omega _3b(cos\theta _3+jsin\theta _3)-j\omega _3c(cos\theta _4+jsin\theta _4)=0 \end{gather} $$
   Real Component

   $$ -a\omega _2 sin(\theta _2)-b\omega _3 sin(\theta _3) + c\omega _4 sin(\theta _4) = 0 $$
   Imaginary Component

   $$ a\omega _2 cos(\theta _2)+b\omega _3 cos(\theta _3) - c\omega _4 cos(\theta _4) = 0 $$
4. Solve for the unknown velocities. Using trigonometric identities, we can use the two above equations to solve for \( \omega_3 \) and \( \omega_4 \) using an equation solver.
+
Warning: Computing Velocities #warn1
Note that it is possible to solve for for \( \omega_3 \) and \( \omega_4 \) analytically. However this can be time consuming and depends on the exact nature of the mechanism that is being analyzed. For a standard 4-bar linkage with pin joints the solution is:
Velocity Solutions for 4-bar Linkage with Pin Joints
$$ \begin{gather} \omega_3 = \frac{a\omega _2 sin(\theta _4-\theta _2)}{bsin(\theta _3-\theta _4)} \\ \omega_4 = \frac{aomega _2 sin(\theta _2-\theta _3)}{bsin(\theta _4-\theta _3)} \end{gather} $$

Acceleration Analysis

1. Identify known and unknown variables. In addition to the known values from the position and velocity analyses, \( \alpha_2 \) is given. That leaves \( \alpha_3 \) and \( \alpha_4 \) as unknowns. Also, recall the velocity vector loop.
Velocity Vector Loop
$$ j \omega _2ae^{j\theta _2} + j\omega _3be^{j\theta _3}-j\omega _3ce^{j\theta _4}=0 $$

2. Take time derivative of velocity vector loop to get acceleration vector loop.
   Acceleration Vector Loop Equations

   $$ \begin{gather} \frac{d}{dt}(j\omega _2ae^{j\theta _2} + j\omega _3be^{j\theta _3}-j\omega _3ce^{j\theta _4})=0 \\ \\ j\frac{d\omega _2}{dt}ae^{j\theta _2} + j^2\omega _2\frac{d\theta _2}{dt}ae^{j\theta _2} +j\frac{d\omega _3}{dt}be^{j\theta _3}+ \\ j^2\omega _3\frac{d\theta _3}{dt}be^{j\theta _3} - j\frac{d\omega _3}{dt}ce^{j\theta _4} - j^2\omega _4\frac{d\theta _4}{dt}ce^{j\theta _4}=0 \\ \\ j\alpha _2 ae^{j\theta _2} - \omega _2^2ae^{j\theta _2} +j\alpha _3be^{j\theta _3} - \omega _3^2be^{j\theta _3} - j\alpha _4ce^{j\theta _4} + \omega _3^2ce^{j\theta _4}=0\end{gather} $$

   Now we can see terms with \( \alpha \) and \( \omega^2 \) appearing for each linkage. The angular acceleration term \( \alpha \) appears due to tangential, acceleration of the linkage while the \( \omega^2 \) term appears due to the centripetal acceleration of the linkage. In total, four different types of acceleration can appear in acceleration analysis. For a refresher on these terms, refer to the particle kinematics page on the dynamics reference page. With this information, we can simplify the equation further to focus on the different acceleration terms:

   Simplified Acceleration Loop Equation

   $$ (j\alpha _2\overrightarrow R_2 - \omega _2^2 \overrightarrow R_2) + (j\alpha _3 \overrightarrow R_3 - \omega _3^2\overrightarrow R_3) - (j\alpha _4\overrightarrow R_4 - \omega _4^2\overrightarrow R_4) = 0 $$

   Or more specifically,

   Reduced Acceleration Loop Equation

   $$ \begin{gather}(\overrightarrow A_A^t + \overrightarrow A_A^n) + (\overrightarrow A_{BA}^t+\overrightarrow A_{BA}^n) - (\overrightarrow A_B^t+ \overrightarrow A_B^n) = 0 \\ \overrightarrow A_A + \overrightarrow A_{BA} - \overrightarrow A_B = 0 \end{gather} $$
   
3. Break vector loop equation into real and imaginary parts.
   Acceleration Vector Loop Equations in Cartesian Form

   $$ \begin{gather} j\alpha _2 ae^{j\theta _2} - \omega _2^2ae^{j\theta _2} +j\alpha _3be^{j\theta _3} - \omega _3^2be^{j\theta _3} - j\alpha _4ce^{j\theta _4} + \omega _3^2ce^{j\theta _4}=0 \\ \\(j\alpha _2-\omega _2^2)ae^{j\theta _2} +(j\alpha _3-\omega _3^2)be^{j\theta _3}-(j\alpha _4-\omega _4^2)ce^{j\theta _4} =0 \\ \\ (j\alpha _2-\omega _2^2)a(cos\theta _2+jsin\theta _2) +(j\alpha _3-\omega _3^2)b(cos\theta _3+jsin\theta _3)- \\ (j\alpha _4-\omega _4^2)c(cos\theta _4+jsin\theta _4) =0 \end{gather} $$
   Real Component

   $$ -a\alpha _2sin\theta _2-a\omega _2^2cos\theta _2-b\alpha _3sin\theta _3-b\omega _3^2cos\theta _3+c\alpha _4sin\theta _4+c\omega _4^2cos\theta _4 = 0 $$
   Imaginary Component

   $$ a\alpha _2cos\theta _2-a\omega _2^2sin\theta _2+b\alpha _3cos\theta _3-b\omega _3^2sin\theta _3-c\alpha _4cos\theta _4+c\omega _4^2sin\theta _4 = 0 $$
4. Solve for the unknown accelerations. Using trigonometric identities, we can use the two above equations to solve for \( \alpha_3 \) and \( \alpha_4 \) using an equation solver.
+

Warning: Computing Accelerations #warn2

Note that it is possible to solve for for \( \alpha_3 \) and \( \alpha_4 \) analytically. However this can be time consuming and depends on the exact nature of the mechanism that is being analyzed. For a standard 4-bar linkage with pin joints the solution is:

\( \alpha _3= \frac{CD-AF}{AE-BD} \)

\( \alpha _4 = \frac{CE-BF}{AE-BD} \)

Where

Acceleration Analytic Solution

$$ \begin{gather} A = csin\theta _4 \\ B = bsin\theta _3 \\ C = a\alpha _2sin\theta _2 + \alpha \omega _2^2 cos\theta _2+b\omega _3^2 cos\theta _3 - c\omega _4^2 cos\theta _4 \\ D = ccos\theta _4 \\ E = bcos\theta _3 \\ F = a\alpha _2cos\theta _2 - a\omega _2^2sin\theta _2 - b\omega _3^2sin\theta _3+c\omega _4^2sin \theta _4 \end{gather} $$

While acceleration analysis is possible to do by hand, it is highly recommended that code or a computer based solver is used to assist solving for the unknown variables.

Application Notes

If you have more unknowns than equations, you can likely create more vector loops. If this is the case, think carefully about the constraints on your problems.

For example, take a look at this slider-crank example. If you create a position loop with three vectors, this leads to three unknowns (\( \theta_3 \), \( s(t) \), and \( \beta (t) \)).

By breaking up the position loop into four position vectors, it makes it clear that there are only two unknowns (\( \theta_3 \) and \( d(t) \)).

This leads to the following position vector loop:

Since \( \theta_4 \) equals 90 degrees and \( \theta_1 \) equals zero degrees, the equation can be simplified to:

When taking the time derivative of this equation, keep in mind that \( d=d(t) \) is still a value that changes with respect to time. This will result in the terms \( \dot(d) \ddot(d) \), which represents the change and rate of change in length of the link with respect to time.

Similarly, if you set up a three-position vector loop for an inverted crank-slider, you will see three unknowns (\( \theta_3 \), \( s(t) \), and \( \beta (t) \)).

By breaking up the inverted crank-slider into four position vector loops, there are still three unknown variables (\( \theta_3 \), \( \beta (t) \), and \( \theta_4 \)), but another equation appears due to geometry of the four-bar linkage. The angle \( \gamma \) is constant because link four and the slider are fixed in an inverted crank-slider. This reveals the equation:

With three unknown variables and three equations, we can use a solver to obtain the values of all unknown variables.