Position Analysis

Types of Coordinate Systems

1. Global coordinate system
2. Inertial reference frame
3. Local coordinate system

Global Coordinate System

• The (X, Y) coordinate system
• Is attached to earth typically but could be attached to other objects
• Absolute positions, velocities, accelerations are with respect ot the GCS

Inertial Reference Frame

• The system that this coordinate system is attached to has no acceleration itself

Local Coordinate System

• Denoted as (X, Y)
• Attached to a point of interest
• Can be rotating or non rotating (LRCS or LNCS)
• Common points of interest are pin joints and centre of gravity (connection points between things)

Kinematics

We make use of Kinematics here.

Translation

All points of a body have the same displacement, no rotation is involved in pure translation.
This includes:

• Rectilinear translation
• Curvilinear translation

Rotation

Different points on a body undergo different displacements, the centre of rotation undergoes no motion with respect to the frame of reference.

Euler’s Theorem

The general displacement of a rigid body with one point fixed is a rotation about some axis.

General Plane (Complex) Motion

This is a type of rigid body motion and is a combination of translation and rotation.

Chasles’ Theorem

Any displacement of a rigid body is equivalent to the sum of a translation of any one point on that body and a rotation of the body about an axis through that point.

Analysis of Four Bar Linkages

Four Bar Linkages

The following analysis techniques are used exclusively for four bar linkages (theres one vector loop analysis applied to a crank-slider as well). Therefore, the equations given should be general and when applying, you should just need to identify each joint and linkage to use each technique.

Graphical Position Analysis

This requires precise accurate drawings of links and their positions and can be useful for a quick approximate solution for one position. You can predict the trajectory of links and how angles will move using your protractor.

Four bar linkage positions

Crossed: Two links adjacent to the shortest link cross each other
Open: Two links adjacent to the shortest link do not cross each other

Algebraic Position Analysis

This is solved based on system geometry and does not require drawings.

1. Start with deriving equations based on geometry of different joints and links (start with joints first and model links based off of joint position like vectors)
   1. $A_x=a\cos {\theta }_2$
   2. $A_y=a\sin {\theta }_2$
   3. $b^2={\left(B_x-A_x\right)}^2+{\left(B_y+A_y\right)}^2$
2. You can define parameters to help create geometric equations for joints by using the cosine law and through geometric manipulation of pre-existing relationships…
3. Solve for $B_x$ and $B_y$
   1. $B_x=S-{\frac{2A_{y}By}{2\left(A_x-d\right)}}_{}$
   2. $B_y=\frac{-Q\pm \sqrt{Q^2-4PR}}{2P}$
   3. $S=\frac{a^2-b^2+c^2-d^2}{2\left(A_x-d\right)}$
   4. $R={\left(d-S\right)}^2-c^2$
   5. $P=\frac{A_y^2}{{\left(A_x-d\right)}^2}+1$
   6. $Q=\frac{2A_y\left(d-S\right)}{A_x-d}$
4. Solve for ${\theta }_3$ and ${\theta }_4$
   1. ${\theta }_3=\arctan \left(\frac{B_y-A_y}{B_x-A_x}\right)$
   2. ${\theta }_4=\arctan \left({\frac{B_y}{B_x-d}}_{}\right)$

Vector Loop Analysis

In this method, links are drawn as position vectors that form a loop. Vectors can be defined in any form.

Eulers Notation

Remember that complex numbers can be represented as
$R{e}^{j\theta }$
$R\cos \theta +jR\sin \theta$

1. Set a reference frame and denote vectors $R_1+R_2-R_3-R_4=0$
2. Draw vector directions to close a loop
3. Calculate the length-dependent parameters (k’s)
   1. $k_1=\frac{d}{a}$
   2. $k_2=\frac{d}{c}$
   3. $k_3=\frac{a^2-b^2+c^2+d^2}{2ac}$
   4. $k_4=\frac{d}{b}$
   5. $k_5=\frac{c^2-d^2-a^2-b^2}{2ab}$
4. Calculate the angle-dependent parameters (ABC’s)
   1. $A=\cos {\theta }_2-k_1-k_2\cos {\theta }_2+k_3$
   2. $B=-2\sin {\theta }_2$
   3. $C=k_1-\left(k_2+1\right)\cos {\theta }_2+k_3$
   4. $D=\cos {\theta }_2-k_1+k_4\cos {\theta }_2+k_5$
   5. $E=B=-2\sin {\theta }_2$
   6. $F=k_1+\left(k_4-1\right)\cos {\theta }_2+k_5$
5. Solve for ${\theta }_3$ and ${\theta }_4$
   1. ${\theta }_3=2\arctan \left(\frac{-E\pm \sqrt{E^2-4DF}}{2D}\right)$
   2. ${\theta }_4=2\arctan \left(\frac{-B\pm \sqrt{B^2-4AC}}{2A}\right)$
   3. Here, positives mean the links are crossed whereas negatives are open, if we get a complex conjugate, the link lengths cannot connect for the given ${\theta }_2$
   4. For some reason, we add 360 when we use the + option (crossed), and subtract 360 when we use the - option (open) Transmission Angles
      ${\mu }_1=\pi -\left|{\theta }_{31}-{\theta }_{41}\right|$ ${\mu }_2=\left|{\theta }_{32}-{\theta }_{42}\right|$

Toggle Conditions

$$\begin{array}{rl}ar{g}_1& =\frac{a^2+d^2-b^2-c^2}{2\cdot a\cdot d}+\frac{b\cdot c}{a\cdot d}\\ ar{g}_2& =\frac{a^2+d^2-b^2-c^2}{2\cdot a\cdot d}-\frac{b\cdot c}{a\cdot d}\\ {\theta }_{2,\text{toggle}}& :=\arccos (ar{g}_2)\end{array}$$

Vector Loop Analysis For a Crank-Slider

1. Denote vectors $R_2-R_3-R_4-R_1=0$ which becomes $a{e}_{}^{j{\theta }_2}-b{e}_{}^{j{\theta }_3}-c{e}_{}^{j{\theta }_4}--c{e}_{}^{j{\theta }_1}=0$
2. $R_1$ should have an angle of zero with respect to the x-axis
3. $R_4$ should be parallel to the y-axis
4. Calculate one possible value for ${\theta }_3$ and d using ${\theta }_3=\arcsin \left(\frac{a\sin {\theta }_2-c}{b}\right)$ and $d=a\cos {\theta }_2-b\cos {\theta }_3$
5. Calculate the other possibility using using ${\theta }_3=\arcsin \left(-\frac{a\sin {\theta }_2-c}{b}\right)+\pi$ and $d=a\cos {\theta }_2-b\cos {\theta }_3$

Example Notes

Choosing abc’s and ABC’s…

• a is always $R_2$ which is usually the input link
• b is always $R_3$
• c is always $R_4$
• d is always $R_1$
• You label the ABC’s CCW from your input joint

Tips from my practice problems

1. Make sure $R_D$ is aligned with the x-axis
2. Calculate ${\theta }_2$ if you have to using the two given angles, where one of those angles might be a change in angle that you need to apply later
3. Solve K’s
4. Solve ABC’s
5. Might have to solve twice if there is a change in angle
6. Solve for ${\theta }_3$ and ${\theta }_4$
7. For the open angles, add 360° for the angle, for crossed angles, subtract 360° twice for some reason
8. If you have a question where you have a change in angle, solve once with the normal ${\theta }_2$ and then once with ${\theta }_{2new}={\theta }_2+\Delta {\theta }_2$ and use the plus option in the quadratic equation
9. You might need to calculate transmission angles to check their differences
10. You may need to check the grashof condition of the mechanism
11. You may need to determine the toggles (formulas are given)