# Example 2.5: Kinematic analysis of a six-bar linkage

## Position analysis

The figure shows a 6-bar linkage. At the instant when the crank forms an angle $\theta_2 = 225^{\circ}$ with the horizontal, calculate the position of the slider, the angle that the segment $O_4B$ forms with the horizontal, and the coordinates of point C.

![fig6B.jpg](./figs/fig6B.jpg)

Data:

Link lengths, in meters: $\overline{O_2 A} = 0.30;\, \overline{AB} = 0.5;\, \overline{O_4B} = \overline{O_4C} = 0.45;\, \overline{C D} = 0.50;$
the angle $\gamma$ between $\overline{O_4B}$ and $\overline{O_4C}$ is $90^\circ$.
$\overline{O_2 O_4} = 0.25;$
The slider moves horizontally on the x-axis.

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## Position analysis: graphical method

The objective is to construct the mechanism to scale to measure the unknowns: the angle $\theta_4$, the coordinates of point C, and the position of slider D.

### Construction Procedure:

1.  **Locate the fixed pivots:** Place $O_2$ at the origin (0,0) and $O_4$ at (0.25, 0).
2.  **Position the crank (Link 2):** From $O_2$, draw link 2 with a length of 0.30 m and an angle $\theta_2 = 225^\circ$ to find point A.
3.  **Close the first loop (4-bar):**
    *   Draw a circle with center A and radius $\overline{AB} = 0.5$ m.
    *   Draw a circle with center $O_4$ and radius $\overline{O_4B} = 0.45$ m.
    *   The intersection of both circles gives the position of point B. Choose the solution that corresponds to the figure.
4.  **Locate point C:**
    *   Draw the line connecting $O_4$ and B.
    *   Rotate this line $90^\circ$ clockwise around $O_4$.
    *   On this new line, measure a distance $\overline{O_4C} = 0.45$ m from $O_4$ to find point C.
5.  **Locate slider D:**
    *   Draw a circle with center C and radius $\overline{CD} = 0.50$ m.
    *   The intersection of this circle with the horizontal axis (x-axis) gives the position of slider D. Choose the solution that corresponds to the figure.
6.  **Measurement of results:** Measure in the graphical software the angle of the bar $O_4B$ ($\theta_4$), the coordinates of C, and the x-coordinate of D.

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## Position analysis: analytical method

The problem is solved in two parts: first the four-bar loop $O_2ABO_4$ and then the dyad $O_4CD$.

### Vector Loop Formulation (Loop 1: $O_2ABO_4$)

The loop equation is $\vec{L}_2 + \vec{L}_3 = \vec{L}_1 + \vec{L}_4$. Decomposing into x and y components:
*   $L_2\cos\theta_2 + L_3\cos\theta_3 = L_1 + L_4\cos\theta_4$
*   $L_2\sin\theta_2 + L_3\sin\theta_3 = L_4\sin\theta_4$

Rearranging and squaring to eliminate $\theta_3$, we arrive at an equation of the form $A\cos\theta_4 + B\sin\theta_4 = C$, which allows solving for $\theta_4$.

### Numerical Calculation

*   **Initial Data:**
    *   $L_1 = \overline{O_2O_4} = 0.25 \text{ m}$
    *   $L_2 = \overline{O_2A} = 0.30 \text{ m}$
    *   $L_3 = \overline{AB} = 0.5 \text{ m}$
    *   $L_4 = \overline{O_4B} = 0.45 \text{ m}$
    *   $\overline{O_4C} = 0.45 \text{ m}$
    *   $\overline{CD} = 0.50 \text{ m}$
    *   $\theta_2 = 225^\circ$
    *   $\gamma = 90^\circ$ (angle $\angle BO_4C$, clockwise)

*   **1. Solve the 4-bar loop ($O_2ABO_4$):**
    Freudenstein's equation is solved for $\theta_4$. The constants are:
    *   $A = 2L_1L_4 - 2L_2L_4\cos\theta_2 = 2(0.25)(0.45) - 2(0.3)(0.45)\cos(225^\circ) = 0.4159$
    *   $B = -2L_2L_4\sin\theta_2 = -2(0.3)(0.45)\sin(225^\circ) = 0.1909$
    *   $C = L_3^2 - L_1^2 - L_2^2 - L_4^2 + 2L_1L_2\cos\theta_2 = 0.5^2 - 0.25^2 - 0.3^2 - 0.45^2 + 2(0.25)(0.3)\cos(225^\circ) = -0.2111$
    
    Solving $0.4159\cos\theta_4 + 0.1909\sin\theta_4 = -0.2111$ yields two solutions. The one corresponding to the figure is:
    *   $\mathbf{\theta_4 = 142.13^\circ}$

*   **2. Calculate the position of point C:**
    The link $\overline{O_4C}$ forms an angle $\theta_C$ with the horizontal. Since the angle $\gamma = \angle BO_4C$ is $90^\circ$ clockwise (subtracting from the angle of bar 4):
    $ \theta_C = \theta_4 - \gamma = 142.13^\circ - 90^\circ = 52.13^\circ $
    The coordinates of C are (relative to $O_4$ which is at (0.25, 0)):
    $ x_C = x_{O_4} + \overline{O_4C}\cos\theta_C = 0.25 + 0.45\cos(52.13^\circ) = 0.25 + 0.276 = 0.526 \text{ m} $
    $ y_C = y_{O_4} + \overline{O_4C}\sin\theta_C = 0 + 0.45\sin(52.13^\circ) = 0.355 \text{ m} $
    *   **Coordinates of C: (0.526, 0.355) m**

*   **3. Calculate the position of slider D:**
    Point D is on the x-axis ($y_D=0$). The distance $\overline{CD}$ is $0.50 \text{ m}$. Using the Pythagorean theorem in the triangle formed by C, D, and the projection of C on the x-axis:
    $ (\overline{CD})^2 = (x_C - x_D)^2 + (y_C - y_D)^2 $
    $ 0.5^2 = (0.526 - x_D)^2 + (0.355 - 0)^2 $
    $ 0.25 = (0.526 - x_D)^2 + 0.126 $
    $ (0.526 - x_D)^2 = 0.124 \implies 0.526 - x_D = \pm\sqrt{0.124} = \pm 0.352 $
    The two possible solutions for $x_D$ are $x_D = 0.526 \mp 0.352$. The one corresponding to the figure (further to the right) is:
    *   **Position of slider D: $x_D = 0.878 \text{ m}$**

*   **Final Position Results:**
    *   Angle of link 4: $\mathbf{\theta_4 = 142.13^\circ}$
    *   Coordinates of C: **(0.526, 0.355) m**
    *   Position of slider D: $\mathbf{x_D = 0.878}$