# Example 2.4: Kinematic analysis of an inverted slider-crank mechanism

## Position analysis

The figure shows an inverted slider-crank mechanism. At the instant when the crank forms an angle $\theta_2 = 60^{\circ}$ with the horizontal, calculate the position of the slider with respect to point A, the angles that links 3 and 4 form with the horizontal, and the coordinates of point C.

![fig_IN.png](./figs/fig_IN.png)

Data:

Link lengths, in meters: $\overline{O_2 A} = 0.45;\, \overline{AC} = 1.7;\, \overline{O_4B} = 0.7855;\, \overline{O_2O_4} = L_1 = 1.30 \text{ m};$
the angle $\gamma$ that links 3 and 4 form is $90^\circ$.

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## Position analysis: graphical method

The objective is to construct the mechanism to scale to measure the unknowns: the distance of the slider to point A ($\overline{AB}$), the angles $\theta_3$ and $\theta_4$, and the coordinates of point C.

### Construction Data:

*   **Link 1 (Frame):** Distance $\overline{O_2O_4} = L_1 = 1.30 \text{ m}$
*   **Link 2 (Crank):** $L_2 = \overline{O_2A} = 0.45 \text{ m}$.
*   **Link 3 (Coupler):** Point C is at a distance $\overline{AC} = 1.7 \text{ m}$ from A.
*   **Link 4 (Rocker):** The perpendicular distance from $O_4$ to the slot is $h = \overline{O_4B} = 0.7855 \text{ m}$.
*   **Input Angle:** $\theta_2 = 60^\circ$.
*   **Geometric Constraint:** The angle between the line $AC$ (link 3) and the line $O_4B$ (link 4) is $90^\circ$.

### Construction Procedure:

1.  **Establish the frame:** Locate pivot $O_2$ at the origin (0,0) and pivot $O_4$ at $(L_1, 0)$.
2.  **Position the crank:** Draw link 2 from $O_2$ with length $L_2=0.45$ and angle $\theta_2=60^\circ$ to find point A.
3.  **Draw the slot (Link 3):**
    *   Draw a circle with center $O_4$ and radius $h = 0.7855$.
    *   Draw a straight line starting from A and tangent to this circle. This defines the direction of link 3 and its angle $\theta_3$. There are two possible tangents, choose the one corresponding to the figure.
4.  **Locate point B:** Point B is the point of tangency between the line and the circle.
5.  **Locate point C:** On the line passing through A and B, measure a distance $\overline{AC} = 1.7$ from A to find C.
6.  **Determine link 4:** Draw the line connecting $O_4$ and B. Its angle is $\theta_4$.
7.  **Measurement of results:** Measure from the drawing the distance $\overline{AB}$, the angles $\theta_3$ and $\theta_4$, and the coordinates of C.

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## Position analysis: analytical method

### Vector Loop Formulation

We consider the triangle formed by points $O_2, O_4, A$. The distance between $O_4$ and A, $d_{O_4A}$, can be calculated with the law of cosines:
$ d_{O_4A}^2 = L_1^2 + L_2^2 - 2L_1L_2\cos\theta_2 $
In the right triangle $O_4BA$, we have:
$ d_{O_4A}^2 = (\overline{O_4B})^2 + (\overline{AB})^2 = h^2 + (\overline{AB})^2 $
Equating both expressions, we can solve for the distance $\overline{AB}$ (position of the slider with respect to A):
$ \overline{AB} = \sqrt{L_1^2 + L_2^2 - 2L_1L_2\cos\theta_2 - h^2} $

### Solving for $\theta_3$ and $\theta_4$

1.  **Calculate angle $\alpha = \angle AO_4O_2$:** Using the law of sines in triangle $O_2O_4A$:
    $ \frac{\sin\alpha}{L_2} = \frac{\sin\theta_2}{d_{O_4A}} \implies \alpha = \arcsin\left(\frac{L_2\sin\theta_2}{d_{O_4A}}\right) $
2.  **Calculate angle $\beta = \angle AO_4B$:** Using the right triangle $O_4BA$:
    $ \cos\beta = \frac{\overline{O_4B}}{d_{O_4A}} = \frac{h}{d_{O_4A}} \implies \beta = \arccos\left(\frac{h}{d_{O_4A}}\right) $
3.  **Calculate $\theta_3$ and $\theta_4$:**
    Observing the geometry of the mechanism:
    $ \theta_3 = 180^\circ - (\alpha + \beta) $
    Since $\vec{O_4B}$ is perpendicular to the line AC (link 3):
    $ \theta_4 = \theta_3 - 90^\circ $

### Calculation of the Coordinates of Point C

$ \vec{R}_C = \vec{R}_A + \vec{R}_{AC} $
*   $x_C = L_2\cos\theta_2 + \overline{AC}\cos\theta_3$
*   $y_C = L_2\sin\theta_2 + \overline{AC}\sin\theta_3$

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### Numerical Calculation

*   **Initial Data:**
    *   $L_1 = 1.3 \text{ m}$
    *   $L_2 = 0.45 \text{ m}$
    *   $\overline{AC} = 1.7 \text{ m}$
    *   $h = \overline{O_4B} = 0.7855 \text{ m}$
    *   $\theta_2 = 60^\circ$

*   **1. Calculate distance $d_{O_4A}$ and slider position $\overline{AB}$:**
    $ d_{O_4A}^2 = 1.3^2 + 0.45^2 - 2(1.3)(0.45)\cos(60^\circ) = 1.69 + 0.2025 - 0.585 = 1.3075 $
    $ d_{O_4A} = \sqrt{1.3075} \approx 1.143 \text{ m} $
    $ \overline{AB} = \sqrt{d_{O_4A}^2 - h^2} = \sqrt{1.3075 - 0.7855^2} = \sqrt{0.6904} \approx 0.831 \text{ m} $

*   **2. Calculate angle of vector $\vec{R}_{O_4A}$ ($\phi$) and angle $\beta$:**
    The angle $\phi$ is that of the vector from $O_4$ to A.
    $ \phi = \arctan\left(\frac{y_A - y_{O_4}}{x_A - x_{O_4}}\right) = \arctan\left(\frac{L_2\sin\theta_2}{L_2\cos\theta_2 - L_1}\right) = \arctan\left(\frac{0.45\sin(60^\circ)}{0.45\cos(60^\circ) - 1.3}\right) $
    $ \phi = \arctan\left(\frac{0.3897}{-1.075}\right) \approx 160.05^\circ $
    The angle $\beta$ is from the right triangle $O_4BA$.
    $ \beta = \arccos\left(\frac{h}{d_{O_4A}}\right) = \arccos\left(\frac{0.7855}{1.143}\right) \approx 46.59^\circ $

*   **3. Calculate $\theta_3$ and $\theta_4$:**
    From the geometry, it is observed that the angle of link 4, $\theta_4$, is the sum of angles $\phi$ and $\beta$ (considering the quadrant). The configuration in the figure corresponds to:
    $ \theta_4 = \phi - \beta = 160.05^\circ - 46.59^\circ = 113.46^\circ $
    And the angle of link 3 (the slot) is perpendicular to link 4:
    $ \theta_3 = \theta_4 - 90^\circ = 113.46^\circ - 90^\circ = 23.46^\circ $

*   **4. Calculate Coordinates of Point C:**
    $ x_C = L_2\cos\theta_2 + \overline{AC}\cos\theta_3 = 0.45\cos(60^\circ) + 1.7\cos(23.46^\circ) = 0.225 + 1.559 = 1.784 \text{ m} $
    $ y_C = L_2\sin\theta_2 + \overline{AC}\sin\theta_3 = 0.45\sin(60^\circ) + 1.7\sin(23.46^\circ) = 0.3897 + 0.677 = 1.067 \text{ m} $

*   **Final Position Results:**
    *   Slider position: $\mathbf{\overline{AB} = 0.831 \text{ m}}$
    *   Angle of link 3: $\mathbf{\theta_3 = 23.46^\circ}$
    *   Angle of link 4: $\mathbf{\theta_4 = 113.46^\circ}$
    *   Coordinates of C: **(1.784, 1.067) m**