# Example 2.2: Kinematic analysis of a slider-crank mechanism

## Position analysis

The figure shows a slider-crank mechanism. For the instant when the crank forms an angle $\theta_2 = 30^{\circ}$ with the horizontal, calculate the position of the slider and the angle that link 3 forms with the horizontal.

![figBM.jpeg](./figs/figBM.jpeg)

Data:

Link lengths, in meters: $\overline{O_2 A} = 0.25;\,\overline{AB} = 0.6;\,y_B = 0.25;$

---

## Position analysis: graphical method

The objective is to construct the mechanism's configuration to scale for the instant $\theta_2 = 30^\circ$ and measure the position unknowns: the horizontal coordinate of the slider ($x_B$) and the angle of the coupler ($\theta_3$).

### Construction Data:

*   **Link 2 (Crank):** $L_2 = \overline{O_2A} = 0.25 \text{ m}$
*   **Link 3 (Coupler):** $L_3 = \overline{AB} = 0.6 \text{ m}$
*   **Slider Axis:** The slider moves on a horizontal line at a constant height $y_B = 0.25 \text{ m}$.
*   **Input Angle:** $\theta_2 = 30^\circ$.

### Construction Procedure:

1.  **Establish the origin:** The fixed pivot $O_2$ is placed at the coordinate origin (0,0).
2.  **Position the crank (Link 2):** From $O_2$, link 2 is drawn with its length $L_2 = 0.25 \text{ m}$ and the input angle $\theta_2 = 30^\circ$. The end of this vector is the position of joint A.
3.  **Draw the slider guide:** The horizontal line that defines the path of slider B is drawn. This line is at a constant height $y = 0.25 \text{ m}$.
4.  **Locate joint B (Loop closure):**
    *   **Coupler Constraint:** Point B must be at a distance $L_3 = 0.6 \text{ m}$ from A. Geometrically, this is a circle with center A and radius 0.6.
    *   **Slider Constraint:** Point B must lie on the line $y = 0.25$.
5.  **Resolve configuration ambiguity:** The intersection of the circle (center A, radius $L_3$) and the line ($y=0.25$) yields two possible solutions for point B. The one corresponding to the configuration shown in the problem figure must be chosen.
6.  **Measurement of results:** With the mechanism fully drawn, the desired quantities are measured directly from the graphical software:
    *   Coordinates of point B: $(x_B, y_B)$.
    *   Angle of the coupler, $\theta_3$: The angle of the segment AB with the horizontal.

---

## Position analysis: analytical method

The objective is to find expressions for the unknowns ($x_B$ and $\theta_3$) as a function of the known parameters.

### Vector Loop Formulation

We model the mechanism as an open vector chain. The position of point B can be expressed starting from the origin $O_2$:
$ \vec{R}_B = \vec{R}_A + \vec{R}_{AB} $
Where:
*   $\vec{R}_A = L_2(\cos\theta_2 \hat{i} + \sin\theta_2 \hat{j})$
*   $\vec{R}_{AB} = L_3(\cos\theta_3 \hat{i} + \sin\theta_3 \hat{j})$
*   $\vec{R}_B = x_B \hat{i} + y_B \hat{j}$ (with $y_B$ constant)

We decompose the vector equation into its scalar components (x and y):

*   **x-component:**
    $ x_B = L_2\cos\theta_2 + L_3\cos\theta_3 \quad (\text{Equation 1}) $
*   **y-component:**
    $ y_B = L_2\sin\theta_2 + L_3\sin\theta_3 \quad (\text{Equation 2}) $

We have a system of two equations with two unknowns: $\theta_3$ and $x_B$.

### Solving for $\theta_3$ and $x_B$

1.  **Solve for $\theta_3$:**
    From Equation 2, we can solve for $\sin\theta_3$:
    $ \sin\theta_3 = \frac{y_B - L_2\sin\theta_2}{L_3} $
    This will give us two possible values for $\theta_3$, since $\theta_{3,2} = 180^\circ - \theta_{3,1}$. We must choose the solution that corresponds to the configuration in the drawing. Once $\sin\theta_3$ is known, we can find $\cos\theta_3 = \pm\sqrt{1 - \sin^2\theta_3}$, where the sign depends on the quadrant in which $\theta_3$ lies.

2.  **Solve for $x_B$:**
    Once we have the correct value of $\theta_3$ (and therefore, of $\cos\theta_3$), we substitute it into Equation 1 to find the position of the slider:
    $ x_B = L_2\cos\theta_2 + L_3\cos\theta_3 $

---

### Numerical Calculation for the Example

The calculation is performed for the two possible configurations.

*   **Initial Data:**
    *   Lengths: $L_2=0.25 \text{ m}, L_3=0.6 \text{ m}$
    *   Vertical position of the slider: $y_B = 0.25 \text{ m}$
    *   Input angle: $\theta_2 = 30^\circ$

*   **Calculation of $\theta_3$:**
    Using Equation 2:
    $ \sin\theta_3 = \frac{0.25 - 0.25\sin(30^\circ)}{0.6} = \frac{0.25 - 0.25 \cdot 0.5}{0.6} = \frac{0.125}{0.6} \approx 0.2083 $
    This gives two possible solutions for $\theta_3$:
    *   **Solution 1:** $\theta_{3,1} = \arcsin(0.2083) \approx 12.02^\circ$. This corresponds to the configuration in the figure.
    *   **Solution 2:** $\theta_{3,2} = 180^\circ - 12.02^\circ = 167.98^\circ$. This corresponds to the other possible configuration of the mechanism.

*   **Calculation of $x_B$ (for Solution 2):**
    The second solution for $\theta_3$ is chosen. First, we calculate the cosine for this angle:
    $ \cos(167.98^\circ) \approx -0.9781 $
    Now, we use Equation 1:
    $ x_B = 0.25\cos(30^\circ) + 0.6\cos(167.98^\circ) = 0.25 \cdot 0.866 + 0.6 \cdot (-0.9781) $
    $ x_B = 0.2165 - 0.5868 = -0.3703 \text{ m} $

*   **Final Position Results (Solution 2):**
    *   Coupler angle: $\mathbf{\theta_3 = 167.98^\circ}$
    *   Slider position: $\mathbf{x_B = -0.3703 \text{ m}}$

## Velocity analysis

At the instant when the crank forms an angle $\theta_2 = 30^{\circ}$ with the horizontal, its angular velocity is $\vec{\omega}_2 = 1\hat{k}$ rad/s, and its acceleration is $\vec{\alpha}_2 = 1\hat{k}$ rad/s². Calculate the velocity of the slider and the angular velocity of link 3.

Data:

Link lengths, in meters: $\overline{O_2 A} = 0.25;\, \overline{AB} = 0.6;\, y_B = 0.25$.
Angle of link 3 with the horizontal, from the position analysis: $\theta_3 = 167.98^{\circ}$.

### Resolution (Calculations in Matlab)

For this example, we will perform the calculations using Matlab. We start by creating variables for the given data:

```matlab
clear, clc
O2A = 0.25; AB = 0.6;
th2 = 30; th3 = 167.98;
w2 = [0 0 1]; af2 = [0 0 1];
```

The mechanism has one degree of freedom, corresponding to $\vec{\omega}_2$. We use this known data to calculate the velocity of point A:

$\vec{v}_A = \vec{\omega}_2 \times \vec{r}_{O2A}$

From the geometric data, we calculate $\vec{r}_{O2A}$:

```matlab
rO2A = O2A * [cosd(th2) sind(th2) 0];
```

Therefore,

$\vec{v}_A = 1\hat{k} \times (0.2165\hat{i} + 0.125\hat{j}) = -0.125\hat{i} + 0.2165\hat{j}$
```matlab
vA = cross(w2, rO2A);
```

Next, we focus on point B. Analyzing it as part of link 3, we can establish the relative velocity equation between A and B:

$\vec{v}_{B3} = \vec{v}_A + \vec{v}_{B/A}$

with $\vec{v}_A$ known and $\vec{v}_{B/A} = \vec{\omega}_3 \times \vec{r}_{AB}$, where we know the direction but not the magnitude of $\vec{\omega}_3$.

```matlab
syms w3k real
w3 = [0 0 w3k];
rAB = AB * [cosd(th3) sind(th3) 0];
vB_A = cross(w3, rAB);
vB3 = vA + vB_A;
```

Now we analyze point B as part of link 4 (the slider). Since it's a slider, we know its direction must be along the sliding axis, which is horizontal in this case:

$\vec{v}_{B4} = v_B\hat{i}$
```matlab
syms vBi real
vB4 = [vBi 0 0];
```

Since there is a revolute joint at B, we can equate the two expressions:

```matlab
sol_v1 = solve(vB3 == vB4, w3k, vBi);
w3k = double(sol_v1.w3k);
vBi = double(sol_v1.vBi);
fprintf('w3k=%0.3f vBi=%0.3f\n', w3k, vBi)
```

```matlabTextOutput
w3k=0.369 vBi=-0.171
```

### Velocity Cinema

The velocity cinema for this example is not provided, but it can be created using the same principles demonstrated in previous examples.

## Acceleration analysis

At the instant when the crank forms an angle $\theta_2 = 30^{\circ}$ with the horizontal, its angular velocity is $\vec{\omega}_2 = 1\hat{k}$ rad/s, and its acceleration is $\vec{\alpha}_2 = 1\hat{k}$ rad/s². Calculate the acceleration of the slider and the angular acceleration of link 3.

Data:

Link lengths, in meters: $\overline{O_2 A} = 0.25;\, \overline{AB} = 0.6;\, y_B = 0.25$.
Angle of link 3 with the horizontal (from position analysis): $\theta_3 = 167.98^{\circ}$, and its angular velocity $\vec{\omega}_3 = 0.369$ rad/s.

### Resolution (Calculations in Matlab)

The procedure for calculating the mechanism's accelerations is similar to that for velocities. We start with point A as we have information about it. We calculate $\vec{a}_A$ from its normal and tangential components:

$\vec{a}_A = \vec{a}_A^n + \vec{a}_A^t = \vec{\omega}_2 \times (\vec{\omega}_2 \times \vec{r}_{O2A}) + \vec{\alpha}_2 \times \vec{r}_{O2A}$
```matlab
aA = cross(w2, cross(w2, rO2A)) + cross(af2, rO2A);
```

Now, analyzing point B as part of link 3, we have:

$\vec{a}_{B3} = \vec{a}_A + \vec{a}_{B/A}$

which, by decomposing, gives:

$\vec{a}_{B3} = \vec{a}_A + \vec{a}_{B/A}^n + \vec{a}_{B/A}^t$

where the acceleration of A is known, and we can express the components of the relative acceleration from the available information:

- For $\vec{a}_{B/A}^n$, everything is known. Therefore:
$\vec{a}_{B/A}^n = \vec{\omega}_3 \times (\vec{\omega}_3 \times \vec{r}_{AB})$
```matlab
aB_An = cross(w3, cross(w3, rAB));
```

- For $\vec{a}_{B/A}^t$, we can operate and express it in terms of the unknown $\vec{\alpha}_3$, of which we only know its direction:
$\vec{a}_{B/A}^t = \vec{\alpha}_3 \times \vec{r}_{AB}$
```matlab
syms af3k real
af3 = [0 0 af3k];
aB_At = cross(af3, rAB);
```

Which leaves us with:

```matlab
aB3 = aA + aB_An + aB_At;
```

On the other hand, if we consider point B as part of link 4, we have:

$\vec{a}_{B4} = a_B\hat{i}$
```matlab
syms aBi real
aB4 = [aBi 0 0];
```

Finally, if we equate both expressions, we have:

```matlab
sol_ac1 = solve(aB3 == aB4, af3k, aBi);
af3k = double(subs(sol_ac1.af3k));
aBi = double(subs(sol_ac1.aBi));
fprintf('af3k=%0.3f aBi=%0.3f\n', af3k, aBi)
```

```matlabTextOutput
af3k=0.127 aBi=-0.277
```

### Acceleration Cinema

The acceleration cinema for this example is not provided, but it can be created using the same principles demonstrated in previous examples.