# Example 2.15: Bicycle suspension (graphical method)
```{raw} html
```
The mechanism in the photo (dimensions in centimeters) shows the rear suspension of a bicycle (January 2023 exam). It is the same mechanism solved analytically in {doc}`L02_13_example9`. Find:
1. All the absolute instantaneous centers of rotation (ICRs).
2. The velocity of point $A$, by the graphical method.
3. From the acceleration polygon given below, the angular accelerations of all the bodies.
::::{grid} 2
:::{grid-item}
:::
:::{grid-item}
:::
::::
Data: $\omega_4 = 1$ rad/s counterclockwise, $\alpha_4 = 2$ rad/s² clockwise. Any other dimension can be measured with the scale bars.
:::{dropdown} Solution
**1. Absolute ICRs**
The mechanism has $N = 6$ bodies: 1 frame (ground), 2 rocker $DO_2A$, 3 swingarm $ABC$, 4 link $O_4B$, 5 shock cylinder and 6 shock rod. So there are $6 \cdot 5 / 2 = 15$ ICRs, of which the absolute ones are required: $I_{12}$, $I_{13}$, $I_{14}$, $I_{15}$ and $I_{16}$.
First, all the ICRs that can be seen directly: the pin joints between pairs of bodies in contact ($I_{12}$, $I_{23}$, $I_{34}$, $I_{14}$, $I_{15}$ and $I_{26}$), and $I_{56}$, which is at infinity in the direction perpendicular to the shock axis, since the rod slides inside the cylinder.
Of the absolute ICRs, only $I_{13}$ and $I_{16}$ are still unknown.
To find $I_{13}$, apply Kennedy's theorem twice (see {doc}`L02_18_example14` for a detailed explanation), with the pairs $I_{12}$-$I_{23}$ and $I_{14}$-$I_{34}$. Each pair defines a line, and the ICR is at their intersection:
Finally, $I_{16}$ is found with the pairs $I_{15}$-$I_{56}$ (the line through $O_5$ perpendicular to the shock axis) and $I_{12}$-$I_{26}$:
**2. Velocity of A**
Even with the graphical method, we must first write the equations that determine the velocities, following a loop that goes through all the bodies of the kinematic chain to analyze. Here, the loop is $O_4$-$B$-$A$-$O_2$, which gives the following relative velocity equations:
$$
\begin{aligned}
\vec{v}_B &= \cancel{\vec{v}_{O_4}} + \omega_4\,\hat{k} \times \vec{r}_{O_4B} && \text{(Body 4)} \\
\vec{v}_A &= \vec{v}_B + \omega_3\,\hat{k} \times \vec{r}_{BA} && \text{(Body 3)} \\
\vec{v}_A &= \cancel{\vec{v}_{O_2}} + \omega_2\,\hat{k} \times \vec{r}_{O_2A} && \text{(Body 2)}
\end{aligned}
$$
For the graphical method it is convenient to combine them into a single vector equation: substitute $\vec{v}_B$ from the first equation into the second, and equate the right-hand sides of the second and third, since both are $\vec{v}_A$:
$$
\underbrace{\omega_2\,\hat{k} \times \vec{r}_{O_2A}}_{\perp\, O_2A}
=
\underbrace{\omega_4\,\hat{k} \times \vec{r}_{O_4B}}_{\text{known}}
+
\underbrace{\omega_3\,\hat{k} \times \vec{r}_{BA}}_{\perp\, AB}
$$
where we have marked which term is fully known (magnitude and direction), and which are known only in direction (perpendicular to their links) but not in magnitude.
From the data, $|\omega_4|\,\overline{O_4B} = 1~\text{rad/s} \cdot \sqrt{22.25^2 + 2.3^2}~\text{cm} = 22.35$ cm/s.
Drawing this equation to scale gives the following velocity polygon:
from which we measure:
* $|\vec{v}_{AO_2}| = |\vec{v}_A| = 13.92$ cm/s, the requested value.
* $|\vec{v}_{BA}| = 12.45$ cm/s, from which $|\omega_3| = |\vec{v}_{BA}| / \overline{BA} = 12.45 / 26.91 = 0.46$ rad/s, counterclockwise.
**3. Angular accelerations**
As for the velocities, it is convenient to write the equations explicitly, to identify each vector of the polygon:
$$
\begin{aligned}
\vec{a}_B &= \cancel{\vec{a}_{O_4}} + \alpha_4\,\hat{k} \times \vec{r}_{O_4B} - \omega_4^2\,\vec{r}_{O_4B} && \text{(Body 4)} \\
\vec{a}_A &= \vec{a}_B + \alpha_3\,\hat{k} \times \vec{r}_{BA} - \omega_3^2\,\vec{r}_{BA} && \text{(Body 3)} \\
\vec{a}_A &= \cancel{\vec{a}_{O_2}} + \alpha_2\,\hat{k} \times \vec{r}_{O_2A} - \omega_2^2\,\vec{r}_{O_2A} && \text{(Body 2)}
\end{aligned}
$$
which, combined into a single equation:
$$
\underbrace{\alpha_2\,\hat{k} \times \vec{r}_{O_2A}}_{\vec{a}^t_A}
\underbrace{- \,\omega_2^2\,\vec{r}_{O_2A}}_{\vec{a}^n_A}
=
\underbrace{\alpha_4\,\hat{k} \times \vec{r}_{O_4B}}_{\vec{a}^t_B}
\underbrace{- \,\omega_4^2\,\vec{r}_{O_4B}}_{\vec{a}^n_B}
+
\underbrace{\alpha_3\,\hat{k} \times \vec{r}_{BA}}_{\vec{a}^t_{BA}}
\underbrace{- \,\omega_3^2\,\vec{r}_{BA}}_{\vec{a}^n_{BA}}
$$
allows identifying each vector of the given polygon by its direction (parallel or perpendicular to each link). The normal components are known from the angular velocities, and $\vec{a}^t_B$ from the given $\alpha_4$; the step-by-step construction below shows how the polygon is built.
Measuring on the polygon:
* $|\vec{a}^t_{BA}| \approx 8.1$ cm/s², so $|\alpha_3| = |\vec{a}^t_{BA}| / \overline{BA} \approx 8.1 / 26.91 \approx 0.3$ rad/s², clockwise.
* $|\vec{a}^t_A| \approx 29.15$ cm/s², so $|\alpha_2| = |\vec{a}^t_A| / \overline{O_2A} \approx 29.15 / 6.04 \approx 4.82$ rad/s², clockwise.
These values agree with the analytical solution of {doc}`L02_13_example9`: $\alpha_3 = -0.3016$ rad/s² and $\alpha_2 = -4.803$ rad/s².
:::
## Step-by-step constructions
**Absolute ICRs**, and a check of $\vec{v}_A$ using $I_{13}$:
```{raw} html
```
**Velocity polygon** (part 2):
```{raw} html
```
**Acceleration polygon** (part 3):
```{raw} html
```
Data: $\omega_4 = 1$ rad/s counterclockwise, $\alpha_4 = 2$ rad/s² clockwise. Any other dimension can be measured with the scale bars.
:::{dropdown} Solution
**1. Absolute ICRs**
The mechanism has $N = 6$ bodies: 1 frame (ground), 2 rocker $DO_2A$, 3 swingarm $ABC$, 4 link $O_4B$, 5 shock cylinder and 6 shock rod. So there are $6 \cdot 5 / 2 = 15$ ICRs, of which the absolute ones are required: $I_{12}$, $I_{13}$, $I_{14}$, $I_{15}$ and $I_{16}$.
First, all the ICRs that can be seen directly: the pin joints between pairs of bodies in contact ($I_{12}$, $I_{23}$, $I_{34}$, $I_{14}$, $I_{15}$ and $I_{26}$), and $I_{56}$, which is at infinity in the direction perpendicular to the shock axis, since the rod slides inside the cylinder.
Of the absolute ICRs, only $I_{13}$ and $I_{16}$ are still unknown.
To find $I_{13}$, apply Kennedy's theorem twice (see {doc}`L02_18_example14` for a detailed explanation), with the pairs $I_{12}$-$I_{23}$ and $I_{14}$-$I_{34}$. Each pair defines a line, and the ICR is at their intersection:
Finally, $I_{16}$ is found with the pairs $I_{15}$-$I_{56}$ (the line through $O_5$ perpendicular to the shock axis) and $I_{12}$-$I_{26}$:
**2. Velocity of A**
Even with the graphical method, we must first write the equations that determine the velocities, following a loop that goes through all the bodies of the kinematic chain to analyze. Here, the loop is $O_4$-$B$-$A$-$O_2$, which gives the following relative velocity equations:
$$
\begin{aligned}
\vec{v}_B &= \cancel{\vec{v}_{O_4}} + \omega_4\,\hat{k} \times \vec{r}_{O_4B} && \text{(Body 4)} \\
\vec{v}_A &= \vec{v}_B + \omega_3\,\hat{k} \times \vec{r}_{BA} && \text{(Body 3)} \\
\vec{v}_A &= \cancel{\vec{v}_{O_2}} + \omega_2\,\hat{k} \times \vec{r}_{O_2A} && \text{(Body 2)}
\end{aligned}
$$
For the graphical method it is convenient to combine them into a single vector equation: substitute $\vec{v}_B$ from the first equation into the second, and equate the right-hand sides of the second and third, since both are $\vec{v}_A$:
$$
\underbrace{\omega_2\,\hat{k} \times \vec{r}_{O_2A}}_{\perp\, O_2A}
=
\underbrace{\omega_4\,\hat{k} \times \vec{r}_{O_4B}}_{\text{known}}
+
\underbrace{\omega_3\,\hat{k} \times \vec{r}_{BA}}_{\perp\, AB}
$$
where we have marked which term is fully known (magnitude and direction), and which are known only in direction (perpendicular to their links) but not in magnitude.
From the data, $|\omega_4|\,\overline{O_4B} = 1~\text{rad/s} \cdot \sqrt{22.25^2 + 2.3^2}~\text{cm} = 22.35$ cm/s.
Drawing this equation to scale gives the following velocity polygon:
from which we measure:
* $|\vec{v}_{AO_2}| = |\vec{v}_A| = 13.92$ cm/s, the requested value.
* $|\vec{v}_{BA}| = 12.45$ cm/s, from which $|\omega_3| = |\vec{v}_{BA}| / \overline{BA} = 12.45 / 26.91 = 0.46$ rad/s, counterclockwise.
**3. Angular accelerations**
As for the velocities, it is convenient to write the equations explicitly, to identify each vector of the polygon:
$$
\begin{aligned}
\vec{a}_B &= \cancel{\vec{a}_{O_4}} + \alpha_4\,\hat{k} \times \vec{r}_{O_4B} - \omega_4^2\,\vec{r}_{O_4B} && \text{(Body 4)} \\
\vec{a}_A &= \vec{a}_B + \alpha_3\,\hat{k} \times \vec{r}_{BA} - \omega_3^2\,\vec{r}_{BA} && \text{(Body 3)} \\
\vec{a}_A &= \cancel{\vec{a}_{O_2}} + \alpha_2\,\hat{k} \times \vec{r}_{O_2A} - \omega_2^2\,\vec{r}_{O_2A} && \text{(Body 2)}
\end{aligned}
$$
which, combined into a single equation:
$$
\underbrace{\alpha_2\,\hat{k} \times \vec{r}_{O_2A}}_{\vec{a}^t_A}
\underbrace{- \,\omega_2^2\,\vec{r}_{O_2A}}_{\vec{a}^n_A}
=
\underbrace{\alpha_4\,\hat{k} \times \vec{r}_{O_4B}}_{\vec{a}^t_B}
\underbrace{- \,\omega_4^2\,\vec{r}_{O_4B}}_{\vec{a}^n_B}
+
\underbrace{\alpha_3\,\hat{k} \times \vec{r}_{BA}}_{\vec{a}^t_{BA}}
\underbrace{- \,\omega_3^2\,\vec{r}_{BA}}_{\vec{a}^n_{BA}}
$$
allows identifying each vector of the given polygon by its direction (parallel or perpendicular to each link). The normal components are known from the angular velocities, and $\vec{a}^t_B$ from the given $\alpha_4$; the step-by-step construction below shows how the polygon is built.
Measuring on the polygon:
* $|\vec{a}^t_{BA}| \approx 8.1$ cm/s², so $|\alpha_3| = |\vec{a}^t_{BA}| / \overline{BA} \approx 8.1 / 26.91 \approx 0.3$ rad/s², clockwise.
* $|\vec{a}^t_A| \approx 29.15$ cm/s², so $|\alpha_2| = |\vec{a}^t_A| / \overline{O_2A} \approx 29.15 / 6.04 \approx 4.82$ rad/s², clockwise.
These values agree with the analytical solution of {doc}`L02_13_example9`: $\alpha_3 = -0.3016$ rad/s² and $\alpha_2 = -4.803$ rad/s².
:::
## Step-by-step constructions
**Absolute ICRs**, and a check of $\vec{v}_A$ using $I_{13}$:
```{raw} html
```
**Velocity polygon** (part 2):
```{raw} html
```
**Acceleration polygon** (part 3):
```{raw} html
```