# Example 2.14: Instantaneous centers of a four-bar linkage
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For the following mechanism (February 2016 exam), find:
1. How many instantaneous centers of rotation (ICRs) exist, including relative and absolute ones.
2. The location of all the ICRs.
3. Using these ICRs, the angular velocity of link 3 as a function of the angular velocity of link 2, for the position of the mechanism shown in the figure.
Bar $CO_4$ is built into the wall, so it is part of the ground (link 1), and link 4 is bar $BC$.
:::{dropdown} Solution
**1. Number of ICRs**
There are as many ICRs as pairs of links, including the ground (link 1). They can be counted with a table:
| | **2** | **3** | **4** |
|---|---|---|---|
| **1** | $I_{12}$ | $I_{13}$ | $I_{14}$ |
| **2** | | $I_{23}$ | $I_{24}$ |
| **3** | | | $I_{34}$ |
so there are 6 ICRs, as given by the general formula for $N = 4$ links:
$$
n_{ICR} = \frac{N (N-1)}{2} = 6
$$
Writing the table is advisable anyway, to cross out each ICR as it is found.
**2. Location of all the ICRs**
First, the primary ICRs, at the pin joints between links: $I_{12}$ (at $O_2$), $I_{23}$ (at $A$), $I_{34}$ (at $B$) and $I_{14}$ (at $C$).
The two missing ICRs are found with Kennedy's theorem: the three ICRs of any three bodies lie on a straight line. Applying it twice for each unknown ICR gives two lines, and the ICR is at their intersection.
For $I_{13}$, the auxiliary polygon (one vertex per link, one side per known ICR) shows the two paths that connect 1 and 3 through known ICRs, $1 \rightarrow 2 \rightarrow 3$ and $1 \rightarrow 4 \rightarrow 3$. So $I_{13}$ lies at the intersection of:
* the line through $I_{12}$ and $I_{23}$ (at $45^\circ$), and
* the line through $I_{14}$ and $I_{34}$ (vertical, at $x = 750$ mm).
That is, at $(750, 750)$ mm from $O_2$, above the mechanism.
For $I_{24}$, the paths are $2 \rightarrow 3 \rightarrow 4$ and $2 \rightarrow 1 \rightarrow 4$, so it lies at the intersection of:
* the line through $I_{23}$ and $I_{34}$, and
* the line through $I_{12}$ and $I_{14}$.
Both lines are parallel (horizontal), so $I_{24}$ is a point at infinity, in the horizontal direction. In such cases, what matters is not any particular point of the line of the ICR but its direction, so it can be drawn wherever it is most convenient.
Since $I_{24}$ is at infinity, links 2 and 4 have the same angular velocity at this instant ($\omega_4 = \omega_2$), as in a parallelogram.
**3. $\omega_3$ as a function of $\omega_2$**
The relative ICR $I_{23}$ is a point of both links: as a point of link 2 it rotates about $I_{12}$ with $\omega_2$, and as a point of link 3 it rotates about $I_{13}$ with $\omega_3$. Using magnitudes only:
$$
|\vec{v}_{I_{23}}| = |\omega_2|\,\overline{I_{12}I_{23}} = |\omega_3|\,\overline{I_{13}I_{23}}
\qquad\rightarrow\qquad
|\omega_3| = |\omega_2|\,\frac{\overline{I_{12}I_{23}}}{\overline{I_{13}I_{23}}}
$$
From the figure, $\overline{I_{12}I_{23}} = 400\sqrt{2}$ mm and $\overline{I_{13}I_{23}} = 350\sqrt{2}$ mm.
Since the relative ICR $I_{23}$ lies **between** the two absolute ICRs $I_{12}$ and $I_{13}$, both links rotate in opposite senses: if one rotates clockwise, the other rotates counterclockwise. Adding this sign:
$$
\omega_3 = -\omega_2\,\frac{400\sqrt{2}}{350\sqrt{2}} = -\frac{8}{7}\,\omega_2
$$
:::
## Step-by-step construction
```{raw} html
```
Bar $CO_4$ is built into the wall, so it is part of the ground (link 1), and link 4 is bar $BC$.
:::{dropdown} Solution
**1. Number of ICRs**
There are as many ICRs as pairs of links, including the ground (link 1). They can be counted with a table:
| | **2** | **3** | **4** |
|---|---|---|---|
| **1** | $I_{12}$ | $I_{13}$ | $I_{14}$ |
| **2** | | $I_{23}$ | $I_{24}$ |
| **3** | | | $I_{34}$ |
so there are 6 ICRs, as given by the general formula for $N = 4$ links:
$$
n_{ICR} = \frac{N (N-1)}{2} = 6
$$
Writing the table is advisable anyway, to cross out each ICR as it is found.
**2. Location of all the ICRs**
First, the primary ICRs, at the pin joints between links: $I_{12}$ (at $O_2$), $I_{23}$ (at $A$), $I_{34}$ (at $B$) and $I_{14}$ (at $C$).
The two missing ICRs are found with Kennedy's theorem: the three ICRs of any three bodies lie on a straight line. Applying it twice for each unknown ICR gives two lines, and the ICR is at their intersection.
For $I_{13}$, the auxiliary polygon (one vertex per link, one side per known ICR) shows the two paths that connect 1 and 3 through known ICRs, $1 \rightarrow 2 \rightarrow 3$ and $1 \rightarrow 4 \rightarrow 3$. So $I_{13}$ lies at the intersection of:
* the line through $I_{12}$ and $I_{23}$ (at $45^\circ$), and
* the line through $I_{14}$ and $I_{34}$ (vertical, at $x = 750$ mm).
That is, at $(750, 750)$ mm from $O_2$, above the mechanism.
For $I_{24}$, the paths are $2 \rightarrow 3 \rightarrow 4$ and $2 \rightarrow 1 \rightarrow 4$, so it lies at the intersection of:
* the line through $I_{23}$ and $I_{34}$, and
* the line through $I_{12}$ and $I_{14}$.
Both lines are parallel (horizontal), so $I_{24}$ is a point at infinity, in the horizontal direction. In such cases, what matters is not any particular point of the line of the ICR but its direction, so it can be drawn wherever it is most convenient.
Since $I_{24}$ is at infinity, links 2 and 4 have the same angular velocity at this instant ($\omega_4 = \omega_2$), as in a parallelogram.
**3. $\omega_3$ as a function of $\omega_2$**
The relative ICR $I_{23}$ is a point of both links: as a point of link 2 it rotates about $I_{12}$ with $\omega_2$, and as a point of link 3 it rotates about $I_{13}$ with $\omega_3$. Using magnitudes only:
$$
|\vec{v}_{I_{23}}| = |\omega_2|\,\overline{I_{12}I_{23}} = |\omega_3|\,\overline{I_{13}I_{23}}
\qquad\rightarrow\qquad
|\omega_3| = |\omega_2|\,\frac{\overline{I_{12}I_{23}}}{\overline{I_{13}I_{23}}}
$$
From the figure, $\overline{I_{12}I_{23}} = 400\sqrt{2}$ mm and $\overline{I_{13}I_{23}} = 350\sqrt{2}$ mm.
Since the relative ICR $I_{23}$ lies **between** the two absolute ICRs $I_{12}$ and $I_{13}$, both links rotate in opposite senses: if one rotates clockwise, the other rotates counterclockwise. Adding this sign:
$$
\omega_3 = -\omega_2\,\frac{400\sqrt{2}}{350\sqrt{2}} = -\frac{8}{7}\,\omega_2
$$
:::
## Step-by-step construction
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