Interactive Lecture
Four-Bar Linkages, Velocity and Acceleration Analysis
by Dr Arnaldo Delli Carri
Mechanism kinematics is the study of motion in mechanical systems without considering the forces that cause the motion. You need the position, velocity, and acceleration of every link to check whether the mechanism does what you need it to — and to spot problems before anything is built.
Key Concept: Kinematics analysis allows engineers to:
The position of any point in a mechanism can be determined using vector loop equations. For a mechanism with rigid links, we can write:
Loop closure equation:
\(\sum_{i=1}^{n} \vec{r}_i = 0\)
Where \(\vec{r}_i\) represents the position vector of link \(i\)
For a point A on a rigid body rotating about point O:
\(\vec{r}_A = \vec{r}_O + \vec{r}_{A/O}\)
In component form:
\(x_A = x_O + r_{AO} \cos(\theta)\)
\(y_A = y_O + r_{AO} \sin(\theta)\)
A position diagram fixes the instantaneous configuration of every link and joint in the mechanism. It sets the geometry you differentiate to get velocities and accelerations, and it exposes where links would collide or run out of travel.
Taking the time derivative of the position equation gives us the velocity relationship:
General velocity equation for point A on a rigid body:
\(\vec{v}_A = \vec{v}_O + \vec{\omega} \times \vec{r}_{A/O}\)
In 2D mechanisms, this simplifies to:
\(\vec{v}_A = \vec{v}_O + \omega \hat{k} \times \vec{r}_{A/O}\)
Where:
Velocity diagrams (hodographs) are graphical constructions that represent the velocity vectors of different points in a mechanism. These diagrams follow specific rules:
Instantaneous Centre of Rotation: For any rigid body in plane motion, there exists a point (which may be outside the body) that has zero velocity at that instant. All other points appear to rotate about this instantaneous centre.
Taking the time derivative of the velocity equation yields the acceleration relationship:
General acceleration equation for point A:
\(\vec{a}_A = \vec{a}_O + \vec{\alpha} \times \vec{r}_{A/O} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{A/O})\)
Breaking into components:
\(\vec{a}_A = \vec{a}_O + \vec{a}_{A/O}^{tangential} + \vec{a}_{A/O}^{centripetal}\)
Where:
The total acceleration consists of:
Important: Even at constant angular velocity (\(\alpha = 0\)), points on rotating bodies experience centripetal acceleration directed towards the centre of rotation.
When analysing mechanisms with sliding joints or multiple moving reference frames, we must consider relative motion:
Relative velocity equation:
\(\vec{v}_A = \vec{v}_B + \vec{v}_{A/B}\)
Relative acceleration equation:
\(\vec{a}_A = \vec{a}_B + \vec{a}_{A/B} + \vec{a}_{Coriolis}\)
When a point moves relative to a rotating reference frame, an additional acceleration component appears:
Coriolis acceleration:
\(\vec{a}_{Coriolis} = 2\vec{\omega} \times \vec{v}_{rel}\)
Where:
Physical Interpretation: Coriolis acceleration arises because the direction of relative velocity changes as the reference frame rotates. It's always perpendicular to both the angular velocity vector and the relative velocity vector.
A four-bar linkage is one of the simplest and most versatile mechanisms. Adjust the angular velocity and acceleration to see how the velocity and acceleration vectors change throughout the mechanism.
The slider-crank mechanism converts rotary motion to linear motion (or vice versa). It's the basis of piston engines and many reciprocating machines.
The quick-return mechanism provides different velocities for forward and return strokes, useful in shaping machines where the cutting stroke should be slower than the return stroke.
Kinematic analysis is the foundation of machine design — you'll use it any time a system has linked moving parts:
| Concept | Equation | Description |
|---|---|---|
| Angular Velocity | \(\omega = \frac{2\pi}{60} N\) | Convert RPM to rad/s |
| Linear Velocity | \(\vec{v} = \vec{\omega} \times \vec{r}\) | Velocity of point on rotating body |
| Velocity equation | \(\vec{v}_P =\vec{v}_O + \vec{\omega} \times \vec{r}_{P/O}\) | Velocity of point P wrt O |
| Relative Velocity | \(\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A\) | Velocity of B relative to A |
| Instantaneous Centre | \(v = \omega \times r_{IC}\) | Velocity using instant centre |
| Normal Acceleration | \(a_n = \omega^2 r\) | Centripetal acceleration |
| Tangential Acceleration | \(a_t = \alpha r\) | Due to angular acceleration |
| Acceleration equation | \(\vec{a}_P =\vec{a}_O + \vec{\omega} \times \left(\vec{\omega} \times\vec{r}_{P/O}\right)+\alpha\hat{k}\times\vec{r}_{P/O}\) | Acceleration of point P wrt O |
A four-bar linkage has the following dimensions: Ground link = 100 mm, Crank = 40 mm, Coupler = 80 mm, Rocker = 60 mm. If the crank rotates at 120 rpm clockwise, calculate:
Show Solution
Given: L₁ = 100 mm (ground O₂O₄), L₂ = 40 mm (crank), L₃ = 80 mm (coupler), L₄ = 60 mm (rocker), N = 120 rpm, θ₂ = 45°. Take the open (non-crossed) assembly.
a) Angular velocity of crank:
\(\omega_2 = \frac{2\pi N}{60} = \frac{2\pi \times 120}{60} = 12.57 \text{ rad/s}\)
b) Velocity of coupler point A (crank-coupler joint):
\(v_A = \omega_2 \, L_2 = 12.57 \times 40 = 502.8 \text{ mm/s}\)
Direction: perpendicular to the crank, i.e. at 45° − 90° = −45° to horizontal.
c) Rocker angular velocity by the vector-loop (analytical) method. Place O₂ at the origin and O₄ at (100, 0). The crank pin is
\(A = (L_2\cos\theta_2,\; L_2\sin\theta_2) = (28.28,\; 28.28)\text{ mm}\)
Intersecting the coupler circle (radius L₃ about A) with the rocker circle (radius L₄ about O₄) gives the joint
\(B = (101.7,\; 60.0)\text{ mm}\)
so \(\vec{r}_{B/A} = (73.46,\; 31.69)\) and \(\vec{r}_{B/O_4} = (1.74,\; 59.98)\) mm. The velocity loop \(\vec{v}_B = \vec{v}_A + \omega_3\hat{k}\times\vec{r}_{B/A} = \omega_4\hat{k}\times\vec{r}_{B/O_4}\) with \(\vec{v}_A = \omega_2\hat{k}\times\vec{r}_{A/O_2} = (-355.4,\; 355.4)\) mm/s gives two scalar equations:
\(-59.98\,\omega_4 = -355.4 - 31.69\,\omega_3\)
\(1.74\,\omega_4 = 355.4 + 73.46\,\omega_3\)
Solving: \(\omega_3 = -4.76 \text{ rad/s}\), \(\omega_4 = 3.41 \text{ rad/s}\).
Answer: (a) \(\omega_2 = 12.57\) rad/s; (b) \(v_A = 502.8\) mm/s at −45° to horizontal; (c) \(\omega_4 = 3.41\) rad/s (open assembly).
A slider-crank mechanism has a crank radius of 50 mm and connecting rod length of 150 mm. The crank rotates at a constant 300 rpm. Calculate the acceleration of the piston when the crank angle is 90° from the horizontal.
Show Solution
Given: r = 50 mm, l = 150 mm, N = 300 rpm (constant), θ = 90° (measured from the crank–slider axis, θ = 0 at maximum extension)
Angular velocity:
\(\omega = \frac{2\pi \times 300}{60} = 31.42 \text{ rad/s}\)
\(\alpha = 0\) (constant speed)
At θ = 90°:
Crank pin position: x = 0, y = 50 mm
Connecting rod angle:
\(\beta = \sin^{-1}\left(\frac{r \sin\theta}{l}\right) = \sin^{-1}\left(\frac{50}{150}\right) = 19.47°\)
Piston acceleration (using standard formula):
\(a_p = -r\omega^2[\cos\theta + \frac{r\cos(2\theta)}{l}]\)
\(a_p = -50 \times 31.42^2[0 + \frac{50 \times (-1)}{150}]\)
\(a_p = -50 \times 987.4 \times (-0.333) = 16{,}446 \text{ mm/s}^2 = 16.45 \text{ m/s}^2\)
The result is positive, and with θ measured from the crank centre positive x points away from it — so at this instant the piston accelerates away from the crank centre (towards the outer/head end). The whole term is the connecting-rod obliquity (secondary) contribution, since the primary term \(\cos\theta\) vanishes at θ = 90°.
Answer: \(a_p = 16{,}446\) mm/s² = 16.45 m/s², directed away from the crank centre along the slider axis.
In a quick-return mechanism, the crank rotates at 60 rpm. At a certain instant, the slotted lever has an angular velocity of 2 rad/s and the sliding block is 100 mm from the pivot with a relative velocity of 200 mm/s along the slot. Calculate the Coriolis acceleration of the sliding block.
Show Solution
Given: ω_lever = 2 rad/s, v_rel = 200 mm/s, r = 100 mm
Coriolis acceleration formula:
\(a_{Coriolis} = 2\omega \times v_{rel}\)
Magnitude:
\(|a_{Coriolis}| = 2 \times \omega_{lever} \times v_{rel}\)
\(|a_{Coriolis}| = 2 \times 2 \times 200 = 800 \text{ mm/s}^2\)
Direction: the lever rotates in the vertical working plane, so its ω points out of that plane (horizontal, out of the page). The Coriolis term \(2\vec{\omega}\times\vec{v}_{rel}\) is therefore in the plane of motion and perpendicular to the slot.
Total acceleration of the sliding block also includes:
Answer: Coriolis acceleration = 800 mm/s², perpendicular to the slot (in the plane of lever motion).