13.1 Mechanism Kinematics, Linkages, Cams & Simple Machines

13.1 Mechanism Kinematics, Linkages & Instant Centers

Kinematics is the study of geometry in motion—analyzing position, displacement, velocity, and acceleration of machine components without regard to the forces causing the motion. In machine design and mechanical automation, kinematic synthesis and analysis ensure that linkages, actuators, and transmissions guide tooling through required paths, produce necessary mechanical advantage, and minimize peak inertial accelerations. Understanding mechanism mobility, four-bar linkage dynamics, slider-crank behavior, and instant centers provides the analytical foundation required for machine design problems on the PE exam.

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Planar Mechanism Mobility & Degrees of Freedom

The mobility (or degrees of freedom, $M$) of a mechanical linkage represents the number of independent input parameters (typically position coordinates or actuator inputs) required to completely define the physical configuration and spatial orientation of every link in the mechanism.

Gruebler's (Kutzbach) Criterion for Planar Mechanisms

In an unconstrained planar system, an isolated rigid link possesses 3 independent degrees of freedom: translation in $x$, translation in $y$, and rotation $\theta$ about the $z$-axis. When $n$ rigid links are assembled into a mechanism, one link is selected as the stationary ground frame (leaving $n-1$ moving links with $3(n-1)$ unconstrained degrees of freedom). Connecting these links with kinematic joints removes degrees of freedom.

The degree of freedom of a planar mechanism is evaluated using Gruebler's criterion (also referred to as the Kutzbach mobility equation):

$$M = 3(n - 1) - 2 j_1 - j_2$$

where:

• $n$ = total number of links in the mechanism, including the fixed base/ground frame ($n \ge 2$).
• $j_1$ = total number of single-degree-of-freedom joints (lower pairs). Each $j_1$ joint removes 2 degrees of relative motion, permitting 1 degree of freedom.
• $j_2$ = total number of two-degree-of-freedom joints (higher pairs). Each $j_2$ joint removes 1 degree of relative motion, permitting 2 degrees of freedom.

                             Joint Classifications
  +-------------------------------------------------------------------------+
  | Joint Type             | Allowed Relative Motion | DOF Removed | Symbol |
  |------------------------+-------------------------+-------------+--------|
  | Revolute (Pin)         | 1-DOF Rotation          | 2 DOF       | j_1    |
  | Prismatic (Slider)     | 1-DOF Translation       | 2 DOF       | j_1    |
  | Cam-Follower (Roll/Slip)| Rotation + Tangent Slip | 1 DOF       | j_2    |
  | Gear Tooth Mesh        | Rolling + Sliding       | 1 DOF       | j_2    |
  | Pin-in-Slot            | Sliding + Rotation      | 1 DOF       | j_2    |
  +-------------------------------------------------------------------------+

Physical Significance of Mobility ($M$)

• $M = 1$ (Constrained Mechanism): The system possesses exactly one degree of freedom. A single driver (e.g., an electric motor rotating a crank at $\omega$ or a hydraulic cylinder extending a rod at $v$) produces determinate, repeatable motion in all other links.
• $M > 1$ (Multi-Degree-of-Freedom Mechanism): The mechanism requires $M$ independent actuators to control its configuration (e.g., a two-link robotic arm with $M = 2$ requiring two servo actuators).
• $M = 0$ (Statically Determinate Structure): The links form a rigid frame or truss incapable of gross kinematic motion without elastic deformation.
• $M < 0$ (Statically Indeterminate Structure): The system contains redundant constraints (pre-stressed structure). Assembling the links requires elastic deformation or high manufacturing tolerances.

Joint Multiplicity and Idle Degrees of Freedom

Two practical caveats frequently appear on kinematic exams:

1. Multiple Pin Joints: When $k$ links are pinned together at the same physical pivot point, the junction is counted as $(k - 1)$ separate 1-DOF revolute joints ($j_1$). For example, three links pinned together at a single hinge count as $3 - 1 = 2$ pin joints.
2. Idle (Redundant) Degrees of Freedom ($f_{\text{idle}}$): Certain mechanisms incorporate components that move without altering the primary kinematic relationship between input and output. A standard cam mechanism with a rotating roller follower possesses an idle rotational degree of freedom of the roller about its own axis. The modified mobility equation is: $$M = 3(n - 1) - 2 j_1 - j_2 - f_{\text{idle}}$$

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Four-Bar Linkages & Grashof's Condition

The planar four-bar linkage is the fundamental building block of kinematic machinery. It consists of four rigid links connected by four revolute joints:

• Link 1: Fixed base frame (ground)
• Link 2: Input link (driver / crank)
• Link 3: Coupler link (connecting rod)
• Link 4: Output link (rocker / follower)

                     Four-Bar Linkage Diagram
                           Coupler (Link 3)
                       A +-----------------+ B
                        /                   \
        Input (Link 2) /                     \ Output (Link 4)
                      /                       \
                   O_2 +-----------------------+ O_4
                           Ground (Link 1)

Grashof's Criterion

Grashof's law predicts the rotational behavior of a planar four-bar linkage based entirely on the lengths of its four links:

• Let $s$ = length of the shortest link
• Let $l$ = length of the longest link
• Let $p, q$ = lengths of the two intermediate links

A linkage is Grashof if:

$$s + l \le p + q$$

If this inequality is satisfied, at least one link can execute a continuous $360^\circ$ rotation relative to the other three links.

Classifications of Grashof Linkages ($s + l < p + q$)

Depending on which link is chosen as the fixed ground frame (inversion), three distinct mechanisms arise:

1. Crank-Rocker Mechanism:

   • Ground link is adjacent to the shortest link $s$.
   • The shortest link $s$ serves as the input crank, rotating continuously through $360^\circ$.
   • The opposite grounded link oscillates back and forth over a finite angular stroke as an output rocker.
   • Widely utilized in windshield wipers, sewing machine needle drives, and reciprocating conveyors.

2. Double-Crank Mechanism (Drag-Link):

   • The shortest link $s$ is the stationary ground link.
   • Both links connected to ground (links 2 and 4) are capable of complete $360^\circ$ continuous rotation.
   • When driven at a constant input angular velocity, the output crank rotates with a cyclically varying angular velocity, functioning as a quick-return motion generator.

3. Double-Rocker Mechanism:

   • The link opposite the shortest link $s$ is chosen as the ground link (i.e., shortest link $s$ is the coupler).
   • Both grounded links rock back and forth through bounded angular ranges; neither grounded link can rotate fully through $360^\circ$, though the floating coupler executes full rotations.

Non-Grashof Linkages ($s + l > p + q$)

If $s + l > p + q$, the linkage is Non-Grashof (Triple-Rocker). No link can make a complete $360^\circ$ revolution relative to any other link. All three moving links oscillate between limit stops regardless of which link is grounded.

Change-Point Mechanisms ($s + l = p + q$)

When $s + l = p + q$, the linkage is a change-point mechanism (e.g., a parallelogram or deltoid linkage). At the collinear positions (where all four links become parallel or lie on a single straight line), the output motion becomes indeterminate. The mechanism can "toggle" or fold back on itself unless guided through the change point by flywheel inertia or duplicate out-of-phase links (such as steam locomotive side-rods).

Transmission Angle ($\mu$)

The transmission angle $\mu$ is defined as the acute angle between the coupler link (link 3) and the output rocker link (link 4):

                F_coupler
         Link 3 ---------> * (Joint B)
                          / \
                         /   \  Transmission Angle mu
                        /     \
                       /       \  Link 4 (Output Rocker)
                      v         v
                   F_axial   F_transverse = F_coupler * sin(mu)

The force transmitted from the coupler drives the output rocker through the transverse component $F_{\text{transverse}} = F_{\text{coupler}} \sin\mu$, while the axial component $F_{\text{axial}} = F_{\text{coupler}} \cos\mu$ generates bearing reaction loads on pivot $O_4$.

• Ideal Condition: $\mu = 90^\circ$ (maximum torque transmission, zero axial bearing load).
• Design Criterion: For smooth, reliable operation without mechanical locking or excessive wear, the transmission angle throughout the entire cycle should satisfy: $$45^\circ \le \mu \le 135^\circ \quad (|90^\circ - \mu| \le 45^\circ)$$
• If $\mu < 45^\circ$ or $\mu > 135^\circ$, force transmission efficiency drops significantly, mechanical deflection increases, and the linkage risks binding or jamming.

Using the law of cosines, the extreme transmission angles occur when the input crank is collinear with the fixed ground frame:

• Minimum Transmission Angle $\mu_{\min}$ (at $\theta_2 = 0^\circ$, inner dead center): $$\cos\mu_{\min} = \frac{L_3^2 + L_4^2 - (L_1 - L_2)^2}{2 L_3 L_4}$$
• Maximum Transmission Angle $\mu_{\max}$ (at $\theta_2 = 180^\circ$, outer dead center): $$\cos\mu_{\max} = \frac{L_3^2 + L_4^2 - (L_1 + L_2)^2}{2 L_3 L_4}$$

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Slider-Crank Mechanism Kinematics

The slider-crank mechanism converts rotational motion into reciprocating linear motion (or vice versa), serving as the foundational kinematics for internal combustion engines, reciprocating compressors, and stamping presses.

                      Slider-Crank Kinematics
                       Coupler (Length l)
                     A +-------------------+
                      /                     \ Piston (Slider)
          Crank (r)  /                       +-------+
                    / theta                  | P (x) | ===>
                 O +-------------------------+-------+
                            Stroke Axis

Analytical Displacement, Velocity & Acceleration

Let:

• $r$ = crank radius (length of link 2)
• $l$ = connecting rod length (length of link 3)
• $\lambda = \frac{r}{l}$ = crank-to-rod ratio (typically $0.20 \le \lambda \le 0.30$)
• $\theta = \omega t$ = crank angle measured from Top Dead Center (TDC)
• $\omega$ = constant crank angular velocity (rad/s)

Measuring slider displacement $x$ from the crank center $O$ toward the slider:

$$x(\theta) = r \cos\theta + \sqrt{l^2 - r^2 \sin^2\theta} = r \cos\theta + l \sqrt{1 - \lambda^2 \sin^2\theta}$$

Measuring piston displacement $s$ from Top Dead Center ($s = 0$ when $\theta = 0$):

$$s(\theta) = (r + l) - x(\theta) = r(1 - \cos\theta) + l \left(1 - \sqrt{1 - \lambda^2 \sin^2\theta}\right)$$

Applying the binomial series approximation $\sqrt{1 - u} \approx 1 - \frac{1}{2}u$ (valid because $\lambda^2 \ll 1$):

$$\sqrt{1 - \lambda^2 \sin^2\theta} \approx 1 - \frac{\lambda^2}{2} \sin^2\theta = 1 - \frac{\lambda^2}{4}(1 - \cos 2\theta)$$

This yields the standard engineering approximations widely applied in engine kinematics:

1. Piston Displacement ($s$ from TDC):

   $$s \approx r \left[ (1 - \cos\theta) + \frac{r}{4l} (1 - \cos 2\theta) \right]$$

2. Piston Velocity ($v = \dot{s}$):

   $$v = r \omega \left( \sin\theta + \frac{r}{2l} \sin 2\theta \right)$$

3. Piston Acceleration ($a = \dot{v} = \ddot{s}$):

   $$a = r \omega^2 \left( \cos\theta + \frac{r}{l} \cos 2\theta \right)$$

Primary vs. Secondary Inertia Dynamics

• Primary Term ($\cos\theta$): Oscillates at engine running frequency $\omega$. Represents pure harmonic motion of an infinitely long connecting rod ($l/r \to \infty$).
• Secondary Term ($\frac{r}{l}\cos 2\theta$): Oscillates at twice engine frequency ($2\omega$). Arises from the angular obliquity of the connecting rod. It causes the piston to travel faster and experience higher accelerations during the top half of the stroke than the bottom half.

At Top Dead Center ($\theta = 0^\circ$):

$$a_{\text{TDC}} = r \omega^2 \left(1 + \frac{r}{l}\right)$$

At Bottom Dead Center ($\theta = 180^\circ$):

$$a_{\text{BDC}} = -r \omega^2 \left(1 - \frac{r}{l}\right)$$

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Instant Centers of Velocity & Kennedy's Theorem

An instant center of velocity (IC) between two bodies in planar motion is a point common to both bodies (or their rigid extensions) having identical instantaneous linear velocity vectors in both magnitude and direction.

Notation and Number of Instant Centers

The instant center between body $i$ and body $j$ is denoted $I_{ij}$ (or $IC(i,j)$). Because $I_{ij} \equiv I_{ji}$, the total number of instant centers $N$ for a mechanism containing $n$ links is the number of 2-link combinations:

$$N = \frac{n(n - 1)}{2}$$

For a standard 4-bar linkage or slider-crank ($n = 4$ links):

$$N = \frac{4(4 - 1)}{2} = 6 \text{ instant centers: } I_{12}, I_{23}, I_{34}, I_{14}, I_{13}, I_{24}$$

Locating Primary Instant Centers by Inspection

1. Revolute (Pin) Joint: The instant center $I_{ij}$ between two pinned links is located directly at the physical center of the pin.
2. Prismatic (Slider) Joint: The instant center $I_{ij}$ between a slider and its straight guide lies at infinity ($\infty$) along a line perpendicular to the sliding axis.
3. Pure Rolling Contact: The instant center $I_{ij}$ between two bodies rolling without slip is located at the instantaneous point of contact.
4. Rolling with Sliding: The instant center lies along the common normal at the point of contact.

Kennedy-Aronhold Theorem of Three Centers

Kennedy-Aronhold Theorem: The three instant centers associated with any three bodies undergoing relative planar motion ($I_{ij}, I_{jk}, \text{and } I_{ik}$) must lie on a single straight line.

This theorem provides the geometric tool to find secondary instant centers without calculations:

• To find $I_{13}$ for links 1, 2, and 3: $I_{13}$ must lie on the line passing through $I_{12}$ and $I_{23}$. Simultaneously, considering links 1, 4, and 3, $I_{13}$ must lie on the line passing through $I_{14}$ and $I_{34}$. The intersection of line $(I_{12}-I_{23})$ and line $(I_{14}-I_{34})$ locates $I_{13}$.
• To find $I_{24}$: $I_{24}$ lies at the intersection of line $(I_{12}-I_{14})$ (the ground link line) and line $(I_{23}-I_{34})$ (the coupler link line).

                      Kennedy's Circle Diagram Method
                            1 (Ground)
                           / \ 
                          /   \ 
                (Link 2) 2 --- 4 (Link 4)
                          \   / 
                           \ / 
                            3 (Coupler)
   Intersection of chords (1-2-3) and (1-4-3) locates instant center I_13
   Intersection of chords (2-1-4) and (2-3-4) locates instant center I_24

Velocity Analysis Using Instant Centers

Instant centers allow direct calculation of velocities without setting up trigonometric loop-closure matrices:

1. Linear Velocity of Instant Center $I_{ij}$: Because $I_{ij}$ is common to body $i$ (rotating about ground pivot $I_{1i}$) and body $j$ (rotating about ground pivot $I_{1j}$):

   $$v_{I_{ij}} = \omega_i \cdot r_{I_{ij}/I_{1i}} = \omega_j \cdot r_{I_{ij}/I_{1j}}$$

2. Angular Velocity Ratio Theorem: Rearranging the common velocity relationship gives the angular velocity ratio between any two links directly from the geometric distances to their mutual instant center:

   $$\frac{\omega_j}{\omega_i} = \frac{r_{I_{ij}/I_{1i}}}{r_{I_{ij}/I_{1j}}}$$

   For a four-bar linkage, the output rocker velocity $\omega_4$ relative to input crank velocity $\omega_2$ is:

   $$\frac{\omega_4}{\omega_2} = \frac{O_2 I_{24}}{O_4 I_{24}}$$

   where $O_2 \equiv I_{12}$ and $O_4 \equiv I_{14}$.

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Mechanical Advantage & Toggle Mechanisms

Mechanical Advantage Definition

Assuming an ideal, frictionless mechanism, power input equals power output:

$$P_{\text{in}} = P_{\text{out}} \implies F_{\text{in}} v_{\text{in}} = F_{\text{out}} v_{\text{out}} \quad (\text{or } T_{\text{in}} \omega_{\text{in}} = T_{\text{out}} \omega_{\text{out}})$$

The Mechanical Advantage ($MA$) is defined as the ratio of output force to input force:

$$MA = \frac{F_{\text{out}}}{F_{\text{in}}} = \frac{v_{\text{in}}}{v_{\text{out}}}$$

For rotational input and output:

$$MA_{\text{torque}} = \frac{T_{\text{out}}}{T_{\text{in}}} = \frac{\omega_{\text{in}}}{\omega_{\text{out}}} = \frac{O_4 I_{24}}{O_2 I_{24}}$$

Toggle Mechanisms ($MA \to \infty$)

A toggle mechanism occurs when two moving links become collinear (the included angle between them approaches $180^\circ$ or $0^\circ$).

                         Toggle Mechanism Alignment
                                F_in
                                 |
                                 v
                     Link 2      A      Link 3
            Fixed O +-----------*-----------+ B (Slider / Output Tool)
                    <----------------------->
                     Collinear Alignment (180 deg)
                     Output velocity v_B -> 0
                     Mechanical Advantage MA -> Infinity

As the links approach collinearity:

$$v_{\text{out}} = v_{\text{in}} \tan\alpha \implies \lim_{\alpha \to 0} v_{\text{out}} = 0$$ $$MA = \frac{F_{\text{out}}}{F_{\text{in}}} = \lim_{v_{\text{out}} \to 0} \frac{v_{\text{in}}}{v_{\text{out}}} \to \infty$$

In actual machinery, the output force is finite, limited only by the structural elasticity of the links, frame rigidity, and pin joint clearances. Toggle positions are widely used in:

• Rock crushers and toggle presses (generating high crushing loads from modest actuator inputs)
• Manual clamping fixtures (holding workpieces securely without continuous clamping force)
• Aircraft landing gear locks (over-center toggle geometry ensures positive mechanical lock under landing impact loads)

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Simple Machines, Cams & Intermittent-Motion Devices

Simple Machines and Ideal Mechanical Advantage

Every mechanism trades force for distance. The ideal mechanical advantage (IMA) follows from geometry alone; the actual advantage is $MA = \eta \times IMA$.

Mini-example: A block and tackle with four rope parts supporting the moving block lifts an $8\,\text{kN}$ load at $85\%$ efficiency. The pull required is $8{,}000/(4 \times 0.85) = 2{,}353\,\text{N}$, and $4\,\text{m}$ of rope must be pulled for each $1\,\text{m}$ of lift. Friction makes many wedges and screws self-locking (see power screws in Section 10.4).

Cams and Followers

A cam converts rotation into a programmed follower motion (rise, dwell, return). Followers may be knife-edge, roller or flat-faced, translating or oscillating, and are held in contact by a spring (force closure) or a groove (form closure). The motion law matters because acceleration drives inertia forces and vibration:

For a cycloidal rise of height $h$ over cam angle $\beta$ at cam speed $\omega$, the follower displacement is $y = h\left[\theta/\beta - \tfrac{1}{2\pi}\sin(2\pi\theta/\beta)\right]$, the peak velocity is $2h\omega/\beta$, and the peak acceleration is $2\pi h \omega^2/\beta^2$. For a simple harmonic rise, the peak acceleration is $\pi^2 h \omega^2/(2\beta^2)$.

The pressure angle is the angle between the follower's direction of motion and the common normal at the contact point. A common rule of thumb keeps it below about $30^\circ$ for translating roller followers; a larger base circle reduces it at the cost of a bigger cam. The return spring must exceed the follower inertia force $m\,|a_{\min}|$ during deceleration, or the follower will jump off the cam.

Intermittent-Motion Devices

• An external Geneva wheel with $n$ slots advances $360^\circ/n$ per driver revolution. It moves only while the driver turns through $180^\circ - 360^\circ/n$ and dwells for the rest: a 4-slot Geneva moves during $90^\circ$ of driver rotation and dwells for $270^\circ$.
• Ratchets and pawls permit motion in one direction only, for indexing and for holding loads on hoists and winches.

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Step-by-Step Worked Kinematic Problem

Problem Statement

A planar four-bar mechanism $O_2 A B O_4$ drives an industrial sorting feeder. The mechanism has the following link lengths:

• Ground link $L_1 = O_2 O_4 = 400\,\text{mm}$
• Input crank $L_2 = O_2 A = 120\,\text{mm}$
• Coupler link $L_3 = A B = 350\,\text{mm}$
• Output rocker $L_4 = O_4 B = 280\,\text{mm}$

The input crank $O_2 A$ rotates at a constant angular velocity $\omega_2 = 25.0\,\text{rad/s}$ counterclockwise.

Calculate:

1. Verify Grashof's condition and identify the mechanism classification.
2. Compute the minimum and maximum transmission angles ($\mu_{\min}$ and $\mu_{\max}$) and evaluate whether the linkage operates within the recommended transmission window ($45^\circ \le \mu \le 135^\circ$).
3. At the configuration where crank angle $\theta_2 = 90.0^\circ$ (crank perpendicular to ground line $O_2 O_4$), the open configuration (pin $B$ above the ground line) places $I_{24}$ on the extended ground line $258.2\,\text{mm}$ to the left of $O_2$, so $O_2 I_{24} = 258.2\,\text{mm}$ and $O_4 I_{24} = 658.2\,\text{mm}$ (solve the triangle for $B$, extend line $AB$ to the ground line, and measure). Determine the instantaneous angular velocity of the output rocker $\omega_4$, the linear velocity of pin $A$ ($v_A$), and the instantaneous mechanical advantage $MA_{\text{torque}} = T_{\text{out}} / T_{\text{in}}$.

Step-by-Step Solution

Step 1: Grashof Condition & Linkage Classification

Identify link lengths:

• Shortest link: $s = L_2 = 120\,\text{mm}$
• Longest link: $l = L_1 = 400\,\text{mm}$
• Intermediate links: $p = L_4 = 280\,\text{mm}$, $q = L_3 = 350\,\text{mm}$

Check Grashof's inequality:

$$s + l = 120 + 400 = 520\,\text{mm}$$ $$p + q = 280 + 350 = 630\,\text{mm}$$

Since $s + l = 520 < p + q = 630$, the linkage satisfies Grashof's condition ($s + l < p + q$).

The shortest link $s$ is link 2 (the input crank), which is pinned directly to the fixed ground frame (link 1). Therefore, by Grashof's inversion criteria, this mechanism is a Grashof Crank-Rocker Mechanism. The input crank $O_2 A$ can rotate continuously through $360^\circ$, while the output link $O_4 B$ rocks back and forth.

Step 2: Transmission Angle Analysis

The transmission angle $\mu$ is the angle between coupler $A B$ and output rocker $O_4 B$. The extreme values occur when the crank is collinear with the ground line $O_2 O_4$:

1. Inner Dead Center ($\,\theta_2 = 0^\circ\,$): The effective base of triangle $A B O_4$ is $L_1 - L_2 = 400 - 120 = 280\,\text{mm}$. Applying the law of cosines:

   $$\cos\mu_{\min} = \frac{L_3^2 + L_4^2 - (L_1 - L_2)^2}{2 L_3 L_4} = \frac{350^2 + 280^2 - 280^2}{2 \times 350 \times 280} = \frac{350^2}{2 \times 350 \times 280} = \frac{350}{560} = 0.625$$ $$\mu_{\min} = \arccos(0.625) \approx 51.32^\circ$$

2. Outer Dead Center ($\,\theta_2 = 180^\circ\,$): The effective base of triangle $A B O_4$ is $L_1 + L_2 = 400 + 120 = 520\,\text{mm}$. Applying the law of cosines:

   $$\cos\mu_{\max} = \frac{L_3^2 + L_4^2 - (L_1 + L_2)^2}{2 L_3 L_4} = \frac{350^2 + 280^2 - 520^2}{2 \times 350 \times 280} = \frac{122,\!500 + 78,\!400 - 270,\!400}{196,\!000} = \frac{-69,\!500}{196,\!000} \approx -0.35459$$ $$\mu_{\max} = \arccos(-0.35459) \approx 110.77^\circ$$

3. Evaluation:

   $$45^\circ \le \mu_{\min} = 51.3^\circ < \mu_{\max} = 110.8^\circ \le 135^\circ$$

   The entire transmission angle excursion ($51.3^\circ$ to $110.8^\circ$) falls well within the recommended $45^\circ \le \mu \le 135^\circ$ operational window. The linkage will operate smoothly with low pin friction and no danger of binding.

Step 3: Velocity and Mechanical Advantage Analysis

Compute the linear velocity of pin $A$:

$$v_A = \omega_2 L_2 = (25.0\,\text{rad/s}) \times (0.120\,\text{m}) = 3.00\,\text{m/s}$$

Using the instant center $I_{24}$, the angular velocity ratio between output link 4 and input link 2 is:

$$\frac{\omega_4}{\omega_2} = \frac{O_2 I_{24}}{O_4 I_{24}} = \frac{258.2\,\text{mm}}{658.2\,\text{mm}} = 0.3923$$

Compute output rocker angular velocity $\omega_4$:

$$\omega_4 = \omega_2 \times 0.3923 = 25.0\,\text{rad/s} \times 0.3923 = 9.81\,\text{rad/s}$$

Compute ideal torque mechanical advantage:

$$MA_{\text{torque}} = \frac{T_{\text{out}}}{T_{\text{in}}} = \frac{\omega_2}{\omega_4} = \frac{1}{0.3923} = \frac{658.2}{258.2} \approx 2.549$$

Because $I_{24}$ lies outside the two fixed pivots, the rocker turns in the same direction as the crank. The linkage amplifies the input torque by a factor of about 2.55 at this configuration.