Section 1: Introduction to Linkage Dynamics

1.1 The High-Speed Challenge in Planar Linkages

In the field of high-speed mechanism design, automation, and robotics, the operating frequencies of planar linkages have risen significantly. Planar mechanisms such as four-bar linkages, slider-cranks, and multi-loop configurations are frequently operated at speeds exceeding $15 \text{ Hz}$ to $30 \text{ Hz}$ ($900 \text{ to } 1800 \text{ RPM}$) in pick-and-place packaging, textile weaving, and engine valve gear trains. At these frequencies, the assumption of rigid-body, quasi-static operation is no longer valid. Instead, the design is dominated by the dynamic forces generated by the accelerating masses of the moving links.

Under quasi-static conditions, the forces acting on the joints and linkages are primarily determined by external workloads. However, because inertial forces scale quadratically with the operating speed ($\mathbf{F}_{in} \propto \omega^2$), a tenfold increase in speed yields a hundredfold increase in dynamic forces. These massive, cyclically varying forces and moments are transmitted through the joints and bearings directly to the machine frame and foundation, leading to a host of engineering challenges:

• Structural Vibrations and Foundation Fatigue: The periodic forces transmitted to the machine foundation can excite the natural frequencies of the mounting structure, leading to resonance. This results in high vibration amplitudes, structure-borne noise, and fatigue crack propagation in the mounting frame.
• Accelerated Joint Bearing Wear: Joint bearings are subjected to severe, fluctuating dynamic loads. According to the classical Lundberg-Palmgren fatigue model, the nominal life of a rolling element bearing, measured in millions of revolutions ($L_{10}$), is given by:
  $$ L_{10} = \left( \frac{C}{P} \right)^p $$
  where $C$ is the basic dynamic load rating, $P$ is the equivalent dynamic load, and the exponent $p$ is $3$ for ball bearings and $10/3$ for roller bearings. Because the dynamic load $P$ is dominated by inertial forces at high speeds ($P \propto \omega^2$), the bearing life scales as:
  $$ L_{10} \propto \omega^{-2p} $$
  For a ball bearing ($p=3$), the bearing life is inversely proportional to the sixth power of the speed ($L_{10} \propto \omega^{-6}$). A twofold increase in speed reduces the bearing life by a factor of $64$. Conversely, reducing the transmitted dynamic forces by balancing the mechanism yields a substantial increase in bearing service life.
• Loss of Kinematic Accuracy (Elastodynamics): High-speed linkages cannot be treated as perfectly rigid bodies. The cyclic inertial forces introduce bending moments and axial forces, causing elastic deflections in the links. These deflections lead to path generation errors and positional deviations at the end-effector.
• Acoustic Radiation: High-frequency structural vibrations are transmitted to the surrounding air as acoustic noise. Industrial noise regulations place strict limits on permissible decibel levels to prevent hearing damage and occupational hazards.
• Input Torque Fluctuations: The dynamic acceleration and deceleration of the links require a continuous transfer of kinetic energy to and from the drive motor. This results in large fluctuations in the input torque. The drive motor must be sized to handle these peak torques, leading to larger, less efficient actuators and requiring large flywheels to maintain a constant operating speed.

To address these challenges, mechanical engineers must perform dynamic balancing. Dynamic balancing involves redistributing the mass of the moving links—often through the addition of counterweights or auxiliary links—so that the net force and moment transmitted to the ground are either completely eliminated or minimized.

1.2 Kinematics of the Planar Four-Bar Linkage

A rigorous dynamic analysis requires a complete kinematic model of the linkage. We focus on the planar four-bar linkage, which is the fundamental building block of more complex mechanisms.

Let the linkage consist of four links, numbered sequentially:

• Link 1: The ground frame (fixed).
• Link 2: The input crank, which is connected to the frame at the fixed pivot $O_2$.
• Link 3: The coupler link, which connects joint $A$ (on the crank) to joint $B$ (on the rocker).
• Link 4: The output rocker, which is connected to the frame at the fixed pivot $O_4$.

We define a Cartesian coordinate system with its origin at the fixed pivot $O_2$. The $x$-axis is aligned along the line of centers connecting the two fixed pivots $O_2$ and $O_4$. The fixed pivot $O_4$ is located at the coordinates $(a_1, 0)$, where $a_1$ represents the ground link length. The lengths of the moving links are denoted by $a_2$ (crank $O_2 A$), $a_3$ (coupler $AB$), and $a_4$ (rocker $O_4 B$).

The angular positions of the crank, coupler, and rocker links with respect to the positive $x$-axis are denoted by $\theta_2$, $\theta_3$, and $\theta_4$, respectively. The input crank angle $\theta_2$ is the independent coordinate, while the coupler angle $\theta_3$ and rocker angle $\theta_4$ are the dependent coordinates.

Using complex numbers to represent the link vectors, the loop-closure equation for the mechanism is formulated as:

In complex exponential form, this is written as:

By separating Eq. 1.2 into real and imaginary parts using Euler's identity ($e^{i\theta} = \cos\theta + i\sin\theta$), we obtain:

1.3 Complete Position Analysis

To determine the dependent angles $\theta_3$ and $\theta_4$ for a given crank angle $\theta_2$, we isolate the coupler angle $\theta_3$ in Equations 1.3a and 1.3b:

Squaring both sides of these equations and adding them eliminates the variable $\theta_3$ using the trigonometric identity $\cos^2\theta_3 + \sin^2\theta_3 = 1$:

Expanding Eq. 1.4 yields:

Applying the identities $\cos^2\theta_4 + \sin^2\theta_4 = 1$ and $\cos^2\theta_2 + \sin^2\theta_2 = 1$ simplifies the equation to:

We rewrite Eq. 1.5 in a standard form:

where:

To solve Eq. 1.6, we use the tangent half-angle substitution:

Substituting these into Eq. 1.6 and multiplying by $(1+t_4^2)$ yields:

Rearranging the terms results in a quadratic equation in $t_4$:

The roots of Eq. 1.7 are:

This gives the analytical solution for the rocker angle $\theta_4$:

The two signs ($\pm$) in Eq. 1.9 correspond to the two different assembly modes of the linkage (open and crossed configurations).

We can eliminate $\theta_4$ in a similar manner to solve for the coupler angle $\theta_3$:

where:

Once $\theta_4$ is computed, we can also determine $\theta_3$ directly using the Cartesian components:

1.4 Velocity Analysis

To find the relationship between the angular velocities of the links, we differentiate the complex loop-closure equation (Eq. 1.2) with respect to time:

Using the chain rule, this yields:

where $\omega_j = \dot{\theta}_j$ is the angular velocity of link $j$. Dividing Eq. 1.12 by the imaginary unit $i$ gives:

Separating Eq. 1.13 into real and imaginary parts:

We can express this system of linear equations in matrix form, grouping the dependent angular velocities $\omega_3$ and $\omega_4$ on the left-hand side:

The coefficient matrix on the left-hand side is the kinematic Jacobian matrix $\mathbf{J}$ of the linkage:

The determinant of this matrix is:

Using Cramer's rule, we solve for the coupler angular velocity $\omega_3$:

Similarly, we solve for the rocker angular velocity $\omega_4$:

1.5 Acceleration Analysis

To find the angular accelerations, we differentiate the velocity equation (Eq. 1.12) with respect to time:

Applying the product and chain rules yields:

where $\alpha_j = \dot{\omega}_j$ is the angular acceleration of link $j$. Separating Eq. 1.16 into real and imaginary parts:

Rearranging terms to place the dependent angular accelerations $\alpha_3$ and $\alpha_4$ on the left-hand side:

where:

Solving this system using Cramer's rule yields:

1.6 Mass Center Kinematics

Next, we establish the position, velocity, and acceleration vectors for the centers of mass of the three moving links.

Let $G_k$ be the center of mass of Link $k$ (for $k=2,3,4$). The position of each CoM is defined relative to its respective link axis:

• Link 2 (Crank): The CoM $G_2$ is located at a distance $r_{G2}$ from the pivot $O_2$ at an angular offset $\phi_2$ relative to the link vector $\mathbf{r}_2$.
• Link 3 (Coupler): The CoM $G_3$ is located at a distance $r_{G3}$ from joint $A$ at an angular offset $\phi_3$ relative to the link vector $\mathbf{r}_3$.
• Link 4 (Rocker): The CoM $G_4$ is located at a distance $r_{G4}$ from the pivot $O_4$ at an angular offset $\phi_4$ relative to the link vector $\mathbf{r}_4$.

The position vectors in the complex plane are:

Differentiating these positions with respect to time yields the velocity vectors of the centers of mass:

Differentiating again yields the acceleration vectors:

1.7 Shaking Force and Shaking Moment Definitions

By applying D'Alembert's principle, we define the Shaking Force ($\mathbf{F}_s$) as the net dynamic reaction force transmitted from the moving links to the ground frame. Physically, this is the vector sum of the forces acting on the fixed ground pivots $O_2$ and $O_4$:

where $\mathbf{F}_{21}$ is the force exerted by the crank on the ground, and $\mathbf{F}_{41}$ is the force exerted by the rocker on the ground.

Applying a global force balance to the entire system of moving links shows that the ground reactions must balance the sum of the link inertia forces:

The Shaking Moment ($M_s$) is the net dynamic moment transmitted to the ground frame. Selecting the crank pivot $O_2$ as our reference point, the shaking moment is the sum of the link inertia torques and the moments of the link inertia forces about $O_2$:

Expressing the position and acceleration vectors in Cartesian components ($\mathbf{r}_{Gk} = x_{Gk}\mathbf{\hat{i}} + y_{Gk}\mathbf{\hat{j}}$ and $\mathbf{a}_{Gk} = \ddot{x}_{Gk}\mathbf{\hat{i}} + \ddot{y}_{Gk}\mathbf{\hat{j}}$), the scalar shaking moment about $O_2$ is:

1.8 Dynamic Effects on Mounting Structures

The transmission of these high-frequency shaking forces and moments to the mounting structure can lead to several undesirable dynamic effects. If we model the machine frame and foundation as a single-degree-of-freedom mass-spring-damper system, the force transmissibility ratio $TR$ is given by:

where:

• $r = \omega / \omega_n$ is the frequency ratio, representing the operating frequency divided by the natural frequency of the structure.
• $\zeta$ is the damping ratio of the support structure.

If the frequency ratio $r$ is near $1$, the transmissibility ratio increases significantly, magnifying the transmitted forces. In high-speed systems, the operating frequency often approaches or exceeds the first natural frequency of the mounting frame, leading to resonant excitation. This results in high vibration amplitudes, increased noise levels, and accelerated structural fatigue.

1.9 Detailed Dynamic Joint Force Analysis (Kinetostatics)

To understand the internal load pathways and joint reaction forces, we construct a full kinetostatic model of the moving links. We write the force and moment balance equations for each individual link, treating joint reaction forces as unknowns. Let $\mathbf{F}_{ij} = F_{ijx}\mathbf{\hat{i}} + F_{ijy}\mathbf{\hat{j}}$ be the force exerted by link $i$ on link $j$. By Newton's third law, $\mathbf{F}_{ji} = -\mathbf{F}_{ij}$.

Applying D'Alembert's principle, we set up the equations of motion for each link:

Link 2 (Crank):

where $T_d$ is the driving input torque applied at joint $O_2$.

Link 3 (Coupler):

Link 4 (Rocker):

Equations 1.62, 1.63, and 1.64 form a system of 9 linear algebraic equations in 9 unknowns: the joint forces ($F_{12x}$, $F_{12y}$, $F_{32x}$, $F_{32y}$, $F_{43x}$, $F_{43y}$, $F_{14x}$, $F_{14y}$) and the input drive torque $T_d$. We write this system in matrix form:

where:

The coefficient matrix $\mathbf{A}$ is:

where $x_{GB3} = x_B - x_{G3}$, $y_{GB3} = y_B - y_{G3}$, and the coordinate terms refer to positions relative to the ground. The right-hand side vector $\mathbf{B}$ contains the dynamic inertia terms:

By solving this system at each step of the kinematic cycle, we obtain the profile of all joint reaction forces. The ground reactions $\mathbf{F}_{21} = -\mathbf{F}_{12}$ and $\mathbf{F}_{41} = -\mathbf{F}_{14}$ can then be summed to compute the shaking force, matching the result of Eq. 1.23.

Section 2: Shaking Force Balancing Theory

2.1 The Stationary Center of Mass Condition

For the shaking force $\mathbf{F}_s$ to be identically zero at all times, the total center of mass of the mechanism must remain stationary. The total mass of the moving parts of the linkage is:

The position vector representing the center of mass of the entire linkage, $\mathbf{r}_S$, is:

Substituting this definition into the shaking force expression:

If the mass distribution is designed such that the total center of mass is stationary, then:

To achieve this condition, the mass-center vector $\mathbf{r}_S$ must be independent of the time-varying joint angles $\theta_2$, $\theta_3$, and $\theta_4$. We solve this mathematical problem using the method of linearly independent vectors.

2.2 Method of Linearly Independent Vectors

Developed by Berkof and Lowen (1969), the method of linearly independent vectors provides a systematic algebraic technique to balance linkages. The method proceeds by expressing the total mass-center vector $\mathbf{M} \mathbf{r}_S$ as a linear combination of the time-varying unit vectors that describe the angular orientation of each link.

We write the total mass-center vector in terms of the complex coordinates of the centers of mass:

Substituting the CoM position equations (Eq. 1.20a-c):

Grouping the coefficients of the time-varying unit vectors $e^{i\theta_2}$, $e^{i\theta_3}$, and $e^{i\theta_4}$:

Equation 2.2 represents the mass-center vector in terms of three time-varying unit vectors. However, because of the closed-loop nature of the four-bar linkage, these three vectors are not linearly independent. They are constrained by the loop-closure equation (Eq. 1.2):

We can eliminate one of the time-varying vectors from the mass-center expression. Typically, we eliminate the coupler term $e^{i\theta_3}$ because adding heavy counterweights to the coupler (which undergoes complex general planar motion) significantly increases the shaking moment and input torque.

Expressing the unit vector $e^{i\theta_3}$ from the loop-closure equation:

Substituting Eq. 2.3 into Eq. 2.2:

Now, we collect terms to separate the constant term and the coefficients of the remaining time-varying unit vectors, $e^{i\theta_2}$ and $e^{i\theta_4}$:

In Eq. 2.4, the unit vectors $e^{i\theta_2}$ and $e^{i\theta_4}$ are linearly independent for a general four-bar linkage because the angles $\theta_2$ and $\theta_4$ cannot maintain a constant relation across the full cycle of motion. Therefore, for the total center of mass vector $\mathbf{r}_S$ to remain constant throughout the cycle, the coefficients of the time-varying vectors $e^{i\theta_2}$ and $e^{i\theta_4}$ must be identically zero:

If Equations 2.5 and 2.6 are satisfied, the total mass-center vector simplifies to:

Because the position of the total center of mass is stationary, the net shaking force is completely eliminated.

2.3 Derivation of Counterweight Parameters

To satisfy the balancing conditions in Equations 2.5 and 2.6, we must adjust the mass distributions of Link 2 and Link 4. In general, the mass properties of the original coupler link (Link 3) are fixed by operational requirements. Therefore, we introduce counterweights on Link 2 (crank) and Link 4 (rocker).

Let the original mass properties (before balancing) of the crank and rocker be:

• Crank (Link 2): Original mass $m_{20}$, CoM distance $r_{G20}$, and phase angle $\phi_{20}$.
• Rocker (Link 4): Original mass $m_{40}$, CoM distance $r_{G40}$, and phase angle $\phi_{40}$.

We add a counterweight of mass $m_{cw2}$ at a radial distance $r_{cw2}$ and angular position $\theta_{cw2}$ on Link 2. The combined mass properties of Link 2 are:

Similarly, we add a counterweight of mass $m_{cw4}$ at a radial distance $r_{cw4}$ and angular position $\theta_{cw4}$ on Link 4. The combined mass properties of Link 4 are:

Substituting the expressions for the combined links (Eq. 2.8b and Eq. 2.9b) into the force balancing conditions (Eq. 2.5 and Eq. 2.6):

Isolating the counterweight parameters on the left-hand side yields two vector equations in the complex plane:

To find the physical parameters (mass-radius product and phase angle), we resolve Eq. 2.10 and Eq. 2.11 into their real and imaginary components.

For the crank counterweight (Link 2), let:

The mass-radius product of the crank counterweight is:

And the phase angle $\theta_{cw2}$ is:

For the rocker counterweight (Link 4), let:

The mass-radius product of the rocker counterweight is:

And the phase angle $\theta_{cw4}$ is:

Once the designer selects a physically feasible radius ($r_{cw2}$ and $r_{cw4}$), the required counterweight masses ($m_{cw2}$ and $m_{cw4}$) are computed directly from the mass-radius products.

By implementing these counterweight parameters, the net dynamic shaking force transmitted to the frame is theoretically reduced to zero. However, this force balance is achieved at the expense of adding substantial mass to the moving links, which increases the shaking moment and the torque required to drive the linkage.

2.4 Lumped Mass Approximation: Physical Insight into Force Balancing

The method of linearly independent vectors can be understood physically using a lumped mass approximation. In this approach, we replace the distributed mass of the coupler (Link 3) with a statically equivalent system of three concentrated point masses: $m_{3A}$ located at joint $A$, $m_{3B}$ located at joint $B$, and a remaining mass $m_{3G}$ at the coupler's center of mass $G_3$.

To ensure the lumped system is statically equivalent to the original coupler, the concentrated masses must satisfy the following conditions:

where $\mathbf{r}_{A/G3}$ and $\mathbf{r}_{B/G3}$ are the position vectors of joints $A$ and $B$ relative to the center of mass $G_3$.

If we model the coupler by lumping its mass only at the two joint endpoints $A$ and $B$, we set $m_{3G} = 0$. The resulting two-mass model is statically equivalent if:

where $l_{3A}$ is the distance from $A$ to $G_3$, and $l_{3B}$ is the distance from $B$ to $G_3$.

By replacing the coupler with these two lumped masses, we apportion the coupler's mass between the crank tip $A$ and the rocker tip $B$. The mass concentrated at joint $A$ ($m_{3A}$) rotates in a circle of radius $a_2$ and can be balanced by a counterweight on the crank (Link 2). The mass concentrated at joint $B$ ($m_{3B}$) oscillates with the rocker and can be balanced by a counterweight on the rocker (Link 4).

However, this two-mass model is not dynamically equivalent because it does not preserve the coupler's mass moment of inertia:

The difference $\Delta I_3 = I_{G3} - I_{G3}^*$ is the inertia error. This error generates an unbalanced dynamic moment during operation, contributing to the shaking moment and causing input torque fluctuations. This highlights the trade-off in linkage balancing: complete force balancing can be achieved using static lumping, but the dynamic mismatch (inertia error) remains, generating an unbalanced shaking moment.

Section 3: Shaking Moment Minimization

3.1 Analytical Derivation of the Shaking Moment

We now derive the shaking moment equation for a four-bar linkage about the crank pivot $O_2$. As defined in Eq. 1.24, the shaking moment $M_s$ consists of the inertia torques and the moments of the inertia forces:

Let us evaluate the cross-product term $\mathbf{M}_{Gk} = \mathbf{r}_{Gk} \times (m_k \mathbf{a}_{Gk})$ for each link individually.

3.1.1 Crank Moment Contribution (Link 2)

The position and linear acceleration of the crank's center of mass $G_2$ are:

Taking the cross product of these two complex numbers:

Using the cross-product properties of complex vectors in the plane:

This simplifies the expression to:

Adding the direct inertia torque contribution, the total shaking moment contribution from Link 2 is:

where $I_{O2}$ is the mass moment of inertia of Link 2 about the fixed pivot $O_2$.

3.1.2 Rocker Moment Contribution (Link 4)

The position and linear acceleration of the rocker's center of mass $G_4$ are:

Computing the cross product:

Expanding the terms:

The unit vector $\mathbf{\hat{i}}$ corresponds to $e^{i0}$. The planar cross products are:

Substituting these expressions back into the equation:

Adding the direct inertia torque contribution, the total shaking moment contribution from Link 4 is:

where $I_{O4} = I_{G4} + m_4 r_{G4}^2$ is the mass moment of inertia of Link 4 about the fixed pivot $O_4$.

3.1.3 Coupler Moment Contribution (Link 3)

The position and linear acceleration of the coupler's center of mass $G_3$ are:

Computing the cross product:

Expanding this cross product yields four terms:

1. $$ T_1 = a_2^2 \left( e^{i\theta_2} \times \left( i \alpha_2 - \omega_2^2 \right) e^{i\theta_2} \right) = a_2^2 \alpha_2 \mathbf{\hat{k}} $$
2. $$ T_2 = r_{G3}^2 \left( e^{i(\theta_3 + \phi_3)} \times \left( i \alpha_3 - \omega_3^2 \right) e^{i(\theta_3 + \phi_3)} \right) = r_{G3}^2 \alpha_3 \mathbf{\hat{k}} $$
3. $$ T_3 = a_2 r_{G3} \left[ e^{i\theta_2} \times \left( i \alpha_3 - \omega_3^2 \right) e^{i(\theta_3 + \phi_3)} \right] = a_2 r_{G3} \left( \alpha_3 \cos(\theta_3 + \phi_3 - \theta_2) - \omega_3^2 \sin(\theta_3 + \phi_3 - \theta_2) \right) \mathbf{\hat{k}} $$
4. $$ T_4 = a_2 r_{G3} \left[ e^{i(\theta_3 + \phi_3)} \times \left( i \alpha_2 - \omega_2^2 \right) e^{i\theta_2} \right] = a_2 r_{G3} \left( \alpha_2 \cos(\theta_3 + \phi_3 - \theta_2) + \omega_2^2 \sin(\theta_3 + \phi_3 - \theta_2) \right) \mathbf{\hat{k}} $$

Summing these four terms:

Adding the direct inertia torque contribution, the total shaking moment contribution from Link 3 is:

where $I_{O3} = I_{G3} + m_3 r_{G3}^2$ is the moment of inertia of Link 3 about joint $A$.

3.1.4 Total Shaking Moment Equation

The total shaking moment $M_s$ about the crank pivot $O_2$ is the sum of the individual link contributions:

Substituting Eq. 3.3, 3.5, and 3.7:

Equation 3.8 represents the general shaking moment about $O_2$ for any planar four-bar linkage.

If the linkage is already force-balanced by counterweights on the crank and rocker, the force balancing conditions (Eq. 2.5 and Eq. 2.6) must hold. Let us examine how these conditions simplify the shaking moment.

From the rocker force balancing condition (Eq. 2.6):

Equating the real and imaginary parts:

Let us substitute these relations into the rocker component of the shaking moment:

Expanding the trigonometric terms:

Thus:

Substituting these back, the rocker moment contribution reduces to:

Using Eq. 3.9, we express the shaking moment of a force-balanced four-bar linkage in a simplified form that is independent of the rocker's center of mass position $r_{G4}$:

where $I_{O2}^*$, $I_{O3}^*$, and $I_{O4}^*$ are the moments of inertia of the force-balanced links, which now include the contributions of the added counterweights.

3.2 Proof of the Impossibility of Complete Passive Moment Balancing

A fundamental question in linkage design is: Can we design a set of passive counterweights to eliminate the shaking moment $M_s$ completely, just as we did for the shaking force $\mathbf{F}_s$?

To answer this, we apply the method of linearly independent vectors to the shaking moment expression and attempt to set the coefficients of all time-varying terms to zero.

We assume the input crank rotates at a constant angular velocity, so $\omega_2 = \text{constant}$ and $\alpha_2 = 0$. Under this condition, the shaking moment equation (Eq. 3.8) simplifies to:

Observe that Eq. 3.11 contains the dependent angular velocities squared ($\omega_3^2$, $\omega_4^2$) and the dependent angular accelerations ($\alpha_3$, $\alpha_4$). From our kinematic derivations (Eq. 1.60, 1.61, 1.18, and 1.19), these variables are highly non-linear functions of the input angle $\theta_2$.

Because the linkage undergoes cyclic motion, any kinematic variable $f(\theta_2)$ can be represented by a Fourier series in terms of the crank angle $\theta_2$:

The Fourier coefficients in Eq. 3.12 are calculated using the Euler-Fourier integrals:

For the shaking moment to be identically zero for all crank positions, every single Fourier coefficient must vanish:

This represents an infinite set of algebraic constraints. However, the designer has only a finite number of physical design variables to adjust. Even if we allow counterweights on all three moving links, we have only six design variables: the mass-radius products and phase angles of the three counterweights ($m_{cw2}r_{cw2}, \theta_{cw2}, m_{cw3}r_{cw3}, \theta_{cw3}, m_{cw4}r_{cw4}, \theta_{cw4}$).

Mathematically, we cannot satisfy an infinite number of independent Fourier coefficient constraints with a finite number of design variables.

To illustrate this impossibility from a physical perspective, let us analyze the rotational inertia forces. The angular accelerations of the coupler and rocker, $\alpha_3$ and $\alpha_4$, are driven by the changing geometry of the linkage. The inertial torques $I_{O3}\alpha_3$ and $I_{O4}\alpha_4$ vary in a complex, non-harmonic manner.

A passive counterweight attached to a rotating link can only generate inertial forces that are directed radially (centripetal forces) or tangentially (tangential acceleration forces). For Link 2 and Link 4, which pivot about fixed points, their counterweights can only apply moments about $O_2$ and $O_4$ that are proportional to their own angular accelerations and the squares of their own angular velocities. These counterweights cannot generate the non-harmonic, high-frequency force components required to cancel the inertial forces of the coupler (Link 3), which undergoes complex translation and rotation.

If we attempt to set up the method of linearly independent vectors for the shaking moment, we find that the time-varying coefficients cannot be grouped into a set of independent vectors whose coefficients can be set to zero. For example, to eliminate the term containing $\alpha_3$, we would require the moment of inertia of the coupler to be zero ($I_{O3} = 0$), or even negative, which is physically impossible for any real mass distribution.

Thus, we establish the fundamental theorem of linkage balancing:

"It is physically impossible to achieve complete shaking moment balancing in a planar four-bar linkage using passive counterweights alone, without adding auxiliary links or active elements."

3.3 Optimization Criteria for Shaking Moment Reduction

Because complete shaking moment balancing is impossible with passive counterweights, we must treat the design of the counterweights as a optimization problem. The goal is to find the mass distributions of the links that minimize the shaking moment over a full cycle of operation, subject to physical constraints.

Let the vector of design variables be:

3.3.1 Objective Functions

Depending on the application, different mathematical metrics are used to define the "minimization" of the shaking moment:

1. Root-Mean-Square (RMS) Shaking Moment: This metric minimizes the total vibrational energy transmitted over a cycle. It is the most common objective function for vibration and noise reduction:

where $T = 2\pi/\omega_2$ is the period of one operating cycle.

2. Peak-to-Peak Shaking Moment: This metric minimizes the maximum shock or structural load experienced by the foundation:

3.3.2 Multi-Objective Formulations and the Force-Moment Trade-Off

If we enforce complete shaking force balancing (using Equations 2.8 to 2.13), the counterweights add significant mass to the crank and rocker. This increases their moments of inertia ($I_{O2}$ and $I_{O4}$), which directly increases the shaking moment.

To resolve this conflict, we allow partial force balancing. We relax the requirement that the shaking force must be exactly zero, and search for a compromise that reduces both the shaking force and the shaking moment. This is formulated as a weighted multi-objective optimization problem:

where:

• $w_f$, $w_m$, and $w_t$ are weighting factors that represent the designer's priorities.
• $F_{ref}$, $M_{ref}$, and $T_{ref}$ are normalizing scale factors, typically chosen as the peak values of the unbalanced linkage.
• $T_d(t, \mathbf{x})$ is the dynamic driving torque required to maintain constant input speed, and $T_{d, avg}$ is its cycle average.

By varying the weights ($w_f, w_m$), we can map out a Pareto frontier that shows the optimal trade-off between shaking force and shaking moment, as illustrated conceptually in the table below:

Balancing Strategy	Shaking Force Reduction	Shaking Moment Reduction	Added Mass & Inertia	Required Drive Torque
Unbalanced	0% (Base Case)	0% (Base Case)	Minimum	Base Case
Complete Force Balance	100%	Increased (+20% to +50%)	High	High
Compromise (Partial Force)	50% - 80%	Decreased (-30% to -60%)	Moderate	Moderate
Complete Moment (Active)	90% - 100%	100%	Very High	Moderate (with gears)

3.3.3 Geometric and Physical Constraints

The optimization process must satisfy several geometric and physical constraints to ensure the resulting linkage is manufacturable:

• Space Envelope Constraints: The counterweights must not collide with other moving links, fixed joints, or the surrounding machine housing. This restricts the maximum radial distance $r_{cw}$ and the thickness of the weights.
• Mass Budgets: Adding too much mass can overload the main bearings and increase the required starting torque, which can trip the motor drives.
• Link Strength and Stiffness: The centrifugal forces generated by the counterweights introduce large bending moments on the link shafts, requiring thicker structures that must be checked against mechanical stress limits.

In practice, these optimization problems are solved using numerical algorithms such as Sequential Quadratic Programming (SQP) or genetic algorithms. In Part 2 of this post, we will implement these algorithms in MATLAB and Python, demonstrating how to compute the optimal counterweight parameters for a high-speed pick-and-place robot linkage.

3.4 Advanced Mechanical Balancing Techniques

Because passive counterweights cannot completely eliminate the shaking moment, advanced dynamic balancing techniques have been developed. These methods either cancel the shaking moment using auxiliary mechanisms or introduce active components:

• Duplicated Symmetric Mechanisms: One of the most effective ways to balance both shaking force and shaking moment is to use two identical linkages mounted in mirror-image configuration. The two mechanisms are driven in opposition, meaning that when one crank rotates clockwise, the other rotates counterclockwise at the same speed.
  Let $\mathbf{F}_{s1}$ and $M_{s1}$ be the shaking force and moment of the first linkage, and $\mathbf{F}_{s2}$ and $M_{s2}$ be those of the second linkage. Because of the symmetry:
  $$ \mathbf{F}_{s2}(t) = -\mathbf{F}_{s1}(t) \quad \text{and} \quad M_{s2}(t) = -M_{s1}(t) \quad \text{(Eq. 3.70)} $$
  Summing the forces and moments transmitted to the frame:
  $$ \mathbf{F}_{s, total} = \mathbf{F}_{s1} + \mathbf{F}_{s2} = \mathbf{0} \quad \text{and} \quad M_{s, total} = M_{s1} + M_{s2} = 0 \quad \text{(Eq. 3.71)} $$
  This yields complete force and moment balancing across the entire cycle of motion. This approach is widely used in high-end industrial pick-and-place systems and high-speed sorting machines, though it doubles the number of parts, increasing the cost, size, and cost of the machine.
• Gear-Driven Counter-Rotational Shafts (Lanchester Balancer): In internal combustion engines and slider-crank linkages, gear-driven shafts are used to balance the second harmonic ($n=2$) of the shaking force or moment. A pair of counter-rotating shafts with eccentric weights are geared to run at twice the input crank speed. The centrifugal forces generated by these weights cancel the second harmonic of the inertial forces.
• Auxiliary Balancing Links: An auxiliary dyad (a two-link combination) can be added to the mechanism to balance the shaking moment. The dyad is designed to move out-of-phase with the primary linkage, so that its inertial torque cancels the shaking moment of the main mechanism.
• Active Moment Balancers: Active moment balancers use a servomotor to drive a rotating mass in a non-harmonic, pre-programmed speed profile. The motor speed is controlled in real-time to match and cancel the shaking moment of the primary linkage. While highly effective, active balancers require high-bandwidth control systems and consume additional electrical energy.

These advanced techniques provide valuable solutions for applications where passive optimization is insufficient to meet strict vibration and noise limits.

$$ L_{\min} + L_{\max} \le L_a + L_b $$

$$ 100 + 400 \le 350 + 300 \implies 500\text{ mm} \le 650\text{ mm} $$

$$ \omega_1 = \frac{1500 \times 2\pi}{60} = 157.08\text{ rad/s} $$

$$ \mathbf{r}_1 + \mathbf{r}_2 = \mathbf{r}_0 + \mathbf{r}_3 $$

$$ L_1 e^{i\theta_1} + L_2 e^{i\theta_2} = L_0 + L_3 e^{i\theta_3} $$

$$ L_1 \cos\theta_1 + L_2 \cos\theta_2 = L_0 + L_3 \cos\theta_3 $$

$$ L_1 \sin\theta_1 + L_2 \sin\theta_2 = L_3 \sin\theta_3 $$

$$ K_1 \cos\theta_3 - K_2 \cos\theta_1 + K_3 = \cos(\theta_1 - \theta_3) $$

$$ K_1 = \frac{L_0}{L_1} = \frac{0.40}{0.10} = 4.0 $$

$$ K_2 = \frac{L_0}{L_3} = \frac{0.40}{0.30} = 1.3333 $$

$$ K_3 = \frac{L_1^2 - L_2^2 + L_3^2 + L_0^2}{2 L_1 L_3} = \frac{0.10^2 - 0.35^2 + 0.30^2 + 0.40^2}{2 \times 0.10 \times 0.30} = \frac{0.01 - 0.1225 + 0.09 + 0.16}{0.06} = 2.2917 $$

$$ A(\theta_1) \cos\theta_3 + B(\theta_1) \sin\theta_3 = C(\theta_1) $$

$$ A(\theta_1) = K_1 - \cos\theta_1 = 4.0 - \cos\theta_1 $$

$$ B(\theta_1) = -\sin\theta_1 $$

$$ C(\theta_1) = K_2 \cos\theta_1 - K_3 = 1.3333 \cos\theta_1 - 2.2917 $$

$$ (A(\theta_1) + C(\theta_1)) t^2 - 2 B(\theta_1) t + (C(\theta_1) - A(\theta_1)) = 0 $$

$$ a_t t^2 + b_t t + c_t = 0 $$

$$ a_t = K_1 - K_3 + (K_2 - 1)\cos\theta_1 = 4.0 - 2.2917 + (1.3333 - 1)\cos\theta_1 = 1.7083 + 0.3333 \cos\theta_1 $$

$$ b_t = 2 \sin\theta_1 $$

$$ c_t = -K_1 - K_3 + (K_2 + 1)\cos\theta_1 = -4.0 - 2.2917 + (1.3333 + 1)\cos\theta_1 = -6.2917 + 2.3333 \cos\theta_1 $$

$$ t = \frac{-b_t \pm \sqrt{b_t^2 - 4 a_t c_t}}{2 a_t} $$

$$ \theta_3 = 2 \arctan\left(t\right) $$

$$ \theta_2 = \arctan2\left( L_3 \sin\theta_3 - L_1 \sin\theta_1, \ L_0 + L_3 \cos\theta_3 - L_1 \cos\theta_1 \right) $$

$$ \begin{bmatrix} -L_2 \sin\theta_2 & L_3 \sin\theta_3 \\ L_2 \cos\theta_2 & -L_3 \cos\theta_3 \end{bmatrix} \begin{bmatrix} \dot{\theta}_2 \\ \dot{\theta}_3 \end{bmatrix} = \begin{bmatrix} L_1 \sin\theta_1 \\ -L_1 \cos\theta_1 \end{bmatrix} \dot{\theta}_1 $$

$$ \det(\mathbf{J}) = L_2 L_3 \sin(\theta_2 - \theta_3) $$

$$ \dot{\theta}_2 = \frac{L_1 \sin(\theta_1 - \theta_3)}{L_2 \sin(\theta_3 - \theta_2)} \dot{\theta}_1 $$

$$ \dot{\theta}_3 = \frac{L_1 \sin(\theta_1 - \theta_2)}{L_3 \sin(\theta_3 - \theta_2)} \dot{\theta}_1 $$

$$ \begin{bmatrix} -L_2 \sin\theta_2 & L_3 \sin\theta_3 \\ L_2 \cos\theta_2 & -L_3 \cos\theta_3 \end{bmatrix} \begin{bmatrix} \ddot{\theta}_2 \\ \ddot{\theta}_3 \end{bmatrix} = \begin{bmatrix} P \\ Q \end{bmatrix} $$

$$ P = L_1 \cos\theta_1 \dot{\theta}_1^2 + L_2 \cos\theta_2 \dot{\theta}_2^2 - L_3 \cos\theta_3 \dot{\theta}_3^2 $$

$$ Q = L_1 \sin\theta_1 \dot{\theta}_1^2 + L_2 \sin\theta_2 \dot{\theta}_2^2 - L_3 \sin\theta_3 \dot{\theta}_3^2 $$

$$ \ddot{\theta}_2 = \frac{-L_3 \cos\theta_3 P - L_3 \sin\theta_3 Q}{L_2 L_3 \sin(\theta_2 - \theta_3)} $$

$$ \ddot{\theta}_3 = \frac{-L_2 \cos\theta_2 P - L_2 \sin\theta_2 Q}{L_2 L_3 \sin(\theta_2 - \theta_3)} $$

$$ x_{G1} = r_1 \cos\theta_1 $$

$$ y_{G1} = r_1 \sin\theta_1 $$

$$ x_{G2} = L_1 \cos\theta_1 + r_2 \cos\theta_2 $$

$$ y_{G2} = L_1 \sin\theta_1 + r_2 \sin\theta_2 $$

$$ x_{G3} = L_0 + r_3 \cos\theta_3 $$

$$ y_{G3} = r_3 \sin\theta_3 $$

$$ a_{x, G1} = -r_1 \omega_1^2 \cos\theta_1 $$

$$ a_{y, G1} = -r_1 \omega_1^2 \sin\theta_1 $$

$$ a_{x, G2} = -L_1 \omega_1^2 \cos\theta_1 - r_2 \omega_2^2 \cos\theta_2 - r_2 \alpha_2 \sin\theta_2 $$

$$ a_{y, G2} = -L_1 \omega_1^2 \sin\theta_1 - r_2 \omega_2^2 \sin\theta_2 + r_2 \alpha_2 \cos\theta_2 $$

$$ a_{x, G3} = -r_3 \omega_3^2 \cos\theta_3 - r_3 \alpha_3 \sin\theta_3 $$

$$ a_{y, G3} = -r_3 \omega_3^2 \sin\theta_3 + r_3 \alpha_3 \cos\theta_3 $$

$$ \mathbf{F}_{sh} = -\left( m_1 \mathbf{a}_{G1} + m_2 \mathbf{a}_{G2} + m_3 \mathbf{a}_{G3} \right) $$

$$ F_{sh, x} = -\left( m_1 a_{x, G1} + m_2 a_{x, G2} + m_3 a_{x, G3} \right) $$

$$ F_{sh, y} = -\left( m_1 a_{y, G1} + m_2 a_{y, G2} + m_3 a_{y, G3} \right) $$

$$ F_{sh} = \sqrt{F_{sh, x}^2 + F_{sh, y}^2} $$

$$ M_{sh} = -\sum_{i=1}^3 \left( I_i \alpha_i + \mathbf{r}_{Gi} \times m_i \mathbf{a}_{Gi} \right)_z $$

$$ M_{sh} = -\left[ I_1 \alpha_1 + m_1 (x_{G1} a_{y, G1} - y_{G1} a_{x, G1}) + I_2 \alpha_2 + m_2 (x_{G2} a_{y, G2} - y_{G2} a_{x, G2}) + I_3 \alpha_3 + m_3 (x_{G3} a_{y, G3} - y_{G3} a_{x, G3}) \right] $$

$$ m_{2A} = m_2 \left( \frac{L_2 - r_2}{L_2} \right) $$

$$ m_{2B} = m_2 \left( \frac{r_2}{L_2} \right) $$

$$ m_{2A} = 1.50 \times \left( \frac{0.35 - 0.175}{0.35} \right) = 0.75\text{ kg} $$

$$ m_{2B} = 1.50 \times \left( \frac{0.175}{0.35} \right) = 0.75\text{ kg} $$

$$ m_{c1} r_{c1} = m_1 r_1 + m_{2A} L_1 $$

$$ m_{c3} r_{c3} = m_3 r_3 + m_{2B} L_3 $$

$$ m_{c1} r_{c1} = (0.50 \times 0.05) + (0.75 \times 0.10) = 0.025 + 0.075 = 0.100\text{ kg}\cdot\text{m} $$

$$ m_{c3} r_{c3} = (1.20 \times 0.15) + (0.75 \times 0.30) = 0.180 + 0.225 = 0.405\text{ kg}\cdot\text{m} $$

$$ m_{c1} = \frac{0.100}{0.04} = 2.50\text{ kg} $$

$$ m_{c3} = \frac{0.405}{0.10} = 4.05\text{ kg} $$

$$ I_{1, new} = I_1 + m_{c1} r_{c1}^2 = 0.000417 + 2.50 \times (0.04)^2 = 0.004417\text{ kg}\cdot\text{m}^2 $$

$$ I_{3, new} = I_3 + m_{c3} r_{c3}^2 = 0.009000 + 4.05 \times (0.10)^2 = 0.049500\text{ kg}\cdot\text{m}^2 $$

$$ \mathbf{F}_{sh, bal} = -\left( m_1 \mathbf{a}_{G1} + m_2 \mathbf{a}_{G2} + m_3 \mathbf{a}_{G3} + m_{c1} \mathbf{a}_{c1} + m_{c3} \mathbf{a}_{c3} \right) $$

$$ \mathbf{a}_{c1} = r_{c1} \omega_1^2 \left( \cos\theta_1 \hat{\mathbf{i}} + \sin\theta_1 \hat{\mathbf{j}} \right) $$

$$ \mathbf{a}_{c3} = r_{c3} \left( \omega_3^2 \cos\theta_3 + \alpha_3 \sin\theta_3 \right) \hat{\mathbf{i}} + r_{c3} \left( \omega_3^2 \sin\theta_3 - \alpha_3 \cos\theta_3 \right) \hat{\mathbf{j}} $$

$$ F_{sh, bal, x} \equiv 0\text{ N} $$

$$ F_{sh, bal, y} \equiv 0\text{ N} $$

$$ M_{sh, bal} = -\left[ I_{1, new} \alpha_1 + m_1 (x_{G1} a_{y, G1} - y_{G1} a_{x, G1}) + M_{c1} + I_2 \alpha_2 + m_2 (x_{G2} a_{y, G2} - y_{G2} a_{x, G2}) + I_{3, new} \alpha_3 + m_3 (x_{G3} a_{y, G3} - y_{G3} a_{x, G3}) + M_{c3} \right] $$

$$ M_{c3} = m_{c3} \left( x_{c3} a_{y, c3} - y_{c3} a_{x, c3} \right) $$

$$ \mathbf{F}_{1} = m_e r_e \omega^2 \left( \cos(\omega t) \hat{\mathbf{i}} + \sin(\omega t) \hat{\mathbf{j}} \right) $$

$$ \mathbf{F}_{2} = m_e r_e \omega^2 \left( \cos(-\omega t) \hat{\mathbf{i}} + \sin(-\omega t) \hat{\mathbf{j}} \right) = m_e r_e \omega^2 \left( \cos(\omega t) \hat{\mathbf{i}} - \sin(\omega t) \hat{\mathbf{j}} \right) $$

$$ \mathbf{F}_{net} = \mathbf{F}_{1} + \mathbf{F}_{2} = 2 m_e r_e \omega^2 \cos(\omega t) \hat{\mathbf{i}} $$

$$ M_{bal} = F_{1, y} \frac{d}{2} - F_{2, y} \frac{d}{2} = m_e r_e \omega^2 \sin(\omega t) \frac{d}{2} - m_e r_e \omega^2 \sin(-\omega t + \pi) \frac{d}{2} = 2 \left( m_e r_e \frac{d}{2} \right) \omega^2 \sin(\omega t) = m_e r_e d \omega^2 \sin(\omega t) $$

$$ M_{sh}(\theta_1) = M_0 + \sum_{k=1}^{\infty} \left( A_k \cos(k\theta_1) + B_k \sin(k\theta_1) \right) $$

$$ \mathbf{r}'_i(t) = \mathbf{M} \mathbf{r}_i(t) = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} x_i(t) \\ y_i(t) \end{bmatrix} = \begin{bmatrix} x_i(t) \\ -y_i(t) \end{bmatrix} $$

$$ \mathbf{F}_{sh, total} = -\sum_{i=1}^3 m_i \mathbf{a}_{Gi} - \sum_{i=1}^3 m_i \mathbf{a}'_{Gi} = -\sum_{i=1}^3 m_i \begin{bmatrix} a_{x, Gi} \\ a_{y, Gi} \end{bmatrix} - \sum_{i=1}^3 m_i \begin{bmatrix} a_{x, Gi} \\ -a_{y, Gi} \end{bmatrix} = \begin{bmatrix} -2 \sum_{i=1}^3 m_i a_{x, Gi} \\ 0 \end{bmatrix} $$

$$ M_{sh, total} = -\sum_{i=1}^3 \left( I_i \ddot{\theta}_i + m_i (x_{Gi} a_{y, Gi} - y_{Gi} a_{x, Gi}) \right) - \sum_{i=1}^3 \left( I_i (-\ddot{\theta}_i) + m_i (x'_{Gi} a'_{y, Gi} - y'_{Gi} a'_{x, Gi}) \right) $$

$$ x'_{Gi} a'_{y, Gi} - y'_{Gi} a'_{x, Gi} = x_{Gi} (-a_{y, Gi}) - (-y_{Gi}) a_{x, Gi} = -\left( x_{Gi} a_{y, Gi} - y_{Gi} a_{x, Gi} \right) $$

$$ M_{sh, total} = -\sum_{i=1}^3 \left( I_i \ddot{\theta}_i + m_i (x_{Gi} a_{y, Gi} - y_{Gi} a_{x, Gi}) \right) - \sum_{i=1}^3 \left( -I_i \ddot{\theta}_i - m_i (x_{Gi} a_{y, Gi} - y_{Gi} a_{x, Gi}) \right) \equiv 0\text{ N}\cdot\text{m} $$

$$ \mathbf{x} = \begin{bmatrix} m_{c1}, & r_{c1}, & \theta_{c1}, & m_{c3}, & r_{c3}, & \theta_{c3} \end{bmatrix}^T $$

$$ J(\mathbf{x}) = w_f \left( \frac{F_{sh, \max}(\mathbf{x})}{F_{sh, \max}^0} \right)^2 + w_m \left( \frac{M_{sh, \max}(\mathbf{x})}{M_{sh, \max}^0} \right)^2 + w_t \left( \frac{T_{dr, \max}(\mathbf{x})}{T_{dr, \max}^0} \right)^2 $$

$$ g_j(\mathbf{x}) \le 0, \quad j=1, \dots, p $$

$$ \min_{\mathbf{d}} \frac{1}{2} \mathbf{d}^T \mathbf{H}_k \mathbf{d} + \nabla f(\mathbf{x}_k)^T \mathbf{d} $$

$$ \nabla g_j(\mathbf{x}_k)^T \mathbf{d} + g_j(\mathbf{x}_k) \le 0, \quad j=1, \dots, p $$

$$ \mathbf{e} = \mathbf{r}_p - \mathbf{r}_b $$

$$ \delta = ||\mathbf{e}|| - c $$

$$ F_c = K \delta^n + D \dot{\delta} $$

$$ D = \frac{3 K (1 - c_e^2) \delta^n}{4 \dot{\delta}^-} $$

$$ K = \frac{4}{3} \frac{E^*}{\left( \frac{1}{R_p} - \frac{1}{R_b} \right)^{1/2}} $$

$$ \frac{1}{E^*} = \frac{1 - \nu_p^2}{E_p} + \frac{1 - \nu_b^2}{E_b} $$

$$ L_{10} = \left( \frac{C}{P} \right)^p \times 10^6 \text{ revolutions} $$

$$ P = \left( \frac{1}{2\pi} \int_0^{2\pi} \left[ F_j(\theta_1) \right]^3 d\theta_1 \right)^{1/3} $$

$$ L_{10, bal} = \left( \frac{C}{2P} \right)^p \times 10^6 = 2^{-p} L_{10} = 2^{-3.33} L_{10} \approx 0.10 L_{10} $$

$$ M_b \ddot{x}_b + C_b \dot{x}_b + K_b x_b = F_{sh, x}(t) $$

$$ F_{transmitted}(t) = K_b x_b(t) + C_b \dot{x}_b(t) $$

$$ \text{TR} = \frac{F_{transmitted, \max}}{F_0} = \sqrt{\frac{1 + (2\zeta r)^2}{(1 - r^2)^2 + (2\zeta r)^2}} $$

$$ \zeta = \frac{C_b}{2 \sqrt{K_b M_b}} $$