First- and Second-Order Centrodes of Both Coupler Links of Stephenson III Six-Bar Mechanisms

Abstract

The subject of this paper is the formulation of a general algorithm that is aimed to obtain and analyze the first- and second-order centrodes of both coupler links of Stephenson III six-bar mechanisms, along with the corresponding pairs of Bresse’s circles. The position vectors of both pairs of instantaneous centers of rotation and acceleration centers are expressed as vector functions of the driving input angle of the mechanism and apply the fundamental theorems for velocities and accelerations, respectively. Thus, the proposed algorithm has been formulated by using vector algebra and validated through significant numerical and graphical results by devoting particular attention to the cases in which the second coupler link shows an instantaneous stop configuration, along with those where the first four-bar linkage reaches an asymptotic configuration. These results can be useful for designing and analyzing specific mechanisms for practical applications, such as composing the jaw-crusher machine, which has been selected as an example.

1. Introduction

Kinematics provides the theoretical basis for understanding and analyzing the motion of rigid bodies. Classical textbooks [1,2,3,4,5,6,7] have provided comprehensive frameworks for describing planar and spatial mechanisms, with a particular focus on both the geometric and algebraic properties of rigid-body motion. These texts cover a range of topics, including the analysis and synthesis of linkages, the mathematical description of centrodes for the purpose of identifying instantaneous centers of rotation, and the role of Bresse’s circles in characterizing velocity and acceleration fields. The combination of these principles allows the precise study of relative motion, thereby facilitating the acquisition of crucial insights into mechanical systems that encompass everything from simple four-bar linkages to complex multi-body mechanisms. The study of centrodes, which are fundamental in planar kinematics, has been investigated with a view to their potential applications in mechanism analysis. The application of algorithms for tracing first- and second-order centrodes in slider-crank mechanisms has been demonstrated to be a valuable technique for the revelation of mechanism behavior [8]. A specific design procedure of Stephenson III six-bar mechanisms, in order to obtain a coupler link instantaneous stop of the five-bar loop and a crank-driven four-bar loop, was proposed in [9]. The geometric characterization of centrodes has been refined through the development of methods for classifying singularities, such as double points and tacnodes. This has contributed to improvements in mechanism design and optimization [10]. Furthermore, degenerate forms of centrodes have been investigated, elucidating distinctive motion characteristics in planar mechanisms that underscore the significance of these geometric tools in kinematic analysis [11]. The study of higher-order kinematics has facilitated a more comprehensive understanding of rigid-body motion. This has been achieved by employing tensor algebra and multidual systems to describe nth-order acceleration fields with closed-form, coordinate-free solutions [12,13]. Advances in the properties of Bresse’s circles and their intersections have led to the introduction of novel theorems, which have been validated through numerical and graphical results [14,15]. Moreover, graphical techniques have facilitated a more detailed analysis of acceleration invariants in rigid-body motion, thereby enhancing our understanding of acceleration field distributions in multi-body systems [16]. The versatility of Stephenson III six-bar mechanisms in generating specific motion profiles has been demonstrated and their utility extends to achieving up to 11 accuracy points for the generation of functions, such as torque-angle profiles in rehabilitation devices [17]. Furthermore, optimization techniques have been employed to direct points along predefined trajectories [18], while biomechanical engineering has utilized six-bar linkages to replicate the flexion and extension of the human finger. In this process, optimization methods have been used to enhance the fidelity of the trajectory [19]. The utilization of timed curve generators for these mechanisms has further expanded their applications, demonstrating their potential in dynamic motion systems [20]. Planar four-bar mechanisms have retained their position as a fundamental component of kinematic analysis due to their inherent simplicity and versatility. A novel design method for mimicking the human knee joint in the sagittal plane has been proposed, demonstrating the potential of kinematic synthesis in the development of human-like mechanisms [21]. The utilization of graphical techniques for the analysis of acceleration properties has resulted in the streamlined computation of linear and angular accelerations, thereby exemplifying their efficacy in planar mechanisms [22]. Polynomial equations have facilitated insights into the geometric and motion characteristics of acceleration pole loci, elucidating the existence of symmetrical and degenerate forms that contribute to a more profound comprehension of these systems [23]. The concept of acceleration centers has been the subject of extensive investigation in the field of planar kinematics, offering efficient frameworks for the study of rigid-body dynamics and the distribution of acceleration across mechanism components [24]. Furthermore, innovative techniques have employed the instantaneous center of zero acceleration to streamline the examination of four-bar mechanisms, thereby augmenting both theoretical and practical comprehension [25]. In the field of spatial kinematics, the concept of acceleration axes and centers has offered new insights into the dynamics of rigid bodies. The application of planar concepts has been extended to the analysis of spatial systems, with the introduction of acceleration axes as a key tool for the investigation of acceleration distributions, offering closed-form solutions [26,27]. These approaches have led to a more sophisticated theoretical understanding and have enabled enhanced modeling techniques to be developed for multi-body systems [28,29]. Screw theory has emerged as a unifying framework for the description of spatial kinematics. By leveraging the geometry of special Euclidean group 3, screw theory has facilitated the computation of higher-order derivatives, including acceleration and jerk [30]. The application of multidual algebra has broadened its scope to encompass higher-order acceleration fields, thereby providing closed-form solutions with considerable implications for mechanism design and analysis [31]. Furthermore, instantaneous invariants have been employed to optimize linkage configurations, thereby enhancing the functional performance of both planar and spatial mechanisms [32]. Furthermore, higher-order kinematics has concentrated on the comprehensive delineation of acceleration fields across rigid-body systems. The application of tensor algebra and advanced mathematical frameworks has led to the discovery of crucial conditions for the existence of higher-order acceleration centers, including jerk and jounce centers [33]. The extension of planar theories to spatial motions has elucidated the geometric relationships that govern acceleration distributions, thereby offering valuable insights into the design of multi-body systems and control mechanisms [34]. Stephenson III six-bar mechanisms have been utilized in various applications, such as the design and modeling of mechanisms for repetitive tasks with symmetrical end-effector motion [35] and for velocity control in gait rehabilitation robots using advanced techniques like deep reinforcement learning [36]. The Stephenson III six-bar mechanisms, which are investigated in this paper, can be proportioned to achieve specific kinematic characteristics of practical interest, such as those for obtaining dwell configurations of the output link or instantaneous stop configurations of the second coupler link. In fact, these kinematic performances can be required to design automatic machinery for the manufacturing industry, where a specific sequence of repetitive operations is required along with dwells and/or instantaneous stops.

The first- and second-order centrodes provide advanced kinematic information about configurations featuring dwells and/or instantaneous stops, thereby assisting designers in selecting the most suitable solution for the required application.

In this paper, a general algorithm that is aimed to obtain and analyze the first- and second-order centrodes of both coupler links of Stephenson III six-bar mechanisms, along with the corresponding pairs of Bresse’s circles, is proposed by expressing the position vectors of both pairs of velocity and acceleration centers as a function of the driving angle.

This formulation is developed by assuming the well-known and common hypothesis of a rigid body for all links composing the Stephenson III six-bar mechanisms. In fact, the rigid-body hypothesis is very well acceptable for developing the kinematic analysis of closed single- or multi-loop linkages, while the flexibility effects are more evident in open-loop kinematic chains, such as in the case of serial manipulators.

Moreover, at this stage of the research activity, the proposed formulation is focused on the higher-order kinematic performance evaluation of Stephenson III six-bar mechanisms, which show dwell and/or instantaneous stop configurations, while the dynamic analysis is not considered and, consequently, not even the friction force effects and the material resistance. These effects are neglectable when using lubricated plain and/or rolling bearings and steel links, respectively.

In particular, the proposed algorithm has been formulated by using vector algebra and validated through significant numerical and graphical results, which can be useful for designing and analyzing specific mechanisms for practical applications. The main scientific contribution is the formulation of a vector algebra algorithm to determine the second-order centrodes of both coupler links of the mechanism. In particular, this work is organized as follows: Section 2 deals with the kinematic analysis of a generic Stephenson III six-bar mechanism by devoting particular attention to the instantaneous centers of rotation and the acceleration centers; Section 3 introduces both pairs of Bresse’s circles, one for each coupler link, and explains their geometric and kinematic meanings; Section 4 proposes the formulation of the first- and second-order centrodes of both coupler links, describing their main kinematic characteristics; Section 5 shows several numerical examples which have allowed the validation of the proposed algorithm by implementing representative mechanism configurations. Moreover, a practical example regarding the jaw-crusher machine, based on a Stephenson III six-bar mechanism, is developed and analyzed in both cases of simple dwell of the swing-jaw plate and the additional instantaneous stop of the second coupler link. Finally, the conclusions are reported in Section 6.

2. Kinematic Analysis: Velocity and Acceleration Centers

The Stephenson III six-bar mechanism is sketched in Figure 1a, composed of the four-bar linkage A₀ABB₀ and the CDD₀ RR dyad, which are connected between them by means of the coupler triangle ABC. Link lengths are indicated with r_i for i = from 1 to 7, along with the angles θ₁ and θ₇, which give the positions of r₁ and r₇, respectively, with respect to the positive X-axis of the OXY fixed reference frame. The relative position of point C with respect to the coupler link AB is given by the distance r_3C and the angle α₀. In general, the driving crank A₀A rotates counterclockwise with angular position θ₂, velocity ω₂, and acceleration α₂. Referring to the kinematic sketch of Figure 1b, where the fixed frame OXY has been chosen with O = A₀, the position analysis of the Stephenson III six-bar mechanisms is formulated through the following two closed-loop equations:

$${\mathbf{r}}_2+{\mathbf{r}}_3={\mathbf{r}}_1+{\mathbf{r}}_4$$

(1)

$${\mathbf{r}}_2+{\mathbf{r}}_{3C}+{\mathbf{r}}_5={\mathbf{r}}_1+{\mathbf{r}}_7+{\mathbf{r}}_6$$

(2)

where each vector r_i for i = 1, …, 7 are expressed in Cartesian form as follows:

$${\mathbf{r}}_i=r_i\left(cos{\theta }_i\mathbf{i}+sin{\theta }_i\mathbf{j}\right)$$

(3)

r_i and θ_i are the magnitude and the counterclockwise oriented angles of r_i, respectively, while r_3C is the position vector of point C with respect to A. The unit vectors of the X and Y axes are i and j, respectively. Thus, substituting each vector r_i of Equation (3) into Equation (1) and developing for the first loop of the four-bar linkage A₀ABB₀, one has the following:

$${\theta }_3={\tan }^{-1}\left({\displaystyle \frac{r_1\sin {\theta }_1-r_2\sin {\theta }_2+r_4\sin {\theta }_4}{r_1\cos {\theta }_1-r_2\cos {\theta }_2+r_4\cos {\theta }_4}}\right)$$

(4)

where θ₃ is the oriented angle of r₃, θ₁ of r₁ is assigned, θ₂ of r₂ is the driving angle, while the output angle θ₄ of r₄ is given by the following:

$${\theta }_4=2\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}{\displaystyle \frac{-\mathcal{B}+\sigma \sqrt{{\mathcal{B}}^2-{\mathcal{C}}^2+{\mathcal{A}}^2}}{\mathcal{C}-\mathcal{A}}}$$

(5)

which terms $\mathcal{A}$, $\mathcal{B}$, and $\mathcal{C}$ take the following form:

$$\begin{gathered} \mathcal{A}=2r_1{r}_4\cos {\theta }_1-2r_2{r}_4\cos {\theta }_2 \\ \mathcal{B}=2r_1{r}_4 \sin {\theta }_1-2r_2{r}_4\sin {\theta }_2 \\ \mathcal{C}=r_1^2+r_2^2+r_4^2-r_3^2-2r_1{r}_2\cos \left({\theta }_1-{\theta }_2\right) \end{gathered}$$

(6)

and σ is equal to ±1 according to the corresponding assembly mode.

Thus, by determining the first-time derivatives of the X and Y scalar components of Equation (1), both angular velocity vectors ω₃ and ω₄ are obtained as a function of the given angular velocity ω₂ of the driving crank 2:

$${\mathsf{\omega }}_3={\dot{\theta }}_3 \mathbf{k}=\left[{\omega }_2{\displaystyle \frac{r_2}{r_3}}{\displaystyle \frac{\sin \left({\theta }_2-{\theta }_4\right)}{\sin \left({\theta }_4-{\theta }_3\right)}}\right]\mathbf{k}$$

(7)

$${\mathsf{\omega }}_4={\dot{\theta }}_4 \mathbf{k}=\left[{\omega }_2{\displaystyle \frac{r_2}{r_4}}{\displaystyle \frac{\sin \left({\theta }_2-{\theta }_3\right)}{\sin \left({\theta }_4-{\theta }_3\right)}}\right]\mathbf{k}$$

(8)

where k is the unit vector of the Z-axis.

Similarly, by determining the second-time derivatives of the X and Y scalar components of Equation (1), both angular acceleration vectors α₃ and α₄ are obtained as a function of the given angular acceleration α₂ of the driving crank 2:

$${\mathsf{\alpha }}_3={\ddot{\theta }}_3 \mathbf{k}=\left[{\displaystyle \frac{ {\alpha }_2 r_2\sin \left({\theta }_2-{\theta }_4\right)+{\omega }_2^2 r_2\cos \left({\theta }_2-{\theta }_4\right)+ {\omega }_3^2 r_3 \cos \left({\theta }_3-{\theta }_4\right)-{\omega }_4^2 r_4 }{r_3 \sin \left({\theta }_4-{\theta }_3\right)}}\right]\mathbf{k}$$

(9)

$${\mathsf{\alpha }}_4={\ddot{\theta }}_4 \mathbf{k}=\left[{\displaystyle \frac{{\alpha }_2 r_2\sin \left({\theta }_2-{\theta }_3\right)+{\omega }_2^2 r_2\cos \left({\theta }_2-{\theta }_3\right)+{\omega }_3^2 r_3-{\omega }_4^2 r_4 \cos \left({\theta }_4-{\theta }_3\right)}{r_4 \sin \left({\theta }_4-{\theta }_3\right)}}\right]\mathbf{k}$$

(10)

The velocity vectors v_A, v_B, and v_C of points A, B, and C, are obtained by developing the first-time derivatives of the following position vectors:

$$\begin{gathered} {\mathbf{r}}_A={\mathbf{r}}_2 \\ {\mathbf{r}}_B={\mathbf{r}}_1+{\mathbf{r}}_4 \\ {\mathbf{r}}_C={\mathbf{r}}_2+{\mathbf{r}}_{3C} \end{gathered}$$

(11)

thus, obtaining the following:

$${\mathbf{v}}_A={\dot{\mathbf{r}}}_2=-\left({\omega }_2{r}_2\sin {\theta }_2\right)\mathbf{i}+\left({\omega }_2{r}_2\cos {\theta }_2\right)\mathbf{j}$$

(12)

$${\mathbf{v}}_B={\dot{\mathbf{r}}}_4={\dot{\mathbf{r}}}_2+{\mathsf{\omega }}_3\times \left({\mathbf{r}}_4-{\mathbf{r}}_2\right)$$

(13)

$${\mathbf{v}}_C={\dot{\mathbf{r}}}_{3C}={\dot{\mathbf{r}}}_2+{\mathsf{\omega }}_3\times \left({\mathbf{r}}_{3C}-{\mathbf{r}}_2\right)$$

(14)

Referring to Figure 2, the position vector r_P3 of the instantaneous center of rotation P₃ of the coupler link AB is obtained by imposing a zero-velocity to P₃, as follows:

$${\mathbf{v}}_{P3}={\dot{\mathbf{r}}}_{P3}={\dot{\mathbf{r}}}_2+{\mathsf{\omega }}_3\times \left({\mathbf{r}}_{P3}-{\mathbf{r}}_2\right)=\mathbf{0}$$

(15)

Thus, substituting ${\dot{\mathbf{r}}}_2={\mathbf{v}}_A$, ${\mathsf{\omega }}_3$, and ${\mathbf{r}}_2={\mathbf{r}}_A$, which are given by Equations (12) and (7), and by the first of Equation (11), into Equation (15) and developing, one has r_P3 as follows:

$${\mathbf{r}}_{P3}=\left(-{\displaystyle \frac{{\dot{r}}_{Ay}}{{\omega }_3}}+r_{2x}\right)\mathbf{i}+\left({\displaystyle \frac{{\dot{r}}_{Ax}}{{\omega }_3}}+r_{2y}\right)\mathbf{j}$$

(16)

The acceleration vectors a_A, a_B, and a_C of points A, B, and C, are obtained by developing the second-time derivatives of the corresponding position vectors of Equation (1).

Therefore, one has the following:

$${\mathbf{a}}_A={\ddot{\mathbf{r}}}_2=\left(-{\omega }_2^2{r}_2\cos {\theta }_2-{\alpha }_2{r}_2\sin {\theta }_2\right)\mathbf{i}+\left(-{\omega }_2^2{r}_2\sin {\theta }_2+{\alpha }_2{r}_2\cos {\theta }_2\right)\mathbf{j}$$

(17)

$${\mathbf{a}}_B={\ddot{\mathbf{r}}}_4={\ddot{\mathbf{r}}}_2+{\mathsf{\alpha }}_3\times \left({\mathbf{r}}_4-{\mathbf{r}}_2\right)+{\mathsf{\omega }}_3\times \left[{\mathsf{\omega }}_3\times \left({\mathbf{r}}_4-{\mathbf{r}}_2\right)\right]$$

(18)

$${\mathbf{a}}_C={\ddot{\mathbf{r}}}_C={\ddot{\mathbf{r}}}_2+{\mathsf{\alpha }}_3\times \left({\mathbf{r}}_C-{\mathbf{r}}_2\right)+{\mathsf{\omega }}_3\times \left[{\mathsf{\omega }}_3\times \left({\mathbf{r}}_C-{\mathbf{r}}_2\right)\right]$$

(19)

The position vector r_K3 of the acceleration center K₃ of the coupler link AB is obtained by imposing a zero-acceleration to K₃, as follows:

$${\mathbf{a}}_{K3}={\ddot{\mathbf{r}}}_{K3}={\ddot{\mathbf{r}}}_2+{\mathsf{\alpha }}_3\times \left({\mathbf{r}}_{K3}-{\mathbf{r}}_2\right)+{\mathsf{\omega }}_3\times \left[{\mathsf{\omega }}_3\times \left({\mathbf{r}}_{K3}-{\mathbf{r}}_2\right)\right]=\mathbf{0}$$

(20)

Thus, substituting ${\ddot{\mathbf{r}}}_2={\mathbf{a}}_A$, ${\mathsf{\alpha }}_3$ and ${\mathbf{r}}_2={\mathbf{r}}_A$, which are given by Equations (17) and (9), and by the first of Equation (11), into Equation (20), and developing, one has r_k3 as follows:

$${\mathbf{r}}_{K3}=\left(r_{Ax}-{\displaystyle \frac{{\ddot{r}}_{Ay}}{{\alpha }_3}}+{\displaystyle \frac{{\omega }_3^2}{{\alpha }_3}}{\displaystyle \frac{{\alpha }_3 {\ddot{r}}_{Ax}+{\omega }_3^2 {\ddot{r}}_{Ay}}{{\omega }_3^4+{\alpha }_3^2}}\right)\mathbf{i}+\left(r_{Ay}+{\displaystyle \frac{{\alpha }_3 {\ddot{r}}_{Ax}+{\omega }_3^2 {\ddot{r}}_{Ay}}{{\omega }_3^4+{\alpha }_3^2}}\right)\mathbf{j}$$

(21)

Therefore, the position vectors r_P3 and r_K3 of the velocity and acceleration centers P₃ and K₃, which are sketched in Figure 2, are expressed in Cartesian form by Equations (16) and (21) and, with respect to the fixed frame OXY, as a function of the driving crank angle θ₂.

Consequently, Equations (16) and (21) also represent the respective vector functions of the first- and second-order fixed centrodes of the coupler link AB (link 3).

Referring to Figure 1 and taking into account that we are dealing with Stephenson III six-bar mechanisms, the second closed-loop equation of Equation (2) must be developed.

Thus, substituting each vector r_i in the Cartesian form of Equation (3) into Equation (2) and developing for the second loop of the five-bar linkage B₀BCDD₀, one has the following:

$${\theta }_5=2\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}{\displaystyle \frac{-\mathcal{E}+\sigma \sqrt{{\mathcal{E}}^2-{\mathcal{G}}^2+{\mathcal{D}}^2}}{\mathcal{G}-\mathcal{D}}}$$

(22)

where $\mathcal{D}$, $\mathcal{E}$, and $\mathcal{G}$ are given by the following:

$$\begin{gathered} \mathcal{D}=2r_5\left[r_2\cos {\theta }_2+r_{3C}\cos \left({\theta }_3+{\alpha }_0\right)-r_1\cos {\theta }_1-r_7\cos {\theta }_7\right] \\ \mathcal{E}=2r_5\left[r_2\sin {\theta }_2+r_{3C}\sin \left({\theta }_3+{\alpha }_0\right)-r_1\sin {\theta }_1-r_7\sin {\theta }_7\right] \\ \begin{array}{l}\mathcal{G}=r_2^2+r_{3C}^2+r_5^2+r_1^2+r_7^2-r_6^2-2r_1{r}_2\cos \left({\theta }_1-{\theta }_2\right)+2r_1{r}_7\cos \left({\theta }_1-{\theta }_7\right)-2r_1{r}_{3C}\cos \left({\alpha }_0+{\theta }_3-{\theta }_1\right)+\\ -2 r_2{r}_7\cos \left({\theta }_2-{\theta }_7\right)+2r_2{r}_{3C}\cos \left({\alpha }_0+{\theta }_3-{\theta }_2\right)-2r_{3C}{r}_7\cos \left({\alpha }_0+{\theta }_3-{\theta }_7\right)\end{array} \end{gathered}$$

(23)

and, in turn, one has the following:

$${\theta }_6={\tan }^{-1}\left[{\displaystyle \frac{r_2\sin {\theta }_2+r_{3C}\sin \left({\theta }_3+{\alpha }_0\right)+r_5\sin {\theta }_5-r_1\sin {\theta }_1-r_7\sin {\theta }_7}{r_2\cos {\theta }_2+r_{3C}\cos \left({\theta }_3+{\alpha }_0\right)+r_5\cos {\theta }_5-r_1\cos {\theta }_1-r_7\cos {\theta }_7}}\right]$$

(24)

The oriented angles θ₅ and θ₆ only depend on the driving crank angle θ₂.

By determining the first-time derivatives of the X and Y scalar components of Equation (2) and developing them, the angular velocity vectors ω₅ and ω₆ are given by the following:

$${\mathsf{\omega }}_5={\dot{\theta }}_5 \mathbf{k}=\left[{\displaystyle \frac{{\omega }_2 r_2\sin \left({\theta }_2-{\theta }_6\right)+{\omega }_3 r_{3C}\sin \left({\theta }_3+{\alpha }_0-{\theta }_6\right)}{r_5\sin \left({\theta }_6-{\theta }_5\right)}}\right]\mathbf{k}$$

(25)

$${\mathsf{\omega }}_6={\dot{\theta }}_6 \mathbf{k}=\left[{\displaystyle \frac{{\omega }_2 r_2\sin \left({\theta }_2-{\theta }_5\right)+{\omega }_3 r_{3C}\sin \left({\theta }_3+{\alpha }_0-{\theta }_5\right)}{r_6\sin \left({\theta }_6-{\theta }_5\right)}}\right]\mathbf{k}$$

(26)

Likewise, by determining the second-time derivatives of the X and Y scalar components of Equation (2) and developing them, both angular acceleration vectors α₅ and α₆ are obtained as follows:

$$\begin{array}{l}{\mathsf{\alpha }}_5={\ddot{\theta }}_5 \mathbf{k}={\displaystyle \frac{1}{r_5 \sin \left({\theta }_6-{\theta }_5\right)}}\left[{\alpha }_2\right. r_2\sin \left({\theta }_2-{\theta }_6\right)+{\omega }_2^2 r_2\cos \left({\theta }_2-{\theta }_6\right)+\\ {\omega }_3^2 r_{3C}\cos \left({\theta }_3+{\alpha }_0-{\theta }_6\right)+ {\alpha }_3 r_{3C}\sin \left({\theta }_3+{\alpha }_0-{\theta }_6\right)+ {\omega }_5^2 r_5\cos \left({\theta }_5-{\theta }_6\right)-\left.{\omega }_6^2{r}_6\right]\mathbf{k}\end{array}$$

(27)

$$\begin{array}{l}{\mathsf{\alpha }}_6={\ddot{\theta }}_6 \mathbf{k}={\displaystyle \frac{1}{r_6 \sin \left({\theta }_6-{\theta }_5\right)}}\left[{\alpha }_2{r}_2\right. \sin \left({\theta }_2-{\theta }_5\right)+{\omega }_2^2 r_2 \cos \left({\theta }_2-{\theta }_5\right)+\\ +{\omega }_3^2 r_{3C}\cos \left({\theta }_3+{\alpha }_0-{\theta }_5\right)+ {\alpha }_3 r_{3C}\sin \left({\theta }_3+{\alpha }_0-{\theta }_5\right)+ {\omega }_6^2 r_6\cos \left({\theta }_5+{\theta }_6\right)+\left.{\omega }_5^2{r}_5\right]\mathbf{k}\end{array}$$

(28)

Moreover, the velocity vector v_D of point D is obtained by developing the first-time derivative of the following position vector:

$${\mathbf{r}}_D={\mathbf{r}}_1+{\mathbf{r}}_7+{\mathbf{r}}_6$$

(29)

thus, obtaining the following:

$${\mathbf{v}}_D={\dot{\mathbf{r}}}_6=-\left({\omega }_6{r}_6\sin {\theta }_6\right)\mathbf{i}+\left({\omega }_6{r}_6\cos {\theta }_6\right)\mathbf{j}$$

(30)

Referring to Figure 2, the position vector r_P5 of the instantaneous center of rotation P₅ of the coupler link CD is obtained by imposing a zero-velocity to P₅, as follows:

$${\mathbf{v}}_{P5}={\dot{\mathbf{r}}}_{P5}={\dot{\mathbf{r}}}_C+{\mathsf{\omega }}_5\times \left({\mathbf{r}}_{P5}-{\mathbf{r}}_C\right)=\mathbf{0}$$

(31)

Thus, substituting ${\dot{\mathbf{r}}}_C={\mathbf{v}}_C$, ${\mathsf{\omega }}_5$, and ${\mathbf{r}}_C$, which are given by Equations (14) and (25), and by the third part of Equation (11), into Equation (31), and developing, one has r_P5 as follows:

$${\mathbf{r}}_{P5}=\left(-{\displaystyle \frac{{\dot{r}}_{Cy}}{{\omega }_5}}+r_{Cx}\right)\mathbf{i}+\left({\displaystyle \frac{{\dot{r}}_{Cx}}{{\omega }_5}}+r_{Cy}\right)\mathbf{j}$$

(32)

The acceleration vector a_D of point D is obtained by developing the second-time derivative of the corresponding position vector ${\mathbf{r}}_D$ of Equation (2), by which one has the following:

$${\mathbf{a}}_D={\ddot{\mathbf{r}}}_D=\left(-{\omega }_6^2{r}_6\cos {\theta }_6-{\alpha }_6{r}_6\sin {\theta }_6\right)\mathbf{i}+\left(-{\omega }_6^2{r}_6\sin {\theta }_6+{\alpha }_6{r}_6\cos {\theta }_6\right)\mathbf{j}$$

(33)

The position vector r_K5 of the acceleration center K₅ of the coupler link CD is obtained by imposing a zero-acceleration to K₅, as follows:

$${\mathbf{a}}_{K5}={\ddot{\mathbf{r}}}_{K5}={\ddot{\mathbf{r}}}_C+{\mathsf{\alpha }}_5\times \left({\mathbf{r}}_{K5}-{\mathbf{r}}_C\right)+{\mathsf{\omega }}_5\times \left[{\mathsf{\omega }}_5\times \left({\mathbf{r}}_{K5}-{\mathbf{r}}_C\right)\right]=\mathbf{0}$$

(34)

Thus, substituting ${\ddot{\mathbf{r}}}_C={\mathbf{a}}_C$, ${\mathsf{\alpha }}_5$ and ${\mathbf{r}}_C$, which are given by Equations (19) and (10), and by the third of Equation (11), into Equation (34), and developing, one has r_k5 as follows:

$${\mathbf{r}}_{K5}=\left(r_{Cx}-{\displaystyle \frac{{\ddot{r}}_{Cy}}{{\alpha }_5}}+{\displaystyle \frac{{\omega }_5^2}{{\alpha }_5}}{\displaystyle \frac{{\alpha }_5 {\ddot{r}}_{Cx}+{\omega }_5^2 {\ddot{r}}_{Cy}}{{\omega }_5^4+{\alpha }_5^2}}\right)\mathbf{i}+\left(r_{Cy}+{\displaystyle \frac{{\alpha }_5 {\ddot{r}}_{Cx}+{\omega }_5^2 {\ddot{r}}_{Cy}}{{\omega }_5^4+{\alpha }_5^2}}\right)\mathbf{j}$$

(35)

Therefore, the position vectors r_P5 and r_K5 of the velocity and acceleration centers P₅ and K₅, which are sketched in Figure 2, are expressed in Cartesian form by Equations (32) and (35) and with respect to the fixed frame OXY, as a function of the driving crank angle θ₂.

Consequently, Equations (32) and (35) also represent the respective vector functions of the first- and second-order fixed centrodes of the coupler link CD (link 5).

3. Bresse’s Circles

Bresse’s circles consist of the well-known inflection and stationarity circles, which intersect at both the instantaneous center of rotation P₃ and the acceleration center K₃. The inflection circle I3 is the geometric locus of all coupler points, which show an inflection point in their paths and I3 is always tangent to both centrodes at the P₃ point. The stationarity circle S3 is the locus of all coupler points which show a pure normal acceleration.

Thus, both Bresse’s circles are now derived by using vector algebra, and, in particular, the equation of the inflection circle I3 is obtained as the locus of all the coupler points whose velocity and acceleration vectors are parallel to each other.

This condition can be expressed by imposing a zero-cross product, as follows:

$${\mathbf{v}}_{M_3}\times {\mathbf{a}}_{M_3}=\mathbf{0}$$

(36)

where M₃ is a generic coupler point of link 3. In particular, vectors ${\mathbf{v}}_{M_3}$ and ${\mathbf{a}}_{M_3}$ are obtained by Equations (13) and (18) for B = M₃ and refer to the velocity and acceleration vectors of point A. Thus, substituting ${\mathsf{\omega }}_3$ and ${\mathsf{\alpha }}_3$ from Equations (7) and (9) into Equations (13) and (18), one has ${\mathbf{v}}_{M_3}$ and ${\mathbf{a}}_{M_3}$ in matrix form, as follows:

$${\mathbf{v}}_{M_3}=\left[\begin{array}{c}{r}_{Ay}\left({\omega }_3 -{\omega }_2 \right)-{\omega }_{3}y\\ r_{Ax}\left({\omega }_2 -{\omega }_3 \right)+{\omega }_{3}x\\ 0\end{array}\right]$$

(37)

$${\mathbf{a}}_{M_3}=\left[\begin{array}{c}{r}_{Ax}\left({\omega }_3^2 -{\omega }_2^2 \right)+r_{Ay}\left({\alpha }_3 -{\alpha }_2 \right)-{\omega }_3^{2}x -{\alpha }_{3}y\\ r_{Ax}\left({\alpha }_2 -{\alpha }_3 \right)+r_{Ay}\left({\omega }_3^2 -{\omega }_2^2 \right)+{\alpha }_{3}x -{\omega }_3^{2}y\\ 0\end{array}\right]$$

(38)

where $r_{Ax}$ and $r_{Ay}$, along with x and y, are the Cartesian components of ${\mathbf{r}}_A$ and ${\mathbf{r}}_{M3}$, respectively, and the symbol T indicates the transpose matrix.

Therefore, by substituting Equations (37) and (38) into Equation (36) and developing it, one has the following algebraic equation of the inflection circle I3:

$$\begin{array}{l}\left[r_{Ay}\left({\omega }_3 -{\omega }_2 \right)-{\omega }_{3}y\right]\cdot \left[r_{Ax}\left({\alpha }_2 -{\alpha }_3 \right)+r_{Ay}\left({\omega }_3^2 -{\omega }_2^2 \right)+{\alpha }_{3}x -{\omega }_3^{2}y\right]-\\ \left[r_{Ax}\left({\omega }_2 -{\omega }_3 \right)+{\omega }_{3}x\right]\cdot \left[r_{Ax}\left({\omega }_3^2 -{\omega }_2^2 \right)+r_{Ay}\left({\alpha }_3 -{\alpha }_2 \right)-{\omega }_3^{2}x -{\alpha }_{3}y\right]=0\end{array}$$

(39)

the compact version of which is reported below:

$$x^2+y^2+{\mathcal{A}}_{I3}x+{\mathcal{B}}_{I3}y+{\mathcal{C}}_{I3}=0$$

(40)

where the following is true:

$$\begin{gathered} {\mathcal{A}}_{I3}={\displaystyle \frac{r_{Ax}\left({\omega }_3^2{\omega }_2-2{\omega }_3^3+{\omega }_2^2{\omega }_3\right)+r_{Ay}\left({\omega }_3{\alpha }_2-{\omega }_2{\alpha }_3\right)}{{\omega }_3^3}} \\ {\mathcal{B}}_{I3}={\displaystyle \frac{r_{Ax}\left(-{\omega }_3{\alpha }_2+{\omega }_2{\alpha }_3\right)+r_{Ay}\left({\omega }_3^2{\omega }_2-2{\omega }_3^3+{\omega }_2^2{\omega }_3\right)}{{\omega }_3^3}} \\ {\mathcal{C}}_{I3}={\displaystyle \frac{\left({\omega }_2-{\omega }_3\right)\left\{r_{Ax}\left[r_{Ax}\left({\omega }_2^2-{\omega }_3^2\right)+r_{Ay}\left({\alpha }_2-{\alpha }_3\right)\right]-r_{Ay}\left[r_{Ax}\left({\alpha }_2-{\alpha }_3\right)-r_{Ay}\left({\omega }_3^2-{\omega }_2^2\right)\right]\right\}}{{\omega }_3^3}} \end{gathered}$$

(41)

Likewise, the stationarity circle S3 is the geometric locus of all coupler points at which the velocity and acceleration vectors ${\mathbf{v}}_{M_3}$ and ${\mathbf{a}}_{M_3}$ are normal to each other, which means that their dot product equals zero, as follows:

$${\mathbf{v}}_{M_3}\cdot {\mathbf{a}}_{M_3}=0$$

(42)

Therefore, substituting Equations (37) and (38) into Equation (42), and developing it, one has the following algebraic equation of the stationarity circle S3:

$$\begin{array}{l}\left[r_{Ay}\left({\omega }_3 -{\omega }_2 \right)-{\omega }_{3}y\right]\cdot \left[r_{Ax}\left({\omega }_3^2 -{\omega }_2^2 \right)+r_{Ay}\left({\alpha }_3 -{\alpha }_2 \right)-{\omega }_3^{2}x -{\alpha }_{3}y\right]+\\ \left[r_{Ax}\left({\omega }_2 -{\omega }_3 \right)+{\omega }_{3}x\right]\left[r_{Ax}\left({\alpha }_2 -{\alpha }_3 \right)+r_{Ay}\left({\omega }_3^2 -{\omega }_2^2 \right)+{\alpha }_{3}x -{\omega }_3^{2}y\right]=0\end{array}$$

(43)

the compact equation of which is reported below:

$$x^2+y^2+{\mathcal{A}}_{S3}x+{\mathcal{B}}_{S3}y+{\mathcal{C}}_{S3}=0$$

(44)

where the following is true:

$$\begin{gathered} {\mathcal{A}}_{S3}={\displaystyle \frac{r_{Ax}\left({\omega }_2{\alpha }_3+{\omega }_3{\alpha }_2-2{\omega }_3{\alpha }_3\right)+r_{Ay}\left({\omega }_2{\omega }_3^2-{\omega }_2^2{\omega }_3\right)}{{\omega }_3{\alpha }_3}} \\ {\mathcal{B}}_{S3}={\displaystyle \frac{r_{2x}\left(-{\omega }_2{\omega }_3^2+{\omega }_2^2{\omega }_3\right)+r_{2y}\left({\omega }_2{\alpha }_3+{\omega }_3{\alpha }_2-2{\omega }_3{\alpha }_3\right)}{{\omega }_3{\alpha }_3}} \\ {\mathcal{C}}_{S3}={\displaystyle \frac{\left({\omega }_2-{\omega }_3\right)\left\{r_{Ax}\left[r_{Ax}\left({\alpha }_2-{\alpha }_3\right)-r_{Ay}\left({\omega }_2^2-{\omega }_3^2\right)\right]+r_{Ay}\left[r_{Ax}\left({\omega }_2^2-{\omega }_3^2\right)+r_{Ay}\left({\alpha }_2-{\alpha }_3\right)\right]\right\}}{{\omega }_3{\alpha }_3}} \end{gathered}$$

(45)

The same approach is now applied to the second coupler link CD of the Stephenson III six-bar mechanism and, thus, for the inflection circle I5, one has the following:

$${\mathbf{v}}_{M_5}\times {\mathbf{a}}_{M_5}=\mathbf{0}$$

(46)

where M₅ is a generic coupler point of link 5 and, consequently, its velocity and acceleration vectors ${\mathbf{v}}_{M_5}$ and ${\mathbf{a}}_{M_5}$ can be derived by Equations (31) and (34) for M₅ = P₅ and M₅ = K₅, respectively, whose reference point is C. Thus, ${\mathbf{v}}_{M_5}$ and ${\mathbf{a}}_{M_5}$ are given in matrix form as follows:

$${\mathbf{v}}_{M_5}=\left[\begin{array}{c}{r}_{Cy}\left({\omega }_5 -{\omega }_3 \right)+{\omega }_3{r}_{P3y}-{\omega }_{5}y\\ r_{Cx}\left({\omega }_3 -{\omega }_5 \right)-{\omega }_3{r}_{P3x}+{\omega }_{5}x\\ 0\end{array}\right]$$

(47)

$${\mathbf{a}}_{M_5}=\left[\begin{array}{c}{r}_{Ax}\left({\omega }_3^2 -{\omega }_2^2 \right)+r_{Ay}\left({\alpha }_3 -{\alpha }_2 \right)+r_{Cx}\left({\omega }_5^2 -{\omega }_3^2 \right)+r_{Cy}\left({\alpha }_5 -{\alpha }_3 \right)-{\omega }_5^{2}x -{\alpha }_{5}y\\ r_{Ax}\left({\alpha }_2 -{\alpha }_3 \right)+r_{Ay}\left({\omega }_3^2 -{\omega }_2^2 \right)+r_{Cx}\left({\alpha }_5 -{\alpha }_3 \right)+r_{Cy}\left({\omega }_5^2 -{\omega }_3^2 \right)+{\alpha }_{5}x -{\omega }_5^{2}y\\ 0\end{array}\right]$$

(48)

where, in addition to the previous terms, $r_{Cx}$ and $r_{Cy}$, along with x and y, are the Cartesian components ${\mathbf{r}}_A$ and ${\mathbf{r}}_{M5}$, respectively.

Therefore, by substituting Equations (47) and (48) into Equation (46) and developing it, one has the following algebraic equation of the inflection circle I5:

$$\begin{array}{l}\left[r_{Cy}\left({\omega }_5 -{\omega }_3 \right)+{\omega }_3{r}_{P3y}-{\omega }_{5}y\right]\cdot \left[r_{Ax}\left({\alpha }_2 -{\alpha }_3 \right)+r_{Ay}\left({\omega }_3^2 -{\omega }_2^2 \right)\right.+r_{Cx}\left({\alpha }_5 -{\alpha }_3 \right)+\\ \left.+r_{Cy}\left({\omega }_5^2 -{\omega }_3^2 \right)+{\alpha }_{5}x -{\omega }_5^{2}y\right]-\left[r_{Cx}\left({\omega }_3 -{\omega }_5 \right)-{\omega }_3{r}_{P3x}+{\omega }_{5}x\right]\cdot \left[r_{Ax}\left({\omega }_3^2 -{\omega }_2^2 \right)+\right.\\ +\left.r_{Ay}\left({\alpha }_3 -{\alpha }_2 \right)+r_{Cx}\left({\omega }_5^2 -{\omega }_3^2 \right)+r_{Cy}\left({\alpha }_5 -{\alpha }_3 \right)-{\omega }_5^{2}x -{\alpha }_{5}y\right]=0\end{array}$$

(49)

the compact form of which is reported below:

$$x^2+y^2+{\mathcal{A}}_{I5}x+{\mathcal{B}}_{I5}y+{\mathcal{C}}_{I5}=0$$

(50)

where the following is true:

$$\begin{gathered} \begin{array}{l}\hfill {\mathcal{A}}_{I5}={\displaystyle \frac{1}{{\omega }_5^3}}\left[r_{2x}{\omega }_5\right.\left({\omega }_2^2-{\omega }_3^2\right)+r_{2y}{\omega }_5\left({\alpha }_2-{\alpha }_3\right)+r_{P3x}{\omega }_5^2{\omega }_3+r_{P3y}{\omega }_3{\alpha }_5+\hfill \\ \hfill +r_{Cx}\left(-2{\omega }_5^3+{\omega }_5^2{\omega }_3+{\omega }_3^2{\omega }_5\right)+r_{Cy}\left.\left({\omega }_5{\alpha }_3-{\omega }_3{\alpha }_5\right)\right]\hfill \end{array} \\ \begin{array}{l}\hfill {\mathcal{B}}_{I5}={\displaystyle \frac{1}{{\omega }_5^3}}\left[r_{2x}{\omega }_5\right.\left({\alpha }_2+{\alpha }_5\right)+r_{2y}{\omega }_5\left({\omega }_2^2-{\omega }_3^2\right)+r_{P3x}{\omega }_3{\alpha }_5-r_{P3y}{\omega }_3{\omega }_5^2+\hfill \\ \hfill +r_{Cx}\left(-{\omega }_5{\alpha }_3-{\omega }_3{\alpha }_5\right)+r_{Cy}\left.\left(-2{\omega }_5^3+{\omega }_3^2{\omega }_5+{\omega }_3{\omega }_5^2\right)\right]\hfill \end{array} \\ \begin{array}{l}{\mathcal{C}}_{I5}={\displaystyle \frac{1}{{\omega }_5^3}}\left\{\left[r_{Cx}\left({\omega }_5-{\omega }_3\right)+r_{P3x}{\omega }_3\right]\right.\cdot \left[r_{2x}\left({\omega }_3^2-{\omega }_2^2\right)\right.+r_{2y}\left({\alpha }_3-{\alpha }_2\right)+r_{Cx}\left({\omega }_5^2-{\omega }_3^2\right)+\hfill \\ +r_{Cy}\left.\left({\alpha }_5-{\alpha }_3\right)\right]+\left[r_{Cy}\left({\omega }_5-{\omega }_3\right)+r_{P3y}{\omega }_3\right]\cdot \left[r_{2x}\left({\alpha }_2-{\alpha }_3\right)+r_{2y}\left({\omega }_3^2-{\omega }_2^2\right)\right.+\hfill \\ +r_{Cx}\left({\alpha }_3-{\alpha }_5\right)+\left.\left.r_{Cy}\left({\omega }_5^2-{\omega }_3^2\right)\right]\right\}\hfill \end{array} \end{gathered}$$

(51)

Finally, the stationarity circle S₅ is obtained by the following:

$${\mathbf{v}}_{M_5}\cdot {\mathbf{a}}_{M_5}=0$$

(52)

and, thus, by substituting Equations (47) and (48) into Equation (52), one has the following:

$$\begin{array}{l}\left[r_{Cy}\left({\omega }_5 -{\omega }_3 \right)+{\omega }_3{r}_{P3y}-{\omega }_{5}y\right]\cdot \left[r_{Ax}\left({\omega }_3^2 -{\omega }_2^2 \right)+r_{Ay}\right.\left({\alpha }_2 -{\alpha }_3 \right)+r_{Cx}\left({\omega }_5^2 -{\omega }_3^2 \right)+\\ \left.+r_{Cy}\left({\alpha }_5 -{\alpha }_3 \right)-{\omega }_5^{2}x -{\alpha }_{5}y\right]+\left[r_{Cx}\left({\omega }_3 -{\omega }_5 \right)-{\omega }_3{r}_{P3x}+{\omega }_{5}x\right]\cdot \left[r_{Ax}\left({\alpha }_3 -{\alpha }_2 \right)+\right.\\ +\left.r_{Ay}\left({\omega }_3^2 -{\omega }_2^2 \right)+r_{Cx}\left({\alpha }_5 -{\alpha }_3 \right)+r_{Cy}\left({\omega }_5^2 -{\omega }_3^2 \right)+{\alpha }_{5}x -{\omega }_5^{2}y\right]=0\end{array}$$

(53)

the compact equation of which is the following:

$$x^2+y^2+{\mathcal{A}}_{S5}x+{\mathcal{B}}_{S5}y+{\mathcal{C}}_{S5}=0$$

(54)

where the following is true:

$$\begin{gathered} \begin{array}{l}\hfill {\mathcal{A}}_{S5}={\displaystyle \frac{1}{{\omega }_5{\alpha }_5}}\left[r_{2x}{\omega }_5\right.\left({\alpha }_2-{\alpha }_3\right)+r_{2y}{\omega }_5\left({\omega }_3^2-{\omega }_2^2\right)-r_{P3x}{\omega }_3{\alpha }_5-r_{P3y}{\omega }_3{\omega }_5^2+\hfill \\ \hfill +r_{Cx}\left({\omega }_5{\alpha }_3-{\omega }_3{\alpha }_5\right)+r_{Cy}\left.\left(-{\omega }_3^2{\omega }_5+{\omega }_3{\omega }_5^2\right)\right]\hfill \end{array} \\ \begin{array}{l}\hfill {\mathcal{B}}_{S5}={\displaystyle \frac{1}{{\omega }_5{\alpha }_5}}\left[r_{2x}{\omega }_5\right.\left({\omega }_2^2-{\omega }_3^2\right)+r_{2y}{\omega }_5\left({\alpha }_2-{\alpha }_3\right)+r_{P3x}{\omega }_3{\omega }_5^2-r_{P3y}{\omega }_3{\alpha }_5+\hfill \\ \hfill +r_{Cx}\left({\omega }_3^2{\omega }_5-{\omega }_3{\omega }_5^2\right)+r_{Cy}\left.\left(-2{\omega }_5{\alpha }_5+{\omega }_5{\alpha }_3-{\omega }_3{\alpha }_5\right)\right]\hfill \end{array} \\ \begin{array}{l}{\mathcal{C}}_{S5}={\displaystyle \frac{1}{{\omega }_5{\alpha }_5}}\left\{\left[r_{Cx}\left({\omega }_5-{\omega }_3\right)+r_{P3x}{\omega }_3\right]\right.\cdot \left[r_{2x}\left({\alpha }_2-{\alpha }_3\right)\right.+r_{2y}\left({\omega }_3^2-{\omega }_2^2\right)+r_{Cx}\left({\alpha }_3-{\alpha }_5\right)+\hfill \\ +\left.r_{Cy}\left({\omega }_5^2-{\omega }_3^2\right)\right]+\left[r_{Cy}\left({\omega }_5-{\omega }_3\right)+r_{P3y}{\omega }_3\right]\cdot \left[r_{2x}\left({\omega }_3^2-{\omega }_2^2\right)\right.+r_{2y}\left({\alpha }_3-{\alpha }_2\right)+\hfill \\ +r_{Cx}\left({\omega }_5^2-{\omega }_3^2\right)+\left.\left.r_{Cy}\left({\alpha }_5-{\alpha }_3\right)\right]\right\}\hfill \end{array} \end{gathered}$$

(55)

Bresse’s circles of both coupler links AB and CD of the Stephenson III six-bar mechanism have been determined in algebraic form with respect to the fixed frame OXY by using vector algebra. In particular, the inflection circles I₃ and I₅, along with the stationarity circles S₃ and S₅, are given by Equations (40), (50), (44), and (54), respectively.

4. First- and Second-Order Centrodes

In general, the relative planar motion between two rigid bodies can be reproduced by the pure rolling of two curves that take the role of the first-order centrodes, which are traced by the instantaneous centers of rotation on their corresponding planes. If one of the two rigid bodies is fixed, a pair of first-order fixed and moving centrodes can be obtained.

In particular, referring to Figure 3a, the fixed centrode λ₃ is the path traced by the instantaneous center of rotation P₃ of the coupler link 3 with respect to the fixed frame OXY, while the moving centrode l₃ is the path traced by P₃ with respect the moving frame O_3×3y₃ that is attached to AB (link 3).

Thus, the first-order fixed centrode λ₃ of the coupler link 3 has been determined and expressed in Equation (16), while that of the corresponding moving centrode l₃ is given by the following:

$${\mathbf{r}}_{P3}^{*}=r_3{\displaystyle \frac{\sin \left({\theta }_4-{\theta }_3\right)}{\sin \left({\theta }_4-{\theta }_2\right)}}{\left[\begin{array}{cc}\cos \left({\theta }_2-{\theta }_3\right),& \sin \left({\theta }_2-{\theta }_3\right), 0\end{array}\right]}^{ T}$$

(56)

where ${\mathbf{r}}_{P3}^{*}$ is the position vector of P₃ with respect to the moving frame O₃x₃y₃.

Referring to Figure 3b, the second-order fixed centrode ${\overline{\lambda }}_3$ of the coupler link 3 has been determined previously and reported in Equation (21), while that of the second-order moving centrode ${\overline{l}}_3$ is given as follows:

$${\mathbf{r}}_{K3}^{*}=\overline{A{K}_3} {\left[\begin{array}{cc}\cos {\varphi }_3,& -\sin {\varphi }_3\end{array},0\right]}^{ T}$$

(57)

where ${\mathbf{r}}_{K3}^{*}$ is the position vector of K₃ with respect to the moving frame O₃x₃y₃, $\overline{A{K}_3}=\sqrt{{\left(x_{K3}-x_A\right)}^2+{\left(y_{K3}-y_A\right)}^2}$, ${\gamma }_3={\tan }^{-1}\left({\displaystyle \frac{{\alpha }_3}{{\omega }_3^2}}\right)$, and ${\varphi }_3=180°-\left({\gamma }_3+{\theta }_2-{\theta }_3\right)$.

Similarly, referring to Figure 4a, the equation of the fixed centrode λ₅ of the coupler link 5 has been determined in Equation (31), while that of the moving centrode l₅ is given by the following:

$${\mathbf{r}}_{P5}^{*}=r_5{\displaystyle \frac{\sin {\theta }_6}{\sin \left({\theta }_6+{\theta }_5-{\theta }_{P5C}\right)}}{\left[\begin{array}{cc}\cos \left({\theta }_{P5C}-{\theta }_5\right),& \sin \left({\theta }_{P5C}-{\theta }_5\right), 0\end{array}\right]}^{ T}$$

(58)

where ${\mathbf{r}}_{P5}^{*}$ is the position vector of P₅ with respect to the moving frame O₅x₅y₅ and ${\theta }_{P5C}$ is its counterclockwise oriented angle, which is given by ${\theta }_{P5C}={\tan }^{-1}\left({\displaystyle \frac{y_{P5}-y_C}{x_{P5}-x_C}}\right)$.

Referring to Figure 4b, the second-order fixed centrode ${\overline{\lambda }}_5$ of the coupler link 5 has been determined and reported in Equation (34), while that of the second-order moving centrode ${\overline{l}}_5$ takes the following form:

$${\mathbf{r}}_{K5}^{*}=\overline{C{K}_5} {\left[\begin{array}{cc}\cos {\varphi }_5,& \sin {\varphi }_5\end{array}, 0\right]}^{ T}$$

(59)

where ${\mathbf{r}}_{K5}^{*}$ is the position vector of K₅ with respect to the moving frame O₅x₅y₅, $\overline{C{K}_5}=\sqrt{{\left(x_{K5}-x_C\right)}^2+{\left(y_{K5}-y_C\right)}^2}$, ${\theta }_{K5C}={\tan }^{-1}\left({\displaystyle \frac{y_{K5}-y_C}{x_{K5}-x_C}}\right)$, and ${\varphi }_5={\theta }_{K5C}-{\theta }_5$.

Therefore, both pairs of moving centrodes, l₃ and ${\overline{l}}_3$, along with l₅ and ${\overline{l}}_5$, of the coupler links 3 and 5 are now expressed with respect to the fixed frame OXY by using the corresponding homogeneous transformation matrices, the general form of which is as follows:

$${\mathbf{T}}_{0j}=\left[\begin{array}{ccc}\cos {\theta }_j& -\sin {\theta }_j& x_{Oj}\\ \sin {\theta }_j& \cos {\theta }_j& y_{Oj}\\ 0& 0& 1\end{array}\right]$$

(60)

where the subscript j has to be equal to 3 and 5 for the coupler links 3 and 5, respectively.

In fact, each moving centrode is attached to the corresponding coupler link and, thus, it performs a planar rigid-body motion which can be decomposed into a rotation and a translation, as Equation (60) does. Consequently, the equation of each moving centrode is transferred to the fixed frame by multiplying the corresponding homogeneous transformation matrix for Equations (56) and (58) of the first-order moving centrodes and for Equations (57) and (59) of the second-order moving centrodes, respectively.

5. Numerical Examples

The proposed formulation is general and, consequently, it could be implemented in any scientific software, such as Matlab, Maple, and Mathematica, or by using different Mathematical Programming Languages (MPLs), such as Python, C/C++, and Fortran. Nevertheless, considering the extended use of Matlab in many worldwide research centers and universities, as well as the availability of several specific tools for engineering applications and the possibility of coworking with other programming languages, the proposed formulation has been implemented in Matlab 2018 (release b) for validation, analysis, and design purposes by running several significant examples in different configurations.

This is performed as shown in Figure 5 and for the following input data: ω₂ = 1 rad/s, r₁ = 244 mm, r₂ = 81 mm, r₃ = 198 mm, r_3C = 288.9 mm, r₄ = 191 mm, r₅ = 170 mm, r₆ = 180 mm, r₇ = 369 mm, α₀ = 29.32°, and θ₂ = 25°.

Figure 5a shows both pairs of Bresse’s circles, along with the velocity and acceleration vectors of the most significant mechanism points, by including both pairs of instantaneous centers of rotation P₃ and P₅, and the acceleration centers K₃ and K₅.

Figure 5b shows both the first- and second-order fixed and moving centrodes of the coupler link 3, along with the corresponding Bresse’s circles. Figure 5c shows both the first- and second-order fixed and moving centrodes of the coupler link 5, along with the corresponding Bresse’s circles.

The first-order centrodes are tangent to each other at the instantaneous center of rotation according to their pure-rolling motion, and they are also tangent to the corresponding inflection circle at the same point, while the second-order centrodes usually intersect with each other at more than one point and, of course, at the acceleration center.

Moreover, it is necessary to point out that the second-order centrodes are kinematic loci and not geometric loci, as the first-order centrodes, which depend on the kinematic input of the mechanism, that have been chosen in all the proposed examples have a constant driving angular velocity of ω₂ = 1 rad/s.

Similarly, Figure 6, Figure 7 and Figure 8 show the same mechanism in different configurations for the driving crank angle θ₂ = 12.2°, θ₂ = 127.15°, and θ₂ = 64°, respectively. In particular, Figure 6a,b shows a case in which the stationarity circle degenerates into the straight line passing through the diameter of the inflection circle I₃ at the P₃ and K₃ centers, and the line at infinity. Consequently, the acceleration center K₃ coincides with the inflection pole and the second-order centrodes are tangent to each other at this point. Figure 6c shows the first- and second-order centrodes of the coupler link 5 in which the first-order centrodes are tangent to each other, as is typical, while the second-order centrodes intersect each other, including the acceleration center K₅. Likewise, Figure 7 shows the particular case for which the instantaneous center of rotation P₃ goes to infinity and both Bresse’s circles degenerate into two orthogonal lines, while Figure 8 reports the example in which the coupler link 5 is at the position of instantaneous stop, since not only the angular velocity ω₅ is equal to zero, as when the instantaneous center of rotation goes to infinity, but also the velocity of any coupler point vanishes. However, the angular acceleration α₅ is different from zero and the acceleration center K₅ coincides with the instantaneous center of rotation P₅. The stationarity circle degenerates into that point, while the inflection circle degenerates into the straight line passing through the points D₀ and D (link 6) and the line at infinity. Moreover, a practical example regarding the jaw-crusher machine of Figure 9, as based on a Stephenson III six-bar mechanism, is developed and analyzed in both cases of simple dwell of the swing-jaw plate and the additional instantaneous stop of the second coupler link. In particular, this machine consists of the driving link A₀A in the form of a suitable eccentric shaft that is attached to a flywheel, the four-bar linkage A₀ABB₀, and the five-bar linkage A₀ACDD₀, the output link D₀D of which is the swing-aw plate, which crushes the stones against the corresponding fixed-jaw plate, as shown schematically in Figure 9. Thus, the Stephenson III six-bar mechanism of the jaw-crusher machine of Figure 9 is sketched as shown in Figure 10 and Figure 11, which refer to the case of simple dwell of the swing-jaw plate D₀D and that of the additional instantaneous stop of the coupler link CD, respectively.

Consequently, the shape of the ternary coupler link ABC is assumed to have the form of an isosceles or right-angled triangle, respectively. In particular, as reported above in the previous examples, the Bresse circles, along with the velocity and acceleration vector fields, the first- and second-order centrodes of the coupler link 3, and finally, the same for the coupler link 5, are shown in Figure 10a and Figure 11a, Figure 10b and Figure 11b, and Figure 10c and Figure 11c, respectively.

In the case of Figure 10, for θ₂ = 270°, the instantaneous center of rotation P₅ coincides with D and is different from the acceleration center K₅, with the swing-jaw plate having a zero angular velocity and, thus, a simple dwell, while Figure 11 shows the case of the instantaneous stop since both the coupler link CD and the swing-jaw plate D₀D have velocities of zero.

In this particular configuration, for θ₂ = 270°, the inflection circle I₅ degenerates into the straight line passing through the points D₀ and D, and the line at infinity; P₃ coincides with C according to the design method proposed by the authors in [9], and P₅ coincides with K₅, while S₅ degenerates into the same point.

The first- and second-order centrodes give useful information in terms of geometry and kinematics, about the planar motion of both pairs of points P₃ and K₃, along with P₅ and K₅.

6. Conclusions

The formulation of a general algorithm for obtaining the first- and second-order centrodes of both coupler links of Stephenson III six-bar mechanisms, along with the corresponding pairs of Bresse’s circles, has been proposed and validated by means of several significant examples. Particular attention has been devoted to the cases in which the second coupler link shows an instantaneous stop configuration, along with those where the first four-bar linkage reaches an asymptotic configuration. The obtained results can be useful to design and analyze specific mechanisms for practical applications, as well as those for obtaining dwell configurations of the output link or instantaneous stop configurations of the second coupler link. These kinematic performances can be required to design mechanisms for automatic machinery for the manufacturing industry, and a practical example regarding the jaw-crusher machine has been developed and analyzed.