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Abstract

To meet various industrial requirements such as ease of motion, scalability, and cost efficiency, it is necessary to innovate the design of robotic platforms. In this research, a novel approach, from mechanical design to control implementation, is introduced for launching a robotic platform using a parallelogram mechanism. First, a reverse engineering process is applied, progressing from kinematics to dynamics. Then, several mechanical computations are conducted to ensure the structural stability of the robot framework. Subsequently, the dynamic performance of the system is analyzed, focusing on the driving torque and moments in each link. Additionally, the electrical design and transfer function of each joint are identified to ensure practical execution. To validate the effectiveness and feasibility of the design, both numerical simulations and experimental tests are performed. Theoretical results show the dynamic response of the proposed method, particularly in terms of the driving moments of the robotic joints. In real-world tests, various trajectories, such as different rectangular paths, are demonstrated to showcase the robot's capabilities. From these results, it is clear that the proposed approach is both feasible and applicable in practical scenarios.

Keywords

Field robot motion control computational mechanics industrial manipulator pick and place

Introduction

Robotic manipulators have transformed numerous industries by offering precision, efficiency, and versatility across a wide range of tasks.^1–3 Among the various mechanisms employed in robotic arm designs, the parallelogram mechanism stands out as an ingenious and highly efficient solution. This mechanism plays a crucial role in enhancing the dexterity and agility of robotic manipulators, making them indispensable in fields such as manufacturing,⁴ automation,⁵ healthcare,⁶ entertainment,⁷ the beverage industry,⁸ and beyond.

The parallelogram mechanism, also known as a parallel linkage or parallel arm, is a mechanical configuration that consists of interconnected links and joints arranged to replicate the geometry of a parallelogram.^9–11 This setup enables the robotic arm to maintain a constant orientation of its end-effector, or tool while manipulating objects. Its inherent stability and precision make it an ideal choice for tasks requiring high accuracy and consistency.

Typically, the term such as parallelogram mechanism refers to a specific type of mechanism used in robotic manipulation and various engineering applications. It is a four-bar linkage that maintains parallelism between two sides of the mechanism as it moves. The most common application of parallelogram mechanisms is to provide stable and parallel motion for end-effectors or tools in robotics.^12,13 Various types and configurations of parallelogram mechanisms exist, each with distinct advantages and applications. Recently, efforts have been made to develop reconfigured versions suitable for robots operating on diverse terrains.^14,15 The contributions of our work are to (i) indicate the theoretical computations of both forward kinematic and inverse kinematic, physical dimensions and displacement validation in mechanical design, (ii) derive the dynamical analysis of the proposed platform to obtain the tracking performance, and (iii) conduct the experimental platform for laboratory robot in several validations.

Related works

A parallelogram mechanism is a mechanical linkage system composed of four rigid members connected to form a closed loop with two pairs of parallel sides. This mechanism is widely used in various engineering applications to create stable, predictable motion or to maintain a consistent orientation of an object. The internal structure of a parallelogram mechanism typically includes the following components: four rigid links, joints or hinges, an actuator, a support structure, a control system, bearings and bushings, linkage design elements, and safety features.^16,17

Four rigid links: The core of the parallelogram mechanism consists of four rigid links or bars. These links are connected at their endpoints to form the closed loop. Two pairs of opposing links are parallel to each other, and their relative lengths and angles determine the behavior of the mechanism.

Joints or hinges: At the endpoints of each rigid link, there are hinge or joint connections. These joints allow the links to pivot or rotate relative to each other, enabling the mechanism to change shape and adapt to different positions or orientations.

Actuators: Parallelogram mechanisms may include actuators such as motors or pistons that are connected to one or more of the links. These actuators provide the necessary force or motion to drive the mechanism and control its movement. The actuators can be positioned at various points in the mechanism, depending on the specific application.

Support structure: In many applications, the parallelogram mechanism is mounted on a support structure or base. This base provides stability and often serves as a reference point for the mechanism's operation.

Control system: Depending on the complexity of the application, a control system may be integrated into the mechanism. This system can include sensors, feedback loops, and controllers to monitor and regulate the motion of the mechanism, ensuring that it operates accurately and precisely.

Bearings and bushings: Bearings and bushings are often used at the joint connections to reduce friction and allow for smooth rotation of the links. These components help improve the efficiency and longevity of the mechanism.

Linkage design: The lengths and angles of the four links are carefully designed to achieve specific types of motion or maintain a particular orientation. The precise design depends on the intended function of the mechanism, such as maintaining a parallel relationship between two objects or creating a specific path of motion.

Safety features: In some applications, safety features such as limit switches, emergency stop mechanisms, or mechanical locks may be incorporated to ensure the safe operation of the parallelogram mechanism and prevent accidents.

The inside structure of a parallelogram mechanism can vary significantly depending on its intended purpose, size, and complexity. Engineers and designers carefully select and configure the components to meet the specific requirements of the application, whether it is for industrial automation, robotics, or any other field where controlled motion or orientation is essential (Table 1).

Table 1.

Summary of the state-of-the-art in related fields.

Application	Publication year	Author(s)	Technique	Advantage(s)	Disadvantage(s)
Calibration	2015	Guo et al.¹⁸	The multilevel calibration technique was developed to improve the absolute accuracy of the manipulator with a parallelogram mechanism.	It is easy to implement and considerably effective	The effect of payload and thermal factor were not taken into account
Calibration	2019	Peng et al.⁹	Kinematic calibration was investigated to enhance the absolute accuracy of the machining system with a parallelogram mechanism.	The computational effectiveness is higher while it maintains almost the same accuracy.	Some nonlinear factors such as the dynamic deformation error, and the cutting force might deteriorate the overall machining accuracy
Exoskeleton	2018	Christensenand Bai¹⁹	A double parallelogram linkage to imitate the motion as a spherical joint was invented	This design gained the motion free of singularity and good manipulability properties.	The drawbacks of a rotating parallelogram stem from its single point of connection and limited stability enhancement options, primarily relying on a larger single connection point.
	2023	Singh and Khatait²⁰	The adjustment concept of remote center-of-motion and a flexure-based compliant mechanism was proposed to minimize these deviations.	It offers a more simplified and economical design.	Advanced equipment for instance laser micromachining and rapid prototyping must be involved
	2019	Bouffard²¹	A three DoFs (degree-of-freedom) exoskeleton with a modified double parallelogram mechanism was innovated.	This device is suitable for any user regardless of biological characters	It should be mounted to a backpack and an arm cuff should be linked to the user's hand
Parking	2019	Zhang et al.²²	A novel compact double-stack parking system with a 1-DoF cam-linkage mechanism aims to solve parking issues, offering space efficiency.	This design enables efficient stack parking without the need to remove ground vehicles, improving space utilization and quickly addressing parking issues in residential areas, campuses, and streets.	The passage does not provide information about any specific disadvantages or drawbacks of the described actuator or parking system. It primarily highlights the advantages and features of the system without mentioning any potential limitations or challenges.
Parking	2021	Fong et al.¹³	The double parallelogram lifting mechanism (DPLM) was studied to obtain static equilibrium at any height and minimize the required torque exertion of motor.	It has a compact and stable multi-linkage design, featuring a low retracted height and a significant working height.	The changes in linkages’ centers of mass which affect DPLM moment balance, tilting effects, and multi-spring optimization, were not considered.

The proposed approach

In this approach, a reverse concept in robotic design is employed to explore advantageous features such as task-centric design, efficiency, and customization. First, both forward and inverse kinematics are applied to determine system parameters, such as the Denavit-Hartenberg (D-H) table. Then, theoretical errors are assessed to predict potential deviations in real-world applications. Second, the mechanical structure of the entire robot is completed to create a functional platform.

Forward kinematic

The structure of this robot is illustrated in Figure 1. Local coordinate systems are assigned to each link, and geometric parameters are introduced to evaluate the shape and length of the robot. These parameters are summarized in Table 2.

Figure 1.

Demonstration of the geometrical parameters in the computation of forward kinematics.

Table 2.

List of Denavit-Hartenberg (D-H) parameters for the proposed design.

Link	$α_{i}$ (degree)	$a$ (mm)	$d$ (mm)	$θ$
1	90	0	$d_{i}$	$θ_{1}$
2	0	$L_{2}$	0	$θ_{2}$
3	0	$L_{3}$	0	$θ_{3}$
4	0	$L_{3}$	0	$θ_{3} + θ_{2}$

In general, matrix transformation from link i to i − 1 is

\begin{aligned} _{i}^{i - 1} T = D_{z, d} \cdot R_{z, θ} \cdot D_{x, a} \cdot R_{x, α} = & [\begin{matrix} \cos (θ_{i}) & - \sin (θ_{i}) \cdot \cos (α_{i}) \\ \sin (θ_{i}) & \cos (θ_{i}) \cdot \cos (α_{i}) \\ 0 & \sin (α_{i}) \\ 0 & 0 \end{matrix} \\ \times & \begin{matrix} \sin (θ_{i}) \cdot \sin (α_{i}) & a_{i} \cdot \cos (α_{i}) \\ - \cos (θ_{i}) \cdot \sin (α_{i}) & a_{i} \cdot \sin (α_{i}) \\ \cos (α_{i}) & d_{i} \\ 0 & 1 \end{matrix}] \end{aligned}

(1)

Hence, from links 0 to 4, matrix transformation is

_{1}^{0} T = [\begin{matrix} \cos (θ_{1}) & 0 & \sin (θ_{1}) & 0 \\ \sin (θ_{1}) & 0 & - \cos (θ_{1}) & 0 \\ 0 & 1 & 0 & d \\ 0 & 0 & 0 & 1 \end{matrix}]

(2)

_{2}^{1} T = [\begin{matrix} \cos (θ_{2}) & - \sin (θ_{2}) & 0 & L_{2} \cos (θ_{2}) \\ \sin (θ_{2}) & \cos (θ_{2}) & 0 & L_{2} \sin (θ_{2}) \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}]

(3)

_{3}^{2} T = [\begin{matrix} \cos (θ_{3}) & - \sin (θ_{3}) & 0 & L_{3} \cos (θ_{3}) \\ \sin (θ_{3}) & \cos (θ_{3}) & 0 & L_{3} \sin (θ_{3}) \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}]

(4)

_{4}^{3} T = [\begin{matrix} \cos (θ_{2} + θ_{3}) & - \sin (θ_{2} + θ_{3}) \\ \sin (θ_{2} + θ_{3}) & \cos (θ_{2} + θ_{3}) \\ 0 & 0 \\ 0 & 0 \end{matrix} \begin{matrix} \begin{matrix} 0 & L_{4} \cos (θ_{2} + θ_{3}) \\ 0 & - L_{4} \sin (θ_{2} + θ_{3}) \\ 1 & 0 \\ 0 & 1 \end{matrix}] \end{matrix}

(5)

or,

_{4}^{0} T =_{1}^{0} T \cdot_{2}^{1} T \cdot_{3}^{2} T \cdot_{4}^{3} T

(6)

_{4}^{0} T = [\begin{matrix} c_{1} (c_{23}^{2} - s_{23}^{2}) & - 2 c_{1} c_{23} s_{23} \\ s_{1} (c_{23}^{2} - s_{23}^{2}) & - 2 s_{1} c_{23} s_{23} \\ 2 c_{23} s_{23} & (c_{23}^{2} - s_{23}^{2}) \\ 0 & 0 \end{matrix} = \begin{matrix} s_{1} & L_{2} c_{1} c_{2} + L_{3} c_{1} c_{23} + L_{4} c_{1} (c_{23}^{2} + s_{23}^{2}) \\ - c_{1} & L_{2} s_{1} c_{2} + L_{3} s_{1} c_{23} + L_{4} s_{1} (c_{23}^{2} + s_{23}^{2}) \\ 0 & d + L_{23} s_{23} + L_{2} s_{2} \\ 0 & 1 \end{matrix}]

where

c_{1} = \cos (θ_{1}) \to c_{2} = \cos (θ_{2})

;

c_{23} = \cos (θ_{2} + θ_{3}) \to s_{1} = \sin (θ_{1})

;

s_{2} = \sin

(θ_{2}) \to s_{23} = \sin (θ_{2} + θ_{3})

Therefore, the location of the end-effector is computed as follows:

p = [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} L_{2} c_{1} c_{2} + L_{3} c_{1} c_{23} + L_{4} c_{1} (c_{23}^{2} + s_{23}^{2}) \\ L_{2} s_{1} c_{2} + L_{3} s_{1} c_{23} + L_{4} s_{1} (c_{23}^{2} + s_{23}^{2}) \\ d + L_{3} s_{23} + L_{2} s_{2} \end{matrix}]

(7)

We get

f (θ_{1}, θ_{2}, θ_{3}) = L_{2} c_{1} c_{2} + L_{3} c_{1} c_{23} + L_{4} c_{1} (c_{23}^{2} + s_{23}^{2})

(8)

h (θ_{1}, θ_{2}, θ_{3}) = L_{2} s_{1} c_{2} + L_{3} s_{1} c_{23} + L_{4} s_{1} (c_{23}^{2} + s_{23}^{2}) g (θ_{1}, θ_{2}, θ_{3})

(9)

= d + L_{3} s_{23} + L_{2} s_{2}

(10)

Then, the matrix Jacobian is

J = [\begin{matrix} \frac{δ f (θ_{1}, θ_{2}, θ_{3})}{δ θ_{1}} & \frac{δ f (θ_{1}, θ_{2}, θ_{3})}{δ θ_{2}} & \frac{δ f (θ_{1}, θ_{2}, θ_{3})}{δ θ_{3}} \\ \frac{δ h (θ_{1}, θ_{2}, θ_{3})}{δ θ_{1}} & \frac{δ h (θ_{1}, θ_{2}, θ_{3})}{δ θ_{2}} & \frac{δ h (θ_{1}, θ_{2}, θ_{3})}{δ θ_{3}} \\ \frac{δ g (θ_{1}, θ_{2}, θ_{3})}{δ θ_{1}} & \frac{δ g (θ_{1}, θ_{2}, θ_{3})}{δ θ_{2}} & \frac{δ g (θ_{1}, θ_{2}, θ_{3})}{δ θ_{3}} \end{matrix}]

(11)

J = [\begin{matrix} J_{11} & J_{12} & J_{13} \\ J_{21} & J_{22} & J_{23} \\ J_{31} & J_{32} & J_{33} \end{matrix}]

with

J_{11} = - s_{1} (L_{2} c_{2} + L_{4} (c_{23}^{2} + s_{23}^{2}) + L_{3} c_{23})

;

J_{12} = - c_{1} (L_{2} s_{2} + L_{3} s_{23})

;

J_{13} = - L_{3} c_{1} s_{23}

;

J_{21} = c_{1} (L_{2} c_{2} + L_{4} (c_{23}^{2} + s_{23}^{2}) + L_{3} c_{23})

;

J_{22} = - s_{1} (L_{2} s_{2} + L_{3} s_{23})

;

J_{23} = - L_{3} s_{1} s_{23};

J_{31} = 0;

J_{32} = L_{3} c_{23} + L_{2} c_{2}

;

J_{33} = L_{3} c_{23} .

Note that the end-effector velocity is related to the joint velocities through an expression

\dot{p} = J \cdot \dot{θ} => \dot{θ} = J^{- 1} \cdot v

(12)

where v is the velocity of the end-effector position in the x, y, z coordinates.

We have a relationship between the end-effector acceleration of the robot and the joint acceleration velocities through a formula

\ddot{p} = \dot{J} \dot{θ} + J \ddot{θ}

(13)

With

\begin{aligned} \dot{J} = [\begin{matrix} \frac{δ J_{11}}{δ θ_{1}} {\dot{θ}}_{1} + \frac{δ J_{11}}{δ θ_{2}} {\dot{θ}}_{2} + \frac{δ J_{11}}{δ θ_{3}} {\dot{θ}}_{3} & \frac{δ J_{12}}{δ θ_{1}} {\dot{θ}}_{1} + \frac{δ J_{12}}{δ θ_{2}} {\dot{θ}}_{2} + \frac{δ J_{12}}{δ θ_{3}} {\dot{θ}}_{3} \\ \frac{δ J_{21}}{δ θ_{1}} {\dot{θ}}_{1} + \frac{δ J_{21}}{δ θ_{2}} {\dot{θ}}_{2} + \frac{δ J_{21}}{δ θ_{3}} {\dot{θ}}_{3} & \frac{δ J_{22}}{δ θ_{1}} {\dot{θ}}_{1} + \frac{δ J_{22}}{δ θ_{2}} {\dot{θ}}_{2} + \frac{δ J_{22}}{δ θ_{3}} {\dot{θ}}_{3} \\ \frac{δ J_{31}}{δ θ_{1}} {\dot{θ}}_{1} + \frac{δ J_{31}}{δ θ_{2}} {\dot{θ}}_{2} + \frac{δ J_{31}}{δ θ_{3}} {\dot{θ}}_{3} & \frac{δ J_{32}}{δ θ_{1}} {\dot{θ}}_{1} + \frac{δ J_{32}}{δ θ_{2}} {\dot{θ}}_{2} + \frac{δ J_{32}}{δ θ_{3}} {\dot{θ}}_{3} \\ 0 & 0 \end{matrix} \\ \times \begin{matrix} s_{1} & \frac{δ J_{13}}{δ θ_{1}} {\dot{θ}}_{1} + \frac{δ J_{13}}{δ θ_{2}} {\dot{θ}}_{2} + \frac{δ J_{13}}{δ θ_{3}} {\dot{θ}}_{3} \\ \begin{matrix} - c_{1} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} & \begin{matrix} \frac{δ J_{23}}{δ θ_{1}} {\dot{θ}}_{1} + \frac{δ J_{23}}{δ θ_{2}} {\dot{θ}}_{2} + \frac{δ J_{23}}{δ θ_{3}} {\dot{θ}}_{3} \\ \begin{matrix} \frac{δ J_{33}}{δ θ_{1}} {\dot{θ}}_{1} + \frac{δ J_{33}}{δ θ_{2}} {\dot{θ}}_{2} + \frac{δ J_{33}}{δ θ_{3}} {\dot{θ}}_{3} \\ 1 \end{matrix} \end{matrix} \end{matrix}] \end{aligned}

Two formulas (12) and (13) would be used to simulate the calculation of velocities and accelerations of the joints when it comes to the end effector.

Inverse kinematic

The meaning of the reverse kinematics problem is to determine the values of the variables that match the given position and orientation of the end-effector. In the context of the palletizing robot, it is responsible for maintaining the orientation of the end-effector. Thus, the computational process would not be overly complex compared to robots that have to control the orientation of the end-effector. The use of geometric methods as Figure 2 is employed to find the joint angles with the desired point D values.

Figure 2.

Demonstration of the geometrical parameters in the computation of reversed kinematics.

We assume that the location of the desired point D is [ $x_{d}$ , $y_{d}$ , $z_{d}$ ]

{\begin{matrix} x_{D} = \sqrt{x_{d}^{2} + y_{d}^{2}} \\ AG = x_{c} = x_{D} - L_{4} \\ CG = z_{D} - d \\ z_{D} = z_{d} \end{matrix}

(14)

In the

ACG

triangle, we have

{\begin{matrix} \begin{matrix} \hat{CAG} = \arctan (\frac{CG}{AG}) \\ = \arctan (\frac{z_{D} - d}{x_{D} - L_{4}}) (with AG, CG > 0) \end{matrix} \\ \begin{matrix} AC = \sqrt{{AG}^{2} + {CG}^{2}} \\ = \sqrt{{(x_{D} - L_{4})}^{2} + {(z_{D} - d)}^{2}} \end{matrix} \end{matrix}

(15)

Similarly, in the

ABC

triangle, we also have

{BC}^{2} = {AC}^{2} + {AB}^{2} - 2 \cdot AB \cdot AC \cdot \cos (\hat{BAC})

(16)

\cos (\hat{BAC}) = \frac{{AC}^{2} + {AB}^{2} - {BC}^{2}}{2 \cdot AB \cdot AC} = \frac{{(x_{D} - L_{4})}^{2} + {(z_{D} - d)}^{2} + L_{2}^{2} - L_{3}^{2}}{2 \cdot L_{2} \cdot \sqrt{{(x_{D} - L_{4})}^{2} + {(z_{D} - d)}^{2}}}

(17)

\hat{BAC} = arcos (\frac{{(x_{D} - L_{4})}^{2} + {(z_{D} - d)}^{2} + L_{2}^{2} - L_{3}^{2}}{2 \cdot L_{2} \cdot \sqrt{{(x_{D} - L_{4})}^{2} + {(z_{D} - d)}^{2}}})

(18)

From (15) to (18),

θ_{2}

could be computed as follows:

θ_{2} = \hat{CAG} + \hat{BAC} = \arctan (\frac{z_{D} - d}{x_{D} - L_{4}}) arcos (\frac{{(x_{D} - L_{4})}^{2} + {(z_{D} - d)}^{2} + L_{2}^{2} - L_{3}^{2}}{2 \cdot L_{2} \cdot \sqrt{{(x_{D} - L_{4})}^{2} + {(z_{D} - d)}^{2}}})

(19)

Consequently,

{AC}^{2} = {AB}^{2} + {BC}^{2} - 2 \cdot AB \cdot BC \cdot \cos (\hat{ABC})

(20)

\cos (\hat{ABC}) = \frac{{AB}^{2} + {BC}^{2} - {AC}^{2}}{2 \cdot AB \cdot BC} = = \frac{L_{2}^{2} + L_{3}^{2} - {(x_{D} - L_{4})}^{2} - {(z_{D} - d)}^{2}}{2 \cdot L_{2} \cdot L_{3}}

(21)

\hat{ABC} = arcos (\frac{L_{2}^{2} + L_{3}^{2} - {(x_{D} - L_{4})}^{2} - {(z_{D} - d)}^{2}}{2 \cdot L_{2} \cdot L_{3}})

(22)

The value of

θ_{3}

θ_{3} = 180 \circ - \hat{ABC} = 180 \circ + arcos (\frac{L_{2}^{2} + L_{3}^{2} - {(x_{D} - L_{4})}^{2} - {(z_{D} - d)}^{2}}{2 \cdot L_{2} \cdot L_{3}})

(23)

And,

θ_{1} = \arctan (\frac{y_{d}}{x_{d}})

(24)

Owing to the matrix Jacobian, the system error in the Descartes coordinate if it is assumed to derive from the positioning sensor, is

d p = J \cdot d θ

(25)

where

d θ

is the rotational angle in each step.

To evaluate the system error, with these inputs consisting of $θ_{2}$ varying from 10 to 90, $θ_{3}$ varying from 0 to 70, it would cause several results as shown in Figure 3.

Figure 3.

Simulation of the system error in (a) x-axis, (b) y-axis, and (c) z-axis.

Mechanical design

In Figure 4, the working range of the robot arm is determined by two parameters, namely $L_{2}$ and $L_{3}$ . The desired reach of the robot arm is 450 mm, which means in an ideal scenario $L = L_{2} + L_{3} = 450$ mm. According to our experience, the optimal ratio r falls within the range of 0.5–0.6, where $r = L_{2} / L$ . Based on the requirements of the task, we choose $L_{2} = 250$ mm, $L_{3} = 230$ mm, and $L_{4} = 20$ mm. In this case, the ratio $r = 0.52$ falls within the necessary optimal range for the design.

Figure 4.

Theoretical diagram for the design parameters of our robot.

With those design parameters, each mechanical component is analyzed as shown in Figure 5 using the ANSYS software tool. The working conditions of the proposed design should be validated by the displacements of robotic links. All links have a displacement of < 0.3 mm when operating at the maximum load as per the specified theory of the assignment. Henceforth, our selection of mechanical parameters ensures the durability of whole links.

Figure 5.

Theoretical analysis of the displacement using ANSYS software for (a) link 1, (b) link 2, (c) link 3, (d) link 7, and (e) link 5.

The parameter values such as $a_{1}$ , $a_{2}$ , $a_{3}$ , $β$ , and $γ$ do not participate in the design of the robot's workspace, but they are values used to design the parallel structure that drives the third link and fourth link of the robot. In Table 3, these values are chosen to be compatible with the compact design of the robot.

Table 3.

List of the system parameters in the Denavit-Hartenberg (D-H) matrix.

$d$	$L_{2}$	$L_{3}$	$L_{4}$	$a_{1}$	$a_{2}$	$a_{3}$	$β$	$γ$
$90$ mm	$250$ mm	$230$ mm	$20$ mm	$76$ mm	$80$ mm	$60$ mm	$30 \circ$	$11 \circ$

Dynamical analysis

Based on the design from the previous sections, our goal is to determine the necessary forces to drive the manipulator. Since the driving motors for each link must be specified, more appropriate specifications are selected based on the results of this section. First, the dynamic equations relating to the existing forces and moments are identified. Then, relevant mathematical terms from these equations are computed. As a result, the required moments at links 1, 2, and 3 can be determined. In the analysis of dynamics, several commonly applied methods are used, including the Lagrange method,²³ Gauss method,²⁴ Kane method,²⁵ and Robertson-Weiden Fort method,²⁶ among others. In this study, the Newton-Euler method²⁷ is utilized. This method combines Newton's force equilibrium with Euler's moment balance to calculate the required torque for moving components at the desired velocity and acceleration. We have,

F = \sum f_{i} = m_{c} a = m_{c} {\dot{v}}_{c}

(26)

with

{\dot{v}}_{c}

is the linear acceleration of the center point C;

\sum f_{i}

is the total force due to the connections of the components (joints);

m_{c}

is the mass of the link

N = \sum n_{i} =_{}^{C} I {\dot{ω}}_{c} + ω_{c} \times_{}^{C} I ω_{c}

(27)

with

{\dot{ω}}_{c}

is the angular acceleration of the center point C;

\sum n_{i}

is the total moment of links (joints);

_{}^{C} I

is the inertial moment of the center point

C_{I} = [\begin{matrix} I_{x x} & - I_{x y} & - I_{x z} \\ - I_{x y} & I_{y y} & - I_{y z} \\ - I_{x z} & - I_{y z} & I_{z z} \end{matrix}]

We have

_{}^{i} ω_{i + 1} =_{}^{i} ω_{i} +_{i + 1}^{i} R \cdot {\dot{θ}}_{i + 1}_{}^{i + 1} {\hat{Z}}_{i + 1}

(28)

with

_{}^{i} ω_{i + 1}

is the angular velocity of link

(i + 1) th

in the coordinate

i th

;

_{i + 1}^{i} R

is the rotational matrix from coordinate

(i + 1) th

i th

{\begin{matrix} _{}^{i} f_{i} =_{}^{i} F_{i} -_{i + 1}^{i} R m_{i} g +_{i + 1}^{i} R_{}^{i + 1} f_{i + 1} \\ _{}^{i} n_{i} =_{}^{i} N_{i} +_{i + 1}^{i} R_{}^{i + 1} n_{i + 1} \\ +_{}^{i} P_{i + 1} \times_{i + 1}^{i} R_{}^{i + 1} f_{i + 1} + P_{C i} \times (_{}^{i} F_{i} -_{i + 1}^{i} R m_{i} g) \\ ζ_{i} =_{}^{i} n_{i}^{T}_{}^{i} {\hat{Z}}_{i} \end{matrix}

(29)

where

_{}^{i} F_{i}

is the acting force of link

i th

in the coordinate

i th

;

_{}^{i} N_{i}

is the inertial moment of link

i th

in the coordinate

i th

;

ζ_{i}

is the torque in the z-axis in order to rotate link

i th

Typically, for open-loop or serial robots with conventional transmission systems, the system can be easily solved because each component experiences two forces. At the end-effector, there is only one force or load, as specified by the problem, resulting in a unique solution for the dynamic analysis. However, the core issue arises when the robot incorporates two parallel mechanisms. Some components, such as the ninth, seventh, and fifth, experience more than two forces and moments. To solve the dynamic analysis of a closed system, it is necessary to reduce the number of forces acting on these complex components.

According to the equations of motion, the system's dynamics depend solely on its motion. Therefore, if the Newton-Euler approach can reduce the excess forces while maintaining the overall motion of the robot, the results from the Newton-Euler and Lagrange methods would remain identical. Simplification or elimination of forces and moments occurs during the calculation process.

As shown in Figure 6, local coordinates are assigned to each link as follows: coordinate 1 is linked to link 1; coordinate 2 to links 2, 4, and 6; coordinate 3 to links 3, 5, and 8; and coordinate 4 to links 7 and 9. Next, we consider link 9 as described below:

{\begin{matrix} _{}^{9} f_{5} =_{}^{9} F_{9} -_{}^{9} f_{8} -_{0}^{9} R m_{9} g \\ _{}^{9} n_{5} =_{}^{9} N_{9} -_{}^{9} n_{8} -_{}^{9} P_{8} \times_{}^{9} f_{8} + P_{C 9} \times (_{}^{9} F_{9} -_{0}^{9} R m_{9} g) \end{matrix}

(30)

With

_{}^{i} f_{j}

is the force of link

j th

affects link

i th

in the coordinate of link

i th

;

_{}^{i} n_{j}

is the moment of link

j th

affects link

i th

in the coordinate of link

i th

;

_{}^{i} F_{i}

is the inertial force of link

i th

in the coordinate of link

i th

;

_{}^{i} N_{i}

is the inertial moment of link

i th

in the coordinate of link

i th

;

_{}^{i} P_{j}

is the position that acting force

_{}^{i} f_{j}

in the coordinate

i th

;

P_{C i}

is the position of center point of link

i th

in the coordinate

i th .

Figure 6.

Theoretical diagram of local coordinates for dynamic analysis.

To simplify, we proceed to classify the moment $_{}^{9} n_{8}$ acting on links 9 and 8. The relative force between links 9 and 5 is $_{}^{9} f_{5}$ , which represents the force $F_{z}$ in the z-direction.

The total force between two joints related to links 8 and 5 is

_{}^{9} f_{8 + 5} =_{}^{9} F_{9} -_{0}^{9} R m_{9} g

(31)

Force on the z-axis is

_{}^{9} f_{8 + 5 Z} = [\begin{matrix} 0 & 0 & 1 \end{matrix}]_{}^{9} f_{8 + 5} \leftrightarrow_{}^{8} f_{9 z} = -_{}^{9} f_{8 + 5 Z}

(32)

On link 8, an inertial moment in the z-direction is

_{}^{8} n_{Z} = [\begin{matrix} 0 & 0 & 1 \end{matrix}] (_{}^{8} N_{8} + P_{C 8} \times (_{}^{8} F_{8} -_{0}^{8} R m_{8} g)) \leftrightarrow_{}^{8} f_{9 X} = \frac{_{}^{8} n_{Z}}{_{}^{8} P_{9 Y}}

(33)

Also, on link 9, the inertial moment in the z-direction is

_{}^{9} n_{Z} = [\begin{matrix} 0 & 0 & 1 \end{matrix}] [_{}^{9} N_{9} + P_{C 9} \times (_{}^{9} F_{9} -_{0}^{9} R m_{9} g)]

As well, we have

_{}^{9} n_{Z} = -_{}^{8} (_{}^{9} P_{8}) \times_{}^{8} f_{9}

; with

_{}^{8} (_{}^{9} P_{8}) =_{9}^{8} R_{}^{9} P_{8}

.Thus

\begin{aligned} _{}^{9} n_{Z} = & - {([\begin{matrix} p x \\ p y \\ p z \end{matrix}] \times_{}^{8} f_{9})}^{T}_{}^{9} {\hat{Z}}_{9} = - (p x_{}^{8} f_{9 Y} - p y_{}^{8} f_{9 X}) \\ \leftrightarrow_{}^{8} f_{8 Y} = \frac{_{}^{9} n_{Z} + p x_{}^{8} f_{9 X}}{p x} \end{aligned}

(34)

The relative force between links 9 and 8 has been solved

_{}^{8} f_{9} = [\begin{matrix} _{}^{8} f_{9 X} \\ _{}^{8} f_{9 Y} \\ _{}^{8} f_{9 Z} \end{matrix}]

(35)

Using the Newton-Euler method, the solution for link 9 is

{\begin{matrix} _{}^{9} f_{5} =_{}^{9} F_{9} +_{8}^{9} R_{}^{8} f_{9} -_{0}^{9} R m_{9} g \\ _{}^{9} n_{5} =_{}^{9} N_{9} -_{}^{9} P_{8} \times_{0}^{9} R_{}^{8} f_{9} \\ + P_{C 9} \times (_{}^{9} F_{9} -_{0}^{9} R m_{9} g) \end{matrix}

(36)

Similarly, the solution for link 8 is

{\begin{matrix} _{}^{8} f_{7} =_{}^{8} F_{8} -_{}^{8} f_{9} -_{0}^{8} R m_{8} g \\ _{}^{8} n_{7} =_{}^{8} N_{8} -_{}^{8} P_{9} \times_{}^{8} f_{9} + P_{C 8} \times (_{}^{8} F_{8} -_{0}^{8} R m_{8} g) \end{matrix}

(37)

On link 7, we obtain

\begin{aligned} {\begin{matrix} _{}^{7} f_{2} =_{}^{7} F_{7} +_{8}^{7} R_{}^{8} f_{7} -_{0}^{7} R m_{7} g +_{6}^{7} R_{}^{6} f_{7} \\ _{}^{7} n_{2} =_{}^{7} N_{7} -_{}^{7} n_{6} +_{8}^{7} R_{}^{8} n_{7} +_{}^{7} P_{6} \times_{6}^{7} R_{}^{6} f_{7} +_{}^{7} P_{8} \times_{8}^{7} R_{}^{8} f_{7} \\ {+ P_{C 9} \times (_{}^{7} F_{7} -_{0}^{7} R m_{7} g) \end{matrix} \end{aligned}

(38)

Correspondingly, we continue proceeding to classify the moment

_{}^{7} n_{6}

acting on links 6 and 7. The relative force between links 6 and 7 is

_{}^{9} f_{5}

, which represents the force

F_{z}

in the z-direction.

The total force between two joints related to links 2 and 6 is

_{}^{7} f_{2 + 6} =_{}^{7} F_{7} -_{0}^{9} R m_{9} g +_{8}^{7} R_{}^{8} f_{7}

(39)

Hence in the z-axis

_{}^{7} f_{2 + 6 Z} = [\begin{matrix} 0 & 0 & 1 \end{matrix}]_{}^{7} f_{2 + 6} \leftrightarrow_{}^{6} f_{7 z} = -_{}^{7} f_{2 + 6 Z}

(40)

On link 6, the inertial moment in the z-direction acting on link 7 is

_{}^{6} n_{Z} = [\begin{matrix} 0 & 0 & 1 \end{matrix}] (_{}^{6} N_{6} + P_{C 6} \times (_{}^{6} F_{6} -_{0}^{6} R m_{6} g)) \leftrightarrow_{}^{6} f_{7 Y} = \frac{_{}^{6} n_{Z}}{_{}^{6} P_{7 x}}

(41)

In the view of link 7, the inertial moment in the z-direction acting on link 6 is

_{}^{7} n_{z} = [\begin{matrix} 0 & 0 & 1 \end{matrix}] (_{}^{7} N_{7} +_{8}^{7} R_{}^{8} n_{7} +_{}^{7} P_{8} \times_{8}^{7} R_{}^{8} f_{7} + P_{C 7} \times (_{}^{7} F_{7} -_{0}^{7} R m_{7} g))

Likewise, we reach $_{}^{7} n_{Z} = -_{}^{6} (_{}^{7} P_{6}) \times_{}^{6} f_{7}$ with $_{}^{6} (_{}^{7} P_{6}) =_{7}^{6} R_{}^{7} P_{6}$ .Therefore

\begin{aligned} _{}^{7} n_{Z} = & - {([\begin{matrix} p x \\ p y \\ p z \end{matrix}] \times_{}^{6} f_{7})}^{T}_{}^{7} {\hat{Z}}_{7} = - (p x_{}^{6} f_{7 Y} - p y_{}^{6} f_{7 X}) \\ \leftrightarrow_{}^{6} f_{7 X} = \frac{_{}^{7} n_{Z} + p x_{}^{6} f_{7 Y}}{p y} \end{aligned}

(42)

We achieve

_{}^{6} f_{7} = [\begin{matrix} _{}^{6} f_{7 X} \\ _{}^{6} f_{7 Y} \\ _{}^{6} f_{7 Z} \end{matrix}]

(43)

For link 7

{\begin{matrix} _{}^{7} f_{2} =_{}^{7} F_{7} +_{8}^{7} R_{}^{8} f_{7} -_{0}^{7} R m_{7} g +_{6}^{7} R_{}^{6} f_{7} \\ _{}^{7} n_{2} =_{}^{7} N_{7} +_{8}^{7} R_{}^{8} n_{7} -_{}^{7} P_{6} \times_{6}^{7} R_{}^{6} f_{7} +_{}^{7} P_{8} \times_{8}^{7} R_{}^{8} f_{7} \end{matrix} {+ P_{C 9} \times (_{}^{7} F_{7} -_{0}^{7} R m_{7} g)

(44)

For link 6

{\begin{matrix} _{}^{6} f_{1} =_{}^{6} F_{6} -_{}^{6} f_{7} -_{0}^{6} R m_{6} g \\ _{}^{6} n_{1} =_{}^{6} N_{6} -_{}^{6} P_{7} \times_{}^{6} f_{7} + P_{C 8} \times (_{}^{8} F_{8} -_{0}^{8} R m_{8} g) \end{matrix}

(45)

For link 5

\begin{aligned} {\begin{matrix} _{}^{5} f_{2} =_{}^{5} F_{5} +_{9}^{5} R_{}^{9} f_{5} -_{0}^{5} R m_{5} g +_{4}^{5} R_{}^{4} f_{5} \\ _{}^{5} n_{2} =_{}^{5} N_{5} -_{}^{5} n_{4} +_{9}^{5} R_{}^{9} n_{5} +_{}^{5} P_{4} \times_{4}^{5} R_{}^{5} f_{4} +_{}^{5} P_{9} \times_{9}^{5} R_{}^{9} f_{5} \\ {+ P_{C 5} \times (_{}^{5} F_{5} -_{0}^{5} R m_{5} g) \end{matrix} \end{aligned}

(46)

In the same way, the moment

_{}^{5} n_{4}

acting on links 4 and 5. The relative force between links 2 and 5 is

_{}^{5} f_{2}

, which represents the force

F_{z}

in the z-direction.

The total force between two joints related to links 2 and 4 is

_{}^{5} f_{2 + 4} =_{}^{5} F_{5} -_{0}^{5} R m_{5} g +_{9}^{5} R_{}^{9} f_{5}

(47)

Then,

_{}^{5} f_{2 + 4 Z} = [\begin{matrix} 0 & 0 & 1 \end{matrix}]_{}^{5} f_{2 + 4} \leftrightarrow_{}^{4} f_{5 z} = -_{}^{5} f_{2 + 4 Z}

(48)

For link 4

_{}^{4} n_{Z} = [\begin{matrix} 0 & 0 & 1 \end{matrix}] [_{}^{4} N_{4} + P_{C 4} \times (_{}^{4} F_{4} -_{0}^{4} R m_{4} g)] \leftrightarrow_{}^{4} f_{5 Y} = \frac{_{}^{4} n_{Z}}{_{}^{4} P_{5 x}}

(49)

For link 5

_{}^{5} n_{Z} = [\begin{matrix} 0 & 0 & 1 \end{matrix}] .

(_{}^{5} N_{5} + +_{9}^{5} R_{}^{9} n_{5} +_{}^{5} P_{9} \times_{9}^{5} R_{}^{9} f_{5} + P_{C 5} \times (_{}^{5} F_{5} -_{0}^{5} R m_{5} g))

It is noted

_{}^{5} n_{Z} = -_{}^{4} (_{}^{5} P_{4}) \times_{}^{4} f_{5}

with

_{}^{4} (_{}^{5} P_{4}) =_{5}^{4} R_{}^{5} P_{4}

Later

\begin{aligned} _{}^{5} n_{Z} = & - {([\begin{matrix} p x \\ p y \\ p z \end{matrix}] \times_{}^{4} f_{5})}^{T}_{}^{5} {\hat{Z}}_{5} = - (p x_{}^{4} f_{5 Y} - p y_{}^{4} f_{5 X}) \\ \leftrightarrow_{}^{4} f_{5 X} = \frac{_{}^{5} n_{Z} + p x_{}^{4} f_{5 Y}}{p y} \end{aligned}

(50)

The relationship between links 4 and 5 is

_{}^{4} f_{5} = [\begin{matrix} _{}^{4} f_{5 X} \\ _{}^{4} f_{5 Y} \\ _{}^{4} f_{5 Z} \end{matrix}]

(51)

The solution for link 5 is discovered by using the Newton-Euler method

{\begin{matrix} _{}^{5} f_{2} =_{}^{5} F_{5} +_{9}^{5} R_{}^{9} f_{5} -_{0}^{5} R m_{5} g +_{4}^{5} R_{}^{4} f_{5} \\ _{}^{5} n_{2} =_{}^{5} N_{5} +_{9}^{5} R_{}^{9} n_{5} +_{}^{5} P_{4} \times_{4}^{5} R_{}^{5} f_{4} +_{}^{5} P_{9} \times_{9}^{5} R_{}^{9} f_{5} \end{matrix} \begin{matrix} {+ P_{C 5} \times (_{}^{5} F_{5} -_{0}^{5} R m_{5} g) \end{matrix}

(52)

The solution for link 4 is discovered by using the Newton-Euler method

{\begin{matrix} _{}^{4} f_{3} =_{}^{4} F_{4} -_{}^{4} f_{5} -_{0}^{4} R m_{4} g \\ _{}^{4} n_{3} =_{}^{4} N_{4} -_{}^{4} P_{5} \times_{}^{4} f_{5} + P_{C 4} \times (_{}^{4} F_{4} -_{0}^{4} R m_{4} g) \end{matrix}

(53)

The solution for link 3 is

{\begin{matrix} _{}^{3} f_{1} =_{}^{3} F_{3} +_{4}^{3} R_{}^{4} f_{3} -_{0}^{3} R m_{3} g \\ _{}^{3} n_{1} =_{}^{3} N_{3} +_{4}^{3} R_{}^{4} n_{3} -_{}^{3} P_{4} \times_{4}^{3} R_{}^{4} f_{3} \\ + P_{C 3} \times (_{}^{3} F_{3} -_{0}^{3} R m_{3} g) \end{matrix}

(54)

The solution for link 2 is

{\begin{matrix} _{}^{2} f_{1} =_{}^{2} F_{2} +_{7}^{2} R_{}^{7} f_{2} +_{5}^{2} R_{}^{5} f_{2} -_{0}^{2} R m_{2} g \\ _{}^{2} n_{1} =_{}^{2} N_{2} +_{7}^{2} R_{}^{7} n_{2} +_{5}^{2} R_{}^{5} n_{2} \\ +_{}^{2} P_{5} \times_{5}^{2} R_{}^{5} f_{2} +_{}^{2} P_{7} \times_{7}^{2} R_{}^{7} f_{2} \end{matrix} {+ P_{C 2} \times (_{}^{2} F_{2} -_{0}^{2} R m_{2} g)

(55)

Link 1 has a solution as

{\begin{matrix} _{}^{1} f_{0} =_{}^{1} F_{1} +_{6}^{1} R_{}^{6} f_{1} +_{3}^{1} R_{}^{3} f_{1} +_{2}^{1} R_{}^{2} f_{1} -_{0}^{1} R m_{1} g \\ _{}^{1} n_{0} =_{}^{1} N_{1} +_{6}^{1} R_{}^{6} n_{1} +_{2}^{1} R_{}^{2} n_{1} \\ +_{3}^{1} R_{}^{3} n_{1} -_{}^{1} P_{2} \times_{}^{1} f_{2} -_{}^{1} P_{3} \times_{}^{1} f_{3} \end{matrix} {-_{}^{1} P_{6} \times_{}^{1} f_{6} + P_{C 1} \times (_{}^{1} F_{1} -_{0}^{1} R m_{1} g)

(56)

The required moments at links 1, 2, and 3 are

{\begin{matrix} ζ_{1} =_{}^{1} n_{0}^{T}_{}^{1} {\hat{Z}}_{1} \\ ζ_{2} =_{}^{2} n_{1}^{T}_{}^{2} {\hat{Z}}_{2} \\ ζ_{3} =_{}^{3} n_{1}^{T}_{}^{3} {\hat{Z}}_{3} \end{matrix}

(57)

Validation of research

Numerical simulation

The proposed model of our robot in 3D space is launched. In each link, a local coordinate is attached to indicate the orientation and dimension of the system parameters. It is assumed that the reference trajectory of the robot is a circular path with the center coordinates $I [x; y; z]$ = $[150; 150; 600]$ , parallel to the $O x y$ plane, with a radius $R = 50$ mm, and the time simulation to track the circle is $T = 5$ seconds, angular velocity is $ω = 2 π / T$ . With these requirements, we have the coordinates of the trajectory, velocity, and acceleration of way points on the circular path in the form of:

From the inverse kinematics computation, the relationship between the end-effector's velocity and the angular velocities of the joints, as well as the relationship between the end-effector's acceleration and the joints’ angular accelerations, is established based on system (58) and the specified requirements. As a result, a sequence of joint angles, joint angular velocities, and joint angular accelerations is obtained, as shown in Figure 7. These reference trajectory values are then inserted into the dynamic equations to obtain the system's response. The entire system is simulated in the Matlab Simulink environment to verify its high performance, as depicted in Figure 8.

{\begin{matrix} x_{k} = x_{i} + R \cdot \cos (ω t) \\ y_{k} = y_{i} + R \cdot \sin (ω t) \\ {\dot{x}}_{k} = - ω R \cdot \sin (ω t) \\ {\dot{y}}_{k} = ω R \cdot \cos (ω t) \\ {\ddot{x}}_{k} = - ω^{2} R \cos (ω t) \\ {\ddot{y}}_{k} = - ω^{2} R \sin (ω t) \end{matrix}

(58)

Figure 7.

Simulation result of the circular reference path for the proposed approach, (a) angular position, (b) angular velocity, and (c) angular acceleration with green line for joint 1, red line for joint 2, and yellow line for joint 3.

Figure 8.

Simulation setup of the dynamic equations in Matlab/Simulink for the proposed robot.

The robot's response, as shown in Figure 9, demonstrates that the driving moments in both the simulation and theoretical computation are consistent. Additionally, the dynamic constraints can be used to select the appropriate driving motor and assess the robot's durability.

Figure 9.

Result of the driving moments in simulation (a) with blue line for joint 1, red line for joint 2, yellow line for joint 3, and in theoretical computation (b) with black line for joint 1, green line for joint 2, and pink line for joint 3.

Experimental validations

To operate this platform, the electrical design, as shown in Figure 10, must be implemented, including power sources and control signals. The system generally consists of four main blocks:

Motor block: Three DC (direct current) motors with positioning encoders, and a driver to interface between the microcontroller and the motors.

Sensor block: Three proximity sensors are used to detect the HOME position.

Control block: A CPU (central processing unit) responsible for receiving signals from the sensors and transmitting control signals to the motor drivers.

Power block: Two primary power sources—a 24 V supply for sensors, drivers, and motors, and a 5 V supply for the microcontroller and positioning encoders.

Figure 10.

Diagram of the electrical wiring system for our approach.

In the next step, the motor control scheme incorporates a proportional–integral–derivative controller. The transfer function of the driving motor is derived from the theoretical model as follows:

T = K_{t} i (t)

(59)

The torque T generated by a DC motor is proportional to the current

i (t)

and the strength of the magnetic field. Since the magnetic field is constant, the motor torque is proportional only to the current

i (t)

with a constant factor

K_{t}

e = K_{e} ω (t) = K_{e} \frac{d θ}{d t} = K_{e} \dot{θ}

(60)

The electromotive force is directly proportional to the angular velocity

ω (t)

of the shaft by a constant called the electromotive force constant

K_{e}

In the SI system, the torque constant and the electromotive force constant are equal, $K_{t} = K_{e}$ , so we can represent these two constants with a single constant K. According to the Newton law and Kirchhoff law, we also have

L \frac{d i}{d t} + R i (t) = v - e

(61)

J \frac{d^{2} θ}{d t^{2}} + B \frac{d θ}{d t} = T

(62)

The transfer function

Ω (s) / V (s)

indicates the model of motor control

(s L + R) I (s) = V (s) - K_{e} Ω (s)

(63)

(s J + B) Ω (s) = K_{t} I (s)

(64)

where

Ω (s)

is the Laplace transformation of

ω (t)

.Hence

\begin{aligned} (s J + B) Ω (s) = & K_{t} \frac{V (s) - K_{e} Ω (s)}{s L + R} \\ \Rightarrow \frac{Ω (s)}{V (s)} = \frac{K}{(s J + B) (s L + R) + K^{2}} \end{aligned}

(65)

Usually, the value of L is much smaller compared to other motor parameters, so we can neglect L in the transfer function. We have the first-order transfer function of the motor as follows:

\frac{Ω (s)}{V (s)} = \frac{K}{(s J + B) R + K^{2}}

(66)

To identify the parameters of the driving motor, the procedure is carried out as follows: (a) after receiving a command from the computer via serial communication, the microcontroller triggers a relay to switch on after 0.1 seconds, driving the motor for 2.9 seconds. The sampling time is set to 1 ms. Once the motor completes its rotation, the microcontroller sends the data to the host computer for further processing.

In theory, three parameters $R, K_{t}, K_{e}$ could be found indirectly

{\begin{matrix} R \cdot I_{24} + K_{e} \cdot ω_{24} = 24 \\ R \cdot I_{12} + K_{e} \cdot ω_{12} = 12 \end{matrix}

(67)

where

I_{i}

is the current passing through the motor when a voltage i is applied;

ω_{i}

is the angular velocity of the motor when a voltage i is applied;

12 and 24 V

are the power supply for the motor.

For the joint 1 with 12 V power, by measuring the average values of current and angular velocity as shown in Figure 11(a) and (b), we have ${\bar{ω}}_{12}^{1} = 3.3594 rad / s$ , ${\bar{I}}_{12}^{1} = 0.3087 A$ . Repeating these steps in 10 times, we have ${\bar{\bar{ω}}}_{12}^{1} = \sum_{i = 1}^{10}_{}^{i} {\bar{ω}}_{12}^{1} / 10 = 3.3512 rad / s$ and ${\bar{\bar{I}}}_{12}^{2} = \sum_{i = 1}^{10}_{}^{i} {\bar{I}}_{12}^{1} / 10 = 0.3017 A$ . Likewise, if the power supply is $24 V$ as shown in Figure 11(c) and (d), then we obtain ${\bar{ω}}_{24}^{1} = 6.7843 rad / s$ , ${\bar{I}}_{24}^{1} = 0.4939 A$ . After 10 validations, we also reach ${\bar{\bar{ω}}}_{24}^{1} = \sum_{i = 1}^{10}_{}^{i} {\bar{ω}}_{24}^{1} / 10 = 6.7761 rad / s$ and ${\bar{\bar{I}}}_{24}^{1} = \sum_{i = 1}^{10}_{}^{i} {\bar{I}}_{24}^{1} / 10 = 0.4926 A$ .

Figure 11.

Experimental response of the driving motor at joint 1 when the power supply is 12 V ((a) current and (b) angular velocity) and when the power supply is 24 V ((c) current and (d) angular velocity).

Then,

{\begin{matrix} R_{1}^{} = 2.2472 Ω \\ K_{t 1}^{} = K_{e 1}^{} = 3.3785 Nm / A \end{matrix}

(68)

In the same way, the joint 2 with $12 V$ power has the average values of current and angular velocity as shown in Figure 12(a) and (b), such ${\bar{ω}}_{12}^{2} = 3.4204 rad / s$ , ${\bar{I}}_{12}^{2} = 0.3187 A$ . After measuring 10 times, these values are ${\bar{\bar{ω}}}_{12}^{2} = \sum_{i = 1}^{10}_{}^{i} {\bar{ω}}_{12}^{2} / 10 = 3.4575 rad / s$ and ${\bar{\bar{I}}}_{12}^{2} = \sum_{i = 1}^{10}_{}^{i} {\bar{I}}_{12}^{2} / 10 = 0.3243 A$ . In the case that power supply is $24 V$ as shown in Figure 12(c) and (d), then ${\bar{ω}}_{24}^{2} = 7.0749 rad / s$ , ${\bar{I}}_{24}^{2} = 0.52472 A$ . After 10 measurements, ${\bar{\bar{ω}}}_{24}^{2} = \sum_{i = 1}^{10}_{}^{i} {\bar{ω}}_{24}^{2} / 10 = 6.9590 rad / s$ and ${\bar{\bar{I}}}_{24}^{2} = \sum_{i = 1}^{10}_{}^{i} {\bar{I}}_{24}^{2} / 10 = 0.5330 A$ .

Figure 12.

Experimental response of the driving motor at joint 2 when the power supply is 12 V ((a) current and (b) angular velocity) and when the power supply is 24 V ((c) current and (d) angular velocity).

From these computations,

{\begin{matrix} R_{2} = 1.2765 Ω \\ K_{t 2} = K_{e 2} = 3.3510 Nm / A \end{matrix}

(69)

For joint 3, we demonstrate the average values of current and angular velocity are shown in Figure 13(a) and (b), with

12 V

power such as

{\bar{ω}}_{12}^{3} = 3.3232 rad / s

{\bar{I}}_{12}^{3} = 0.3868 A

{\bar{\bar{ω}}}_{12}^{3} = \sum_{i = 1}^{10}_{}^{i} {\bar{ω}}_{12}^{3} / 10 = 3.3271 rad / s,

and

{\bar{\bar{I}}}_{12}^{3} = \sum_{i = 1}^{10}_{}^{i} {\bar{I}}_{12}^{3} / 10 = 0.3679 A

. Then, the response of joint 3 with

24 V

the power supply is illustrated in Figure 13(c) and (d), then

{\bar{ω}}_{24}^{3} = 6.6947 rad / s

{\bar{I}}_{24}^{3} = 0.67 A

{\bar{\bar{ω}}}_{24}^{3} = \sum_{i = 1}^{10}_{}^{i} {\bar{ω}}_{24}^{3} / 10 = 6.6873 rad / s

and

{\bar{\bar{I}}}_{24}^{3} = \sum_{i = 1}^{10}_{}^{i} {\bar{I}}_{24}^{3} / 10 = 0.6754 A

Figure 13.

Experimental response of the driving motor at joint 3 when the power supply is 12 V ((a) current and (b) angular velocity) and when the power supply is 24 V ((c) current and (d) angular velocity).

We estimate the gains of the motor model for joint 3 as below:

{\begin{matrix} R_{3} = 1.8652 Ω \\ K_{t 3} = K_{e 3} = 3.4005 Nm / A \end{matrix}

(70)

Since these motors are attached to the gearbox, the driving moment must suffer additional components

T_{m s}

from friction forces. Consequently

{\begin{matrix} T_{m s} + B \cdot ω_{24} = K_{t} \cdot I_{24} \\ T_{m s} + B \cdot ω_{12} = K_{t} \cdot I_{12} \end{matrix}

(71)

The results of the above computations for joint 1 are

{\begin{matrix} B_{1} = 0.1883 N \cdot s / m \\ T_{m s 1} = 0.3883 Nm \end{matrix}

(72)

For joint 2

{\begin{matrix} B_{2} = 0.0995 N \cdot s / m \\ T_{m s 2} = 1.0928 Nm \end{matrix}

(73)

For joint 3

{\begin{matrix} B_{3} = 0.2308 N \cdot s / m \\ T_{m s 3} = 0.7526 Nm \end{matrix}

(74)

In Figure 14, the virtual model of our robot is shown, and the physical platform has been successfully manufactured. During the production process, it is crucial to release detailed drawings for all components. After assembly, the entire robot must be validated in real-world scenarios. Figure 15 illustrates the tracking performance of the hardware. On the host computer, the reference path is generated with multiple points, and the coordinates of each point are transmitted to the microcontroller, while the actual position is fed back via a positioning encoder.

Figure 14.

Description of the proposed robotic platform, (a) the 3D model, and (b) the real-world model.

Figure 15.

Experimental result of the tracking performance for the proposed robotic platform.

To monitor the working performance of the driving motors, Figure 16 demonstrates the practical results of direct currents and rotational angles in each one. From Figure 16(b) and (d) to (f), the tracking velocities are nearly identical, with the most noticeable linear change in this value occurring between 1.0 and 1.5 seconds. Therefore, the rotational angle would be initially observed at 1.5 seconds. In fact, the roles of the three motors are quite different. While both motors 1 and 2 drive the parallelogram mechanism, the third motor drives our base around the z-axis and provides flexible motion for this platform. In this case, when the whole system activates at 1.0 seconds, the currents of the three motors in Figure 16(a), (c) and (e) fluctuate since they try to hold mechanical components in the fixed position. Besides, motor 3 slightly moves to adjust the proper location as our desired command at 2.6 seconds. Whenever it comes to the target location, all motors continue to sustain their performance.

Figure 16.

Experimental result of the direct current (a, c, e) and rotational angle (b, d, f) in motor 1, motor 2, and motor 3 respectively.

Conclusions

In this investigation, a comprehensive solution for robotic design using a parallelogram mechanism is presented. Our objectives are to: (i) demonstrate the theoretical computations of both forward and inverse kinematics, as well as validate dimensions and displacement in the mechanical design; (ii) derive the dynamic analysis of the proposed platform to assess its tracking performance; and (iii) fully develop an experimental platform for a laboratory robot to carry out various validations.

Future work is essential. Mechanical improvements for a specialized gripper should be explored to enable the grasping of various objects. Additionally, advanced algorithms for end-effector manipulation, such as adaptive schemes, model predictive control, or intelligent control, could be considered. Moreover, integrating image processing techniques with artificial intelligence may further enhance the quality of grasping actions.

Footnotes

Acknowledgements

We acknowledge Ho Chi Minh City University of Technology (HCMUT), VNU-HCM for supporting this study.

Author contributions

Thanh Phuong Nguyen: data curation, methodology, and supervision; Hung Nguyen: investigation, formal analysis, and visualization; and Thinh Ngo: conceptualization, validation, and writing–review and editing.

Declaration of conflicting interests

The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.

Funding

The author(s) received no financial support for the research, authorship, and/or publication of this article.

ORCID iD

Thinh Ngo

Appendix

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Design and implementation of a field robot using a parallelogram mechanism

Abstract

Keywords

Introduction

Related works

The proposed approach

Forward kinematic

Inverse kinematic

Mechanical design

Dynamical analysis

Validation of research

Numerical simulation

Experimental validations

Conclusions

Footnotes

Acknowledgements

Author contributions

Declaration of conflicting interests

Funding

ORCID iD

Appendix

References