Structural design and characteristic analysis for a 4-degree-of-freedom parallel manipulator

Abstract

Previous studies on parallel mechanisms have been limited to symmetrical and semi-symmetrical parallel robots. However, there is relatively little research on asymmetrical parallel robot, especially on asymmetrical parallel robots with 4 degrees of freedom. Considering of disadvantages of modern technology, a new type of asymmetric parallel manipulator is proposed based on the theory and method of position and orientation characteristic equations in this article. First, the topology theory of parallel mechanism is expounded, the robot structure is described and the position and orientation characteristic set is calculated to obtain the degrees of freedom and the coupling degree of the parallel manipulator. Then the structure coupling–reducing theory is used to realize the structure decoupling of the robot. Second, the forward kinematics solution of the parallel robot is analyzed by ordered single-open-chain method, the inverse kinematics solution is solved by the algebraic method, and a numerical example is provided to confirm the correctness of the solution procedure. Finally, the workspace of the robot is analyzed by search method. Generally, the above theoretical research provides an important reference value for the industrial design and application of the parallel robot.

Introduction

Parallel robots have attracted more attention of researchers both at home and abroad because of their high strength, high precision, and small cumulative error. Especially, parallel robots with small degree of freedom (DOF) have good application prospects in the industrial field. ¹ In recent years, enumeration method based on DOF formula $\mathrm{Gr}\stackrel{··}{\mathrm{u}}\mathrm{bler}-\mathrm{Kutzbach}$ , ² displacement subgroup synthesis method based on group theory, ³ and combination method based on screw theory have appeared successively.^4,5 The aforementioned methods have exerted a great influence in the industrial field and have synthesized many parallel mechanisms meeting social demand. However, many of the mechanisms synthesized by these methods only have instantaneous characteristic and the non-instantaneous characteristic need to be judged. Professor Yang Ting-li pioneered the method of mechanism synthesis based on the theory of position and orientation characteristic (POC) set, which made many unknown parallel mechanisms springing up like bamboo shoots after a spring rain.^6,7

Among the low-DOF mechanisms, the mechanisms of 4-DOF and 5-DOF robots are difficult problems that the mechanism scientists have been trying to break through, especially the 4-DOF parallel mechanisms with three movements and one rotation. However, Changzhou University has synthesized some 4-DOF mechanisms and made kinematic analysis by using the theory of POC set;^8–11 most robots are asymmetric parallel mechanisms, which can achieve a specific DOF movement, so they have broad market prospects. Fully symmetrical parallel robots refer to the fact that the branching chains connecting the moving and the fixed platforms of the robots are identical, not only in the branching chains but also in the space. Asymmetric parallel robots refer to the branched chains that connect the moving and fixed platform of the robot. At least one branched chain is different from other branched chains in structure or space configuration. Because of the symmetry of structure and space configuration, it is very difficult to synthesize such mechanisms, especially for 4-DOF and 5-DOF parallel robots. Therefore, the syntheses of fully symmetric parallel robots are few in the world. Asymmetric parallel robots can obtain the required DOFs through different branched chains or different combinations of the same branched chains. Generally speaking, a parallel robot can be regarded as an asymmetric parallel robot if one or one of its branches is different from other branches in structure or space configuration.

For parallel manipulators, the mechanism is complex, and the forward solution is difficult. Only a few mechanisms can be solved by the analytical expression. In order to solve the forward solution problem of complex parallel manipulators, the position and attitude of the manipulator can be solved by reducing the coupling degree of the mechanism combined with POC equations. At present, the research of mechanism decoupling method is very few, and Shen and colleagues^12–14 have done some research in this field. When the coupling degree of the mechanism is 1, we can use the ordered single-open-chain (SOC) method to find the direct solution of the mechanism.^15–20

Based on the aforementioned review, it could be found that seldom research has ever given regard to the asymmetrical parallel robots. The thesis, therefore, will contribute to the solution of this problem so that this asymmetric parallel robot can provide some reference value for configuration design. At the same time, the non-instantaneous mechanism can be obtained using POC set method. According to the theory and method based on POC set, a new type of asymmetrical parallel robot is proposed in this article, which can realize three-dimensional (3D) movement and one-dimensional (1D) rotation. The structure of the mechanism is decoupled, and the direct position and attitude of the mechanism are obtained by the method of ordered SOC method, and the workspace of the mechanism is obtained by the inverse solution equations.

An overview of POC set theory

The 12 basic topological structure characteristics are indispensable for analyzing parallel mechanism and are the basis for studying its internal mechanism. These characteristics specifically include the POC set, type and number of DOF, basic kinematic chain (BKC), independent displacement equation number, independent loop number, input–output decoupling, over-constrained number, drive pair selection, and so on.^21,22 The motion of parallel mechanism is the result of each branch chain motion synthesis, meanwhile, is also the result of the intersection of POC set of branch chain. Its basic operation conforms to the inherent law of mechanism.

The terminal POC equations of the mechanism can be written as

$$M_{\mathrm{bi}}=\underset{i=1}{\overset{m}{\cup }}{M}_{\mathrm{Ji}}$$

(1)

$$M_{\mathrm{Pa}}=\underset{i=1}{\overset{n}{\cap }}{M}_{\mathrm{bi}}$$

(2)

$M_{\mathrm{bi}}$ denotes the POC set of the ith branch chain, $M_{\mathrm{Ji}}$ represents the POC set of the ith pair of motion, and $M_{\mathrm{Pa}}$ is the POC set of the moving platform of the parallel mechanism.

The formula of DOF is

$${\xi }_{\mathrm{Lj}}=\dim .\left\{\underset{j=1}{\overset{j}{(\cap M_{\mathrm{bj}})\cup M_{b(j+1)}}}\right\}$$

(3)

$$F=\sum _{i=1}^m{f}_i-\sum _{j=1}^v{\xi }_{\mathrm{Lj}}$$

(4)

where F is the DOF of parallel mechanism, $f_i$ denotes the DOF of the $i\mathrm{th}$ pair. ${\xi }_{\mathrm{Lj}}$ represents the number of independent displacement equations in the j independent loop. $M_{\mathrm{bj}}$ is the POC set of the preceding j branch chain. v is an independent loop number. $M_{b(j+1)}$ is the POC set of the preceding $j+1$ branch chain.

According to the principle of mechanism composition of the ordered SOC method, a mechanism can be decomposed into several BKCs, and a BKC can be decomposed into several SOCs. The constraint degree of the $j\mathrm{th}$ SOC can be expressed as^7,21,22

$$\mathrm{\Delta }=\sum _{i=1}^{m_j}{f}_i-I_j-{\xi }_{\mathrm{Lj}}=\{\begin{array}{c}{\mathrm{\Delta }}_j^{-}=-5,-4,\dots ,-1\hfill \\ \multicolumn{1}{c}{{\mathrm{\Delta }}_j^0=0}\\ \multicolumn{1}{c}{{\mathrm{\Delta }}_j^{+}=1,2,\dots }\end{array}$$

(5)

$m_j$ denotes the number of kinematic pairs of the $j\mathrm{th}$ SOC; and $I_j$ is the number of driving pairs of the $j\mathrm{th}$ SOC.

For a BKC, the following formula must be satisfied

$$\mathrm{\Delta }=\sum _{j=1}^v{\mathrm{\Delta }}_j=0$$

(6)

Therefore, the coupling degree can be written as

$$k={\mathrm{\Delta }}_j^{+}=\left|{\mathrm{\Delta }}_j^{-}\right|=\frac{1}{2}\min \left(\sum _{j=1}^v\left|{\mathrm{\Delta }}_j\right|\right)$$

(7)

The physical meaning of k is that it reveals the intrinsic correlation and dependence between the basic loop variables of the mechanism. The larger the value, the more complex is the mechanism. In addition, its kinematics and dynamics analyses are more complex. In mechanism analysis, we hope to obtain mechanisms with lower coupling degree, especially those with 0 or 1 coupling degree. The forward and backward solutions of such mechanisms are relatively simple and more practical in daily life. Therefore, structural decoupling of complex mechanisms is also an important part of mechanism research.

Topological characteristics of mechanisms

Description of mechanism characteristics

The parallel mechanism was described in this article is shown in Figure 1. It is mainly composed of a moving platform M, a fixed platform N, and four branches I, II, III, and IV. Branched chain I is a series chain which can produce 3D movement and two-dimensional (2D) rotation. Branched chain II is also a series chain which can generate 3D movement and 2D rotation. Branch chain III and branch chain IV are a 6-degree-of-freedom chain with end-output motion, whose DOFs both are 6 and do not impose any restrictions on the moving platform. The four driving pairs of the parallel mechanism should be arranged on the fixed platform and are all revolute pairs in accordance with the composing characteristics of mechanism. According to POC set operation of the parallel mechanism, the compound motion of four branches can be obtained, that is, the output motion of the moving platform M. Finally, the output motion of the moving platform is three translations and one rotation.

Figure 1.

The sketch of robot structure.

The arrangement of parallel mechanism is as follows: the moving platform adopts arbitrary quadrilateral form, the fixed platform adopts square form, and the four rotation pairs on the fixed platform are located in the middle of the square, respectively. In order to ensure that the ball pairs S1, S2, S3, and S4 form parallelograms, it is necessary to add two rotating pairs R_a and R_b with the length equals to the short side of the parallelogram. At the same time, the axis of $R_{11}$ , $R_{12}$ , and $R_{13}$ should be parallel, the axis of $R_{13}$ and $R_{14}$ should be perpendicular, and the axis of $R_{14}$ and $R_{15}$ should be parallel to the axis of $R_{22}$ . Branched chains I and II are located on the adjacent side of the square, and branched chains III and IV can be arranged arbitrarily.

Topological analysis of parallel mechanism

Figure 1 shows the model of the mechanism before decoupling, and the four branches of the mechanism are independent. The topological structure of branched chain I is $\{-R_{11}\parallel R_{12}\parallel R_{13}\perp R_{14}\parallel R_{15}-\}$

The POC set of the branched chain is

$$\begin{array}{cc}\multicolumn{1}{c}{M_{b1}}& =\left[\begin{array}{c}{t}^2(\perp (R_{13})\hfill \\ \multicolumn{1}{c}{r^1(\parallel (R_{13})}\end{array}\right]\cup \left[\begin{array}{c}{t}^1(\perp (R_{14})\cup t^1(\perp (R_{15})\hfill \\ \multicolumn{1}{c}{r^1(\parallel ◇(R_{14})}\end{array}\right]\hfill \\ \multicolumn{1}{c}{}& =\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^2(\parallel ◇(R_{13},R_{14})}\end{array}\right]\hfill \end{array}$$

The topological structure of branched chain I is $\{-R_{21}(-:◇(4S))\perp R_{22}-\}$

The POC set of the branched chain is

$$\begin{array}{cc}\multicolumn{1}{c}{M_{b2}}& =\left[\begin{array}{c}{t}^1(\perp R_{21})\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{21})}\end{array}\right]\cup \left[\begin{array}{c}{t}^2\hfill \\ \multicolumn{1}{c}{r^1}\end{array}\right]\cup \left[\begin{array}{c}{t}^1(\perp R_{22})\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{22})}\end{array}\right]\hfill \\ \multicolumn{1}{c}{}& =\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{21})\cup r^1(\parallel R_{22})}\end{array}\right]\hfill \end{array}$$

The topological structures of branched chains III and IV are $\{-R-S-S-\}$

The POC sets of these two branched chains are

$$M_{b3}=\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]$$

According to formula (2), the POC set of the moving platform can be obtained as follows

$$M_{\mathrm{Pa}}=\underset{i=1}{\overset{4}{\cap }}{M}_{\mathrm{bi}}=\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]$$

Determining the DOF and coupling degree of parallel mechanism

• 1. Determining the DOF and POC set of the first loop.

According to equations (3) and (4), we have

$$\begin{array}{cc}\multicolumn{1}{c}{{\xi }_1}& =\dim .\{M_{b1}\cup M_{b2}\}\hfill \\ \multicolumn{1}{c}{}& =\dim .\{\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^2(\parallel ◇(R_{13},R_{14})}\end{array}\right]\cup \left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{21})\cup r^1(\parallel R_{22})}\end{array}\right]\}\hfill \\ \multicolumn{1}{c}{}& =\dim .\left\{\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]\right\}=6\hfill \end{array}$$

The DOF and POC set of sub parallel mechanisms consisting of branched chains I and II are

$$\begin{array}{c}{F}_{(1-2)}=\sum _{i=1}^m{f}_i-\sum _{j=1}^1{\xi }_{\mathrm{Li}}=10-6=4\hfill \\ M_{\mathrm{pa}(1-2)}=\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^2(\parallel ◇(R_{13},R_{14})}\end{array}\right]\cap \left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{21})\cup r^1(\parallel R_{22})}\end{array}\right]\hfill \\ =\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]\hfill \end{array}$$

• 2. Determining the DOF and POC set of the second loop

$$\begin{array}{c}{\xi }_2=\dim .\{M_{\mathrm{pa}(1-2)}\cup M_{b3}\}\hfill \\ =\dim .\{\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]\cup \left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]\}=\dim .\left\{\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]\right\}=6\hfill \\ F_{(1-2-3)}=\sum _{i=1}^m{f}_i-\sum _{j=1}^1{\xi }_{\mathrm{Li}}=16-(6+6)=4\hfill \\ M_{\mathrm{pa}(1-2-3)}=\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]\cap \left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]=\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]\hfill \end{array}$$

• 3. Determining the DOF and POC set of the third loop

$$\begin{array}{c}{\xi }_3=\dim .\{M_{\mathrm{pa}(1-2-3)}\cup M_{b4}\}\hfill \\ =\dim .\{\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]\cup \left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]\}=\dim .\left\{\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]\right\}=6\hfill \\ F=\sum _{i=1}^m{f}_i-\sum _{j=1}^1{\xi }_{\mathrm{Li}}=22-(6+6+6)=4\hfill \\ M_{\mathrm{pa}}=\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]\cap \left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]=\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]\hfill \end{array}$$

From the above analysis, it can be seen that when the four rotation pairs of the fixed platform are chosen by mechanism as driving pairs, the moving platform can realize the output motion of three translations and one rotation.

Calculation of coupling degree of parallel mechanism

From the above analysis, it is known that the independent displacement equations of the three loops are ${\xi }_{L1}={\xi }_{L2}={\xi }_{L3}=6$ , so that the constraint degrees of each loop are as follows

$$\begin{array}{c}{\mathrm{\Delta }}_1=\sum _{i=1}^{10}{f}_i-I_1-{\xi }_{L1}=10-2-6=2\hfill \\ {\mathrm{\Delta }}_2=\sum _{i=1}^3{f}_i-I_2-{\xi }_{L2}=6-1-6=-1\hfill \\ {\mathrm{\Delta }}_3=\sum _{i=1}^3{f}_i-I_3-{\xi }_{L3}=6-1-6=-1\hfill \end{array}$$

Therefore, the coupling degree of the mechanism can be obtained as follows

$$k=\frac{1}{2}\sum _{j=1}^3\left|{\mathrm{\Delta }}_j\right|=\frac{1}{2}(|-1|+|-1|+2)=2$$

Because the coupling degree of the mechanism is 2, it reflects that the parallel mechanism is very complex. In addition, it is very inconvenient to analyze its kinematics and dynamics. In order to facilitate the simple kinematics calculation and the control, it is necessary to analyze the structural decoupling of the parallel mechanism. When the coupling degree is reduced to 1, the forward solution of the mechanism can be obtained by 1D search method. When the coupling degree is reduced to 0, the analytical solution of the mechanism can be obtained directly.

Structural coupling-reducing analysis

According to the principle and method of structural coupling-reducing,^12–14 the branched chains III and IV are kept unchanged, and the rotation pair axis connected to the moving platform coincide with the branched chains I and II; by doing so, the structural coupling reducing of the mechanism can be achieved, as shown in Figure 2.

Figure 2.

Coupling-reducing design of parallel mechanism.

In order to better understand the connection between the links of the mechanism and the spatial configuration of the kinematic pairs, SolidWorks is used to draw the 3D model of the structure-decoupled parallel robot as shown in Figure 3, which clearly shows the configuration of mobile platform, fixed platform, and four branched chains in space. The output motion of the moving platform represents the output motion of the whole machine. The green part in the picture can realize 3D movement and 1D rotation in space.

Figure 3.

Three-dimensional sketch of decoupling mechanism.

The topological structures of the branched chain after decoupling are

$$\begin{array}{c}\mathrm{HSO}{C}_1\{-R_{11}\parallel R_{12}\parallel R_{13}\perp R_{14}\parallel R_{22}-(◇4S)-R_{21}-\}\hfill \\ \mathrm{SO}{C}_2\{-R_{15}-S42-S41-R41-\}\hfill \\ \mathrm{SO}{C}_3\{-S33-S32-R31-\}\hfill \end{array}$$

• 1. Determining the DOF and POC set of the first loop

$$\begin{array}{c}{\xi }_{L1}=\dim .\{\left[\begin{array}{c}{t}^2\hfill \\ \multicolumn{1}{c}{r^2(\parallel ◇(R_{13},R_{14})}\end{array}\right]\cup \left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{21})\cup r^1(\parallel R_{22})}\end{array}\right]\}\hfill \\ =\dim .\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]=6\hfill \\ F_{(\mathrm{sub}-H)}=\sum _{i=1}^m{f}_i-\sum _{j=1}^1{\xi }_{\mathrm{Li}}=9-6=3\hfill \\ M_{\mathrm{pa}(\mathrm{sub}-H)}=\left[\begin{array}{c}{t}^2(\perp R_{11})\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})\cup r^1(\parallel R_{11})}\end{array}\right]\cap \left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{22})\cup r^1(\parallel R_{21})}\end{array}\right]\hfill \\ =\left[\begin{array}{c}{t}^2(\perp R_{11})\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]\hfill \end{array}$$

• 2. Determining the DOF and POC set of the second loop

$$\begin{array}{c}{\xi }_{L2}=\dim .\{\left[\begin{array}{c}{t}^2(\perp R_{11})\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]\cup \left[\begin{array}{c}{t}^1(\perp R_{15})\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{15})}\end{array}\right]\cup \left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]\}\hfill \\ =\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]=6\hfill \\ M_{\mathrm{pa}(\mathrm{sub}-H-3)}=\{\left[\begin{array}{c}{t}^2(\perp R_{11})\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]\cup \left[\begin{array}{c}{t}^1(\perp R_{15})\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{15})}\end{array}\right]\cap \left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]\}\hfill \\ =\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]\hfill \\ F_{\mathrm{pa}(\mathrm{sub}-H-3)}=\sum _{i=1}^m{f}_i-\sum _{j=1}^2{\xi }_{\mathrm{Li}}=16-(6+6)=4\hfill \end{array}$$

• 3. Determining the DOF and POC set of the third loop

$$\begin{array}{c}{\xi }_{L3}=\dim .\{\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]\cup \left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]\}=\dim .\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^3}\end{array}\right]=6\hfill \\ F=\sum _{i=1}^m{f}_i-\sum _{j=1}^3{\xi }_{\mathrm{Li}}=22-(6+6+6)=4\hfill \\ M_{\mathrm{pa}}=M_{\mathrm{pa}(\mathrm{sub}-H-3)}\cap M_{b3}=\left[\begin{array}{c}{t}^3\hfill \\ \multicolumn{1}{c}{r^1(\parallel R_{14})}\end{array}\right]\hfill \end{array}$$

Therefore, the constraint degrees of each loop are given as follows

$$\begin{array}{c}{\mathrm{\Delta }}_1=\sum _{i=1}^9{f}_i-I_1-{\xi }_{L1}=9-2-6=1\hfill \\ {\mathrm{\Delta }}_2=\sum _{i=1}^4{f}_i-I_2-{\xi }_{L2}=7-1-6=0\hfill \\ {\mathrm{\Delta }}_3=\sum _{i=1}^3{f}_i-I_3-{\xi }_{L3}=6-1-6=-1\hfill \end{array}$$

Therefore, the coupling degree of the coupling-reducing mechanism can be obtained as follows

$$k=\frac{1}{2}\sum _{j=1}^3\left|{\mathrm{\Delta }}_j\right|=\frac{1}{2}(|-1|+|-1|)=1$$

From the above analysis, it can be seen that the coupling degree of the parallel mechanism is 1, and the forward position solution of the mechanism can be obtained by 1D search method.

The forward and inverse kinematics of parallel mechanism

As shown in Figure 4, the moving platform is set as an equilateral triangle, and the fixed platform is a square. A moving coordinate system O’-XYZ is established in the center of the moving platform, where the X-axis is perpendicular to $R_{15}{S}_{33}$ , the Y-axis is parallel to $R_{15}{S}_{33}$ , and the Z-axis is determined by the right-hand rule. A fixed coordinate system O-XYZ is established in the center of the fixed platform, where the X-axis is perpendicular to $R_{21}{R}_{41}$ , the Y-axis is parallel to $R_{11}{R}_{31}$ , and the Z-axis is determined by the right-hand rule.

Figure 4.

Kinematics model of coupling-reducing mechanism.

The structure parameters of the mechanism are as follows: the side length of the fixed platform is $2a$ ; the side length of the moving platform is $2m$ ; ${\alpha }_1$ represents the rotation angle of the rotating pair $R_{11}$ , $\beta$ represents the rotation angle of the rotating pair $R_{21}$ , $\theta$ represents the rotation angle of the rotating pair $R_{31}$ , $\delta$ represents the rotation angle of the rotating pair $R_{41}$ . To facilitate understanding, we draw the top view of the mechanism as illustrated in Figure 5. For the parallel robot, the forward solution problem can be described as when the rotation angles ${\alpha }_1$ , $\beta$ , $\theta$ , and $\delta$ are known, the position $(x,y,z)$ and attitude angle $\gamma$ of the moving platform center point can be obtained.

Figure 5.

Expanding top view of coupling-reducing mechanism.

Forward position analysis

Principle of position forward solution based on ordered SOC method

The constraint degrees of an SOC can be positive, negative, and zero, and their physical meanings are as follows: the $\mathrm{SOC}({\mathrm{\Delta }}_j^{+})$ with a positive degree of constraint increases the DOF of the mechanism by ${\mathrm{\Delta }}_j^{+}$ . To determine the motion of the mechanism, it is necessary to set ${\mathrm{\Delta }}_j^{+}$ virtual variables on the chain. The $\mathrm{SOC}({\mathrm{\Delta }}_j^0)$ with a constraint degree of zero indicates that the mechanism has definite motion and its positive solution can be solved independently. The $\mathrm{SOC}({\mathrm{\Delta }}_j^{-})$ with a negative degree of constraint reduces the DOF of the mechanism by $|{\mathrm{\Delta }}_j^{-}|$ , that is, $|{\mathrm{\Delta }}_j^{-}|$ equations with variables are needed. Especially when ${\mathrm{\Delta }}_j^{+}=|{\mathrm{\Delta }}_j^{-}|=1$ , it shows that the mechanism only needs to introduce a virtual variable, and it is easy to obtain the positive solution by 1D search method.

According to the aforementioned positive solution steps, the ordered SOC method can be used in 1D search to obtain the solution of decoupling mechanism. The specific steps are as follows:

1. For the first SOC, $\mathrm{HSO}{C}_1$ , setting the angle ${\alpha }_2$ of the rotation pair $R_{12}$ as the virtual input variable ${\alpha }_2^{*}$ , then ${\alpha }_3=f({\alpha }_2^{*})$ can be obtained.

2. For the second SOC, $\mathrm{SO}{C}_2$ , the rod length constraint equation $S_{41}{S}_{42}=h_{42}$ is established to get the rotation angle $\gamma =f({\alpha }_2^{*})$ .

3. Similarly, the rod constraint equation $S_{32}{S}_{33}=h_{32}$ is established according to the third branch chain $\mathrm{SO}{C}_3$ , and the objective function equation about ${\alpha }_2^{*}$ is obtained. 1D search method is used to continuously change the initial value of ${\alpha }_2^{*}$ from 0° to 360° until the established objective function is 0.

Solution of SOC with positive degree of constraint

For the branched chains with topological structure $\{-R_{11}\parallel R_{12}\parallel R_{13}\perp R_{14}\parallel R_{22}\parallel R_{15}-\}$ , the coordinates of $R_{11}$ , $R_{12}$ , $R_{13}$ , $R_{14}$ , $R_{22}$ , $R_{15}$ , e, and f can be obtained according to the geometric parameters and position characteristics of the mechanism. The coordinates of the center point $O\prime$ of the moving platform can be obtained by vector operation $\overrightarrow{\mathrm{OO}\prime }=\overrightarrow{O{R}_{15}}-\overrightarrow{O\prime R_{15}}$

$$\begin{array}{c}\left[\begin{array}{c}{x}_{o\prime }\hfill \\ \multicolumn{1}{c}{y_{o\prime }}\\ \multicolumn{1}{c}{z_{o\prime }}\end{array}\right]=\hfill \\ \left[\begin{array}{c}-h_{12}-h_{14}-h_{16}\sin {\alpha }_3-m\sin \gamma -\frac{\sqrt{3}}{3}m\cos \gamma \hfill \\ \multicolumn{1}{c}{-a+h_{11}\cos {\alpha }_1+h_{13}\cos {\alpha }_2+h_{16}\cos {\alpha }_3+m\cos \gamma -\frac{\sqrt{3}}{3}m\sin \gamma }\\ \multicolumn{1}{c}{h_{11}\sin {\alpha }_1+h_{13}\sin {\alpha }_2+h_{15}+h_{17}}\end{array}\right]\hfill \end{array}$$

(8)

According to rod length constraint relation $\mathrm{ef}=h_{22}$ , we have

$$\begin{array}{c}(-h_{12}-h_{14}-h_{16}\sin {\alpha }_3+h_{23}\cos \phi -a-h_{21}\cos \beta {)}^2\hfill \\ +(-a+h_{11}\cos {\alpha }_1+h_{13}\cos {\alpha }_2+h_{16}\cos {\alpha }_3+h_{23}\sin \phi {)}^2\hfill \\ +(h_{11}\sin {\alpha }_1+h_{13}\sin {\alpha }_2+h_{15}-h_{21}\sin \beta {)}^2=h_{22}^2\hfill \end{array}$$

Then, we can obtain the following equation

$${\alpha }_3=f_1({\alpha }_2^{*})$$

(9)

Solution of SOC with zero degree of constraint

The coordinates of $S_{41}$ and $S_{42}$ can be obtained from branched chain IV

$$\begin{array}{c}\left[\begin{array}{c}{x}_{S41}\hfill \\ \multicolumn{1}{c}{y_{S41}}\\ \multicolumn{1}{c}{z_{S41}}\end{array}\right]=\left[\begin{array}{c}-a+h_{41}\cos \delta \hfill \\ \multicolumn{1}{c}{0}\\ \multicolumn{1}{c}{h_{41}\sin \delta }\end{array}\right],\left[\begin{array}{c}{x}_{S42}\hfill \\ \multicolumn{1}{c}{y_{S42}}\\ \multicolumn{1}{c}{z_{S42}}\end{array}\right]\hfill \\ =\left[\begin{array}{c}-h_{12}-h_{14}-h_{16}\sin {\alpha }_3-2m(r+\frac{\pi }{3})\hfill \\ \multicolumn{1}{c}{-a+h_{11}\cos {\alpha }_1+h_{13}\cos {\alpha }_2+h_{16}\cos {\alpha }_3+2m\cos (r+\frac{\pi }{3})}\\ \multicolumn{1}{c}{h_{11}\sin {\alpha }_1+h_{13}\sin {\alpha }_2+h_{15}+h_{17}}\end{array}\right]\hfill \end{array}$$

According to rod length constraint relation $l_{S41}{l}_{S42}=h_{42}$ , we have

$$A_2\sin (\gamma +\frac{\pi }{3})+B_2\cos (\gamma +\frac{\pi }{3})+C_2=0$$

(10)

$$k_2=\tan \frac{\gamma +\frac{\pi }{3}}{2}$$

(11)

$$k_2=\frac{-A_2\pm \sqrt{A_2^2+B_2^2-C_2^2}}{C_2-B_2}$$

(12)

In the above equation

$$A_2=-4m(-h_{12}-h_{14}-h_{16}\sin {\alpha }_3+a-h_{41}\cos \delta )$$

$$B_2=(-a+h_{11}\cos {\alpha }_1+h_{13}\cos {\alpha }_2+h_{16}\cos {\alpha }_3)$$

$$\begin{array}{c}{C}_2=(-h_{12}-h_{14}-h_{16}\sin {\alpha }_3+a-h_{41}\cos \delta {)}^2\hfill \\ +4m^2+(-a+h_{11}\cos {\alpha }_1+h_{13}\cos {\alpha }_2+h_{16}\cos {\alpha }_3{)}^2\hfill \\ +(h_{11}\sin {\alpha }_1+h_{13}\sin {\alpha }_2+h_{15}+h_{17}-h_{41}\sin \delta {)}^2-h_{42}^2\hfill \end{array}$$

Thus, the next relationship can be obtained

$$\gamma =f_2({\alpha }_2^{*})$$

(13)

Solution of SOC with negative degree of constraint

The coordinates of $S_{31}$ and $S_{32}$ can be obtained from branched chain III

$$\begin{array}{c}\left[\begin{array}{c}{x}_{S31}\hfill \\ \multicolumn{1}{c}{y_{S31}}\\ \multicolumn{1}{c}{z_{S31}}\end{array}\right]=\left[\begin{array}{c}{h}_{31}\cos \theta \hfill \\ \multicolumn{1}{c}{a}\\ \multicolumn{1}{c}{h_{31}\sin \theta }\end{array}\right],\left[\begin{array}{c}{x}_{S32}\hfill \\ \multicolumn{1}{c}{y_{S32}}\\ \multicolumn{1}{c}{z_{S32}}\end{array}\right]\hfill \\ =\left[\begin{array}{c}-h_{12}-h_{14}-h_{16}\sin {\alpha }_3-2m\sin r\hfill \\ \multicolumn{1}{c}{-a+h_{11}\cos {\alpha }_1+h_{13}\cos {\alpha }_2+h_{16}\cos {\alpha }_3+2m\cos r}\\ \multicolumn{1}{c}{h_{11}\sin {\alpha }_1+h_{13}\sin {\alpha }_2+h_{15}+h_{17}}\end{array}\right]\hfill \end{array}$$

Similarly, based on the rod length constraint condition $l_{S31}{l}_{S32}=h_{32}$ , the following equation can be gained

$$\begin{array}{c}f({\alpha }_2^{*})=(-h_{12}-h_{14}-h_{16}\sin {\alpha }_3-2m\sin \gamma -h_{31}\cos \theta {)}^2\hfill \\ +(-2a+h_{11}\cos {\alpha }_1+h_{13}\cos {\alpha }_2\hfill \\ +h_{16}\cos {\alpha }_3+2m\cos \gamma {)}^2\hfill \\ +(h_{11}\sin {\alpha }_1+h_{13}\sin {\alpha }_2+h_{15}+h_{17}-h_{31}\sin \theta {)}^2\hfill \\ -h_{32}^2\hfill \end{array}$$

(14)

By changing the value ${\alpha }_2^{*}$ , the function relation $f({\alpha }_2^{*})=0$ can be obtained, and the position and attitude of the center point of the moving platform can be obtained by putting the ${\alpha }_2^{*}$ satisfying $f({\alpha }_2^{*})=0$ into the previous equation.

The inverse solution analysis

For this mechanism, the inverse problem can be summed up as follows: when the position $O\prime (x,y,z)$ and attitude angle $\gamma$ of the moving platform are known, the input angles ${\alpha }_1$ , $\beta$ , $\theta$ , and $\delta$ can be obtained.

Solution of input angle ${\alpha }_1$

According to equation (8), the x coordinates of $O\prime$ can be obtained

$$x=-h_{12}-h_{14}-h_{16}\sin {\alpha }_3-m\sin \gamma -\frac{\sqrt{3}}{3}m\cos \gamma$$

(15)

Then

$${\alpha }_3=\arcsin \frac{-h_{12}-h_{14}-m\sin \gamma -\frac{\sqrt{3}}{3}m\cos \gamma -x}{h_{16}}$$

(16)

According to equation (8), the $y,z$ coordinates of $O\prime$ can be obtained

$$\begin{array}{cc}\multicolumn{1}{c}{y=}& -a+h_{11}\cos {\alpha }_1+h_{13}\cos {\alpha }_2+h_{16}\cos {\alpha }_3\hfill \\ \multicolumn{1}{c}{}& +m\cos \gamma -\frac{\sqrt{3}}{3}m\sin \gamma \hfill \end{array}$$

(17)

$$z=h_{11}\sin {\alpha }_1+h_{13}\sin {\alpha }_2+h_{15}+h_{17}$$

(18)

Then, we have

$$-h_{13}\cos {\alpha }_2=P_0+h_{11}\cos {\alpha }_1$$

(19)

$$-h_{13}\sin {\alpha }_2=P_1+h_{11}\sin {\alpha }_1$$

(20)

where

$$\begin{array}{c}{P}_0=-a+h_{16}\cos {\alpha }_3+m\cos \gamma -\frac{\sqrt{3}}{3}m\sin \gamma -y\hfill \\ P_1=h_{15}+h_{17}-z\hfill \end{array}$$

To eliminate ${\alpha }_2$ can achieve

$$P_2\sin {\alpha }_1+P_3\cos {\alpha }_1+P_4=0$$

(21)

where

$$\begin{array}{c}{P}_2=2P_1{h}_{11},P_3=2P_0{h}_{11},P_4=P_0^2+P_1^2+h_{11}^2-h_{13}^2\hfill \\ k_3=\tan \frac{{\alpha }_1}{2}\hfill \end{array}$$

(22)

We can understand the following formula

$$k_3=\frac{-P_2\pm \sqrt{P_2^2+P_3^2-P_4^2}}{P_4-P_3}$$

(23)

Thus, the input angle ${\alpha }_1$ can be solved.

Solution of input angle $\beta$

According to the topological structure $\{-R_{21}-◇(4S)-R_{22}-R_{15}-\}$ of branched chain II, the coordinates of $R_{21}$ , e, f, $R_{22}$ , and $R_{15}$ can be obtained. The coordinates of the center point $O\prime$ of the moving platform can be obtained by vector operation $\overrightarrow{\mathrm{OO}\prime }=\overrightarrow{O{R}_{15}}-\overrightarrow{O\prime R_{15}}$

$$\begin{array}{c}\left[\begin{array}{c}{x}_{o\prime }\hfill \\ \multicolumn{1}{c}{y_{o\prime }}\\ \multicolumn{1}{c}{z_{o\prime }}\end{array}\right]=\hfill \\ \left[\begin{array}{c}a+h_{21}\cos \beta +h_{22}\cos \sigma \sin \mathrm{\Omega }+h_{23}\sin \phi -m\sin \gamma -\frac{\sqrt{3}}{3}m\cos \gamma \hfill \\ \multicolumn{1}{c}{h_{22}\cos \mathrm{\Omega }-h_{23}\cos \phi +m\cos \gamma -\frac{\sqrt{3}}{3}m\sin \gamma }\\ \multicolumn{1}{c}{h_{21}\sin \beta +h_{22}\sin \sigma \sin \mathrm{\Omega }+h_{17}}\end{array}\right]\hfill \end{array}$$

(24)

According to equation (24), the y coordinate of $O\prime$ can be obtained

$$y=h_{22}\cos \mathrm{\Omega }-h_{23}\cos \phi +m\cos \gamma -\frac{\sqrt{3}}{3}m\sin \gamma$$

(25)

$$\mathrm{\Omega }=\arccos \frac{h_{23}\cos \phi -m\cos \gamma +\frac{\sqrt{3}}{3}m\sin \gamma +y}{h_{22}}$$

(26)

According to equation (24), the $x\mathrm{and}z$ coordinates of $O\prime$ can be obtained

$$\begin{array}{cc}\multicolumn{1}{c}{x=}& a+h_{21}\cos \beta +h_{22}\cos \sigma \sin \mathrm{\Omega }+h_{23}\sin \phi \hfill \\ \multicolumn{1}{c}{}& -m\sin \gamma -\frac{\sqrt{3}}{3}m\cos \gamma \hfill \end{array}$$

(27)

$$z=h_{21}\sin \beta +h_{22}\sin \sigma \sin \mathrm{\Omega }+h_{17}$$

(28)

Then, we have

$$-h_{22}\cos \sigma \sin \mathrm{\Omega }=h_{21}\cos \beta +P_5$$

(29)

$$-h_{22}\sin \sigma \sin \mathrm{\Omega }=h_{21}\sin \beta +P_6$$

(30)

$$\begin{array}{c}{P}_5=a+h_{23}\sin \phi -m\sin \gamma -\frac{\sqrt{3}}{3}m\cos \gamma -x,\hfill \\ P_6=h_{17}-z\hfill \end{array}$$

The elimination of $\sigma$ can be achieved

$$\begin{array}{c}2h_{21}{P}_6\sin \beta +2h_{21}{P}_5\cos \beta +h_{21}^2\hfill \\ +P_5^2+P_6^2-2h_{22}^2\sin \mathrm{\Omega }=0\hfill \end{array}$$

(31)

$$P_7\sin \beta +P_8\cos \beta +P_9=0$$

(32)

In the above equation

$$\begin{array}{c}{P}_7=2h_{21}{P}_6,P_8=2h_{21}{P}_5,\hfill \\ P_9=h_{21}^2+P_5^2+P_6^2-2h_{22}^2\sin \mathrm{\Omega }\hfill \\ k_4=\tan \frac{\beta }{2}\hfill \end{array}$$

(33)

We can understand the following formula

$$k_4=\frac{-P_7\pm \sqrt{P_7^2+P_8^2-P_9^2}}{P_9-P_8}$$

(34)

Thus, the input angle $\beta$ can be solved.

Solution of input angle $\delta$

The coordinates of $S_{41}$ and $S_{42}$ can be obtained from branched chain IV

$$\begin{array}{cc}\multicolumn{1}{c}{\left[\begin{array}{c}{x}_{S41}\hfill \\ \multicolumn{1}{c}{y_{S41}}\\ \multicolumn{1}{c}{z_{S41}}\end{array}\right]}& =\left[\begin{array}{c}-a+h_{41}\cos \delta \hfill \\ \multicolumn{1}{c}{0}\\ \multicolumn{1}{c}{h_{41}\sin \delta }\end{array}\right],\left[\begin{array}{c}{x}_{S42}\hfill \\ \multicolumn{1}{c}{y_{S42}}\\ \multicolumn{1}{c}{z_{S42}}\end{array}\right]\hfill \\ \multicolumn{1}{c}{}& =\left[\begin{array}{c}x-\frac{2\sqrt{3}}{3}m\cos \gamma \hfill \\ \multicolumn{1}{c}{y-\frac{2\sqrt{3}}{3}m\sin \gamma }\\ \multicolumn{1}{c}{z}\end{array}\right]\hfill \end{array}$$

According to rod length constraint relation $l_{S41}{l}_{S42}=h_{42}$ , we have

$$Q_1\sin \delta +Q_2\cos \delta +Q_3=0$$

(35)

where

$$\begin{array}{c}{Q}_1=-2z{h}_{41},Q_2=-2(x-\frac{2\sqrt{3}}{3}m\cos \gamma +a)h_{41}\hfill \\ Q_3={(x-\frac{2\sqrt{3}}{3}m\cos \gamma +a)}^2+{(y-\frac{2\sqrt{3}}{3}m\sin \gamma )}^2\hfill \\ +z^2+h_{41}^2-h_{42}^2\hfill \\ k_5=\tan \frac{\delta }{2}\hfill \end{array}$$

(36)

We can understand the following formula

$$k_5=\frac{-Q_1\pm \sqrt{Q_1^2+Q_2^2-Q_3^2}}{Q_3-Q_2}$$

(37)

Thus, the input angle $\delta$ can be solved.

Solution of input angle $\theta$

The coordinates of $S_{31}$ and $S_{32}$ can be obtained from branched chain IV

$$\begin{array}{cc}\multicolumn{1}{c}{\left[\begin{array}{c}{x}_{S31}\hfill \\ \multicolumn{1}{c}{y_{S31}}\\ \multicolumn{1}{c}{z_{S31}}\end{array}\right]}& =\left[\begin{array}{c}{h}_{41}\cos \theta \hfill \\ \multicolumn{1}{c}{a}\\ \multicolumn{1}{c}{h_{41}\sin \theta }\end{array}\right],\left[\begin{array}{c}{x}_{S42}\hfill \\ \multicolumn{1}{c}{y_{S42}}\\ \multicolumn{1}{c}{z_{S42}}\end{array}\right]\hfill \\ \multicolumn{1}{c}{}& =\left[\begin{array}{c}x+\frac{\sqrt{3}}{3}m\cos \gamma -m\sin \gamma \hfill \\ \multicolumn{1}{c}{0}\\ \multicolumn{1}{c}{y+\frac{\sqrt{3}}{3}m\sin \gamma +m\cos \gamma }\end{array}\right]\hfill \end{array}$$

According to rod length constraint relation $l_{S31}{l}_{S32}=h_{32}$ , we have

$$Q_4\sin \theta +Q_5\cos \theta +Q_6=0$$

(38)

where

$$\begin{array}{c}{Q}_4=-2z{h}_{31},Q_5=-2(x+\frac{\sqrt{3}}{3}m\cos \gamma -m\sin \gamma )h_{31}\hfill \\ Q_6={(x+\frac{\sqrt{3}}{3}m\cos \gamma -m\sin \gamma )}^2\hfill \\ +{(y+\frac{\sqrt{3}}{3}m\sin \gamma +m\cos \gamma -a)}^2+z^2+h_{31}^2-h_{32}^2\hfill \\ k_6=\tan \frac{\theta }{2}\hfill \end{array}$$

(39)

We can understand the following formula

$$k_6=\frac{-Q_4\pm \sqrt{Q_4^2+Q_5^2-Q_6^2}}{Q_6-Q_5}$$

(40)

Thus, the input angle $\theta$ can be solved. According to the above analysis, there are $2^4=16$ sets of inverse solutions.

Verification of forward and inverse solutions

The structural parameters of a given parallel mechanism are $h_{11}=205$ ; $h_{12}=0$ ; $h_{13}=20$ ; $h_{14}=0$ ; $h_{15}=60$ ; $h_{16}=100$ ; $h_{17}=60$ ; $h_{21}=205$ ; $h_{22}=355$ ; $h_{23}=100$ ; $h_{31}=205$ ; $h_{32}=500$ ; $h_{41}=205$ ; $h_{42}=500$ ; $a=200$ ; and $m=150$ , assuming that four driving angles are ${\alpha }_1=127.{4269}^0$ , $\beta =17.{7558}^0$ , $\delta =91.{3969}^0$ , and $\theta =110.{8537}^0$ .

When the virtual variable ${\alpha }_2^{*}$ is searched between $0\mathrm{and}2\pi$ , the forward position solution of the mechanism can be obtained from equations (8)–(14), as shown in Table 1.

Table 1.

Numerical solutions of direct kinematics.

$\mathrm{No}.$	$\mathrm{X}(\mathrm{cm})$	$\mathrm{Y}(\mathrm{cm})$	$\mathrm{Z}(\mathrm{cm})$	$\gamma (°)$
1	−14.7860	−12.0186	14.0088	53.2965
2	10.0000	10.5000	35.0000	5.0000
3	−16.5975	1.0676	35.7259	5.3342
4	−8.1544	5.7695	37.1486	2.5898

The second set of forward solutions is substituted into the inverse equations (15)–(40), and 16 sets of inverse solutions of ${\alpha }_1$ , $\beta$ , $\delta$ , and $\theta$ are obtained, as shown in Table 2.

Table 2.

Numerical solutions of inverse kinematics.

$\mathrm{No}.$	${\alpha }_1(°)$	$\beta (°)$	$\delta (°)$	$\theta (°)$
1	127.4269	−131.6271	91.3969	110.8537
2	127.4269	−131.6271	91.3969	−5.8208
3	127.4269	−131.6271	17.3398	110.8537
4	127.4269	−131.6271	17.3398	−5.8208
5	127.4269	17.7558	91.3969	110.8537
6	127.4269	17.7558	91.3969	−5.8208
7	127.4269	17.7558	17.3398	110.8537
8	127.4269	17.7558	17.3398	−5.8208
9	−19.6492	−131.6271	91.3969	110.8537
10	−19.6492	−131.6271	91.3969	−5.8208
11	−19.6492	−131.6271	17.3398	110.8537
12	−19.6492	−131.6271	17.3398	−5.8208
13	−19.6492	17.7558	91.3969	110.8537
14	−19.6492	17.7558	91.3969	−5.8208
15	−19.6492	17.7558	17.3398	110.8537
16	−19.6492	17.7558	17.3398	−5.8208

It can be seen from the fifth set of data values that it is consistent with the input angle value, which proves the correctness of the direct and inverse solutions. In most cases, the inverse and the forward kinematics are the key to kinematics analysis and control of parallel robots, so the forward and inverse kinematics analysis of robots is not only an important part of the follow-up analysis but also the basic link of the mechanism research.

Workspace analysis

Workspace is an important index for measuring the working characteristics of robots. At present, the methods of workspace analysis mainly include Monte Carlo method, ²³ polar coordinate search method, ²⁴ and graphic method. ²⁵ The workspace analysis of the decoupled mechanism is an indispensable work and an important prerequisite for the study of such mechanism.

The forward solution of parallel robots has great complexity, while the inverse solution is relatively simple. In this article, inverse kinematics is used to solve the workspace of decoupling robot. In solving the workspace of symmetrical and asymmetrical mechanisms, because of the irregularity of the workspace of mechanisms, it is almost impossible to use analytical and graphical methods. The numerical rules provide a very simple solution step.

According to the set of parameters, the searching limits are $0\le z\le 60$ , $-\pi \le \psi \le \pi$ , and $0\le \rho \le 50$ . The constraint conditions are as follows: $-\pi \le {\alpha }_1\le \pi$ , $-\pi \le \beta \le \pi$ , $-\pi \le \delta \le \pi$ , and $-\pi \le \theta \le \pi$ . The workspace sectional view of the decoupling mechanism can be obtained, as shown in Figure 6. This article mainly observes the change of workspace in different sections and the influence of each branch chain on the shape of workspace, so the workspace under four different sections is given.

Figure 6.

Workspace section of coupling-reducing mechanism: (a) $Z=10\mathrm{cm}$ , (b) $Z=20\mathrm{cm}$ , (c) $Z=30\mathrm{cm}$ , and (d) $Z=40\mathrm{cm}$ .

Figure 6 shows that the workspace of parallel mechanism becomes larger and larger with the increase of cross-section, and there is no cavity in the workspace. The area in the lower-right corner of the workspace is basically S-S symmetric because the two unrestricted branches R_SS act on the workspace the same way. In the upper-left corner of the workspace, the S-S asymmetry of the diagonal line is obvious, which is caused by the different structures of branched chains I and II.

In order to evaluate the performance of the 4-DOF parallel robot and other parallel mechanisms with three translations and one rotation, we can use the total workspace size to characterize its working ability. Here, we use the polar coordinate search method to obtain the workspace volume of the mechanism. The basic idea of solving parallel mechanism workspace with numerical method: using the plane that is parallel with XOY, the workspace is cut into a series of micro molecular spaces, with thickness of $\mathrm{\Delta }z$ . If $\mathrm{\Delta }z$ is small enough, each micro molecular space can be regarded as a column with the height of $\mathrm{\Delta }z$ , so that the volume of the mechanism workspace is obtained by superimposing the volume of all cylinders.

Any points on the bottom boundary of each micro molecule space can be expressed in polar coordinates $(p,\theta )$ , where $p=\sqrt{x^2+y^2}$ , $\theta \in (0°,360°)$ , First, assuming that $m_0(p_0,{\theta }_0)$ is the bottom boundary point of micro molecule space, then setting the step length increment $\mathrm{\Delta }\theta$ , we will obtain the next boundary point $m_1(p_1,{\theta }_0+\mathrm{\Delta }\theta )$ . In turn, you can get all the boundary points on that surface.

According to these boundary points, the cylinder can be divided into several small differential regions, which are labeled as $\mathrm{\Delta }{A}_k$ . When the step length increment is small enough, these small regions can be regarded as sector, whose angle is $\mathrm{\Delta }\theta$ . Each sector area $\mathrm{\Delta }{A}_k$ can be obtained by sector area formula

$$\mathrm{\Delta }{A}_k=\frac{1}{2}{p}_k^2\mathrm{\Delta }\theta (k=0,1,2,\dots )$$

(41)

Therefore, the bottom area of the cylinder can be obtained by the superposition of all the differential sectorial areas, and the volume $\mathrm{\Delta }{V}_k$ of the differential body can get by the volume formula of the cylinder

$$\mathrm{\Delta }{V}_i=\sum _i\mathrm{\Delta }{A}_k\mathrm{\Delta }z=\frac{1}{2}\sum _i{p}_{\mathrm{ki}}^2\mathrm{\Delta }\theta \mathrm{\Delta }z(k=0,1,2,\dots )$$

(42)

Therefore, the workspace of parallel mechanism is as follows

$$V_1=\sum _i\mathrm{\Delta }{V}_i=\frac{1}{2}\sum _j\sum _k{p}_{\mathrm{ki}}^2\mathrm{\Delta }\theta \mathrm{\Delta }z$$

(43)

The workspace volume of Cross IV-3 mechanism is 4.4274 × 10⁸ mm³ by obtaining the mechanism parameters of document; ²⁶ however, the workspace volume of coupling-reducing mechanism is 4.831 × 10⁸ mm³. Therefore, the workspace of the coupling-reducing mechanism is 8.35% larger than that of Cross IV-3. It proves that the mechanism has greater flexibility in space.

Conclusion

1. In this article, a new type of asymmetric parallel robot is proposed based on the theory of POC set, which can realize 3D movement and 1D rotation.

2. It is found that the coupling degree of the parallel mechanism is 2, and its kinematics, dynamics, and control are very difficult. Therefore, the structure-decoupling method is used to reducing the coupling degree. The coupling degree of the coupling-reducing mechanism was reduced to 1, which can be easily controlled.

3. Aiming at the difficulty of solving the forward solution of parallel robot, the forward solution of the mechanism is obtained by using ordered SOC method, and the correctness of the forward and inverse solution is verified by an example.

4. Through the inverse equation of the parallel mechanism, the workspace of the robot is plotted. It can be seen that the parallel robot has a larger workspace, and the influence of the branch chain on the workspace of the robot can be observed intuitively. Simultaneously, there is no void phenomenon in the workspace, so that the continuity of machining process can be ensured.

5. The decoupling mechanism consists of two unconstrained branched chains, which are simple in structure and easy to manufacture. Because of this, they have certain research and development value.

6. In order to evaluate the performance of the proposed parallel mechanism, the workspace volume of the mechanism is obtained using the polar coordinate search method, and the rationality of the proposed mechanism is proved by comparing it with the parallel robot of CrossIV-3 under the same parameters.