A New Method for Displacement Modelling of Serial Robots Using Finite Screw

Abstract

Kinematics is a hot topic in robotic research, serving as a foundational step in the synthesis and analysis of robots. Forward kinematics and inverse kinematics are the prerequisite and foundation for motion control, trajectory planning, dynamic simulation, and precision guarantee of robotic manipulators. Both of them depend on the displacement models. Compared with the previous work, finite screw is proven to be the simplest and nonredundant mathematical tool for displacement description. Thus, it is used for displacement modelling of serial robots in this paper. Firstly, a finite-screw-based method for formulating displacement model is proposed, which is applicable for any serial robot. Secondly, the procedures for forward and inverse kinematics by solving the formulated displacement equation are discussed. Then, two typical serial robots with three translations and two rotations are taken as examples to illustrate the proposed method. Finally, through Matlab simulation, the obtained analytical expressions of kinematics are verified. The main contribution of the proposed method is that finite-screw-based displacement model is highly related with instantaneous-screw-based kinematic and dynamic models, providing an integrated modelling and analysis methodology for robotic mechanisms.

1. Introduction

Robots have been widely utilized in industrial application in recent years [1,2]. Kinematics, including forward kinematics and inverse kinematics, is a fundamental problem in the synthesis and analysis of robots. It is the prerequisite for motion control, trajectory planning, dynamic simulation, and precision guarantee of robotic manipulators. Both forward kinematics and inverse kinematics depend on displacement modelling. Through describing displacements of robots by mathematical tools, displacement models can be formulated in an algebraic way [3,4,5]. The two main issues involved in displacement modelling of robots are as follows:

(1) Select a mathematical tool to analytically express the displacements generated by the joints of the robots [6];

(2) Formulate the displacement equations of the robots by compositing the joints’ displacements [7].

Up to now, two main kinds of methods for displacement modelling of serial robots (SRs) have been proposed, i.e., vector chain methods and product of exponentials methods. In vector chain methods, three-dimensional position vectors are employed to formulate the displacement equation by means of building mappings between positions and orientations of the end-effector of an SR at the given pose [8]. In the formulated position equations, all joint parameters are independent and decoupled. In the product of exponentials method, exponential matrices (or dual quaternions) containing joint parameters are used to express pose transformations between adjacent links of an SR [9]. The displacement equation can be formulated by multiplying these matrices or (or dual quaternions) together [10,11].

By employing these methods, displacement models of any SR can be formulated. Forward kinematics can be directly carried out by applying the obtained displacement equation [12]. Additionally, some methods for inverse kinematics have also been put forward, which accompany the displacement modelling methods. The works related to vector chain methods have been reported by elimination and engine value analysis of univariate polynomial equations [13]. The works related to the product of exponentials methods can be found by algebraic and geometric approaches [14].

Compared with the mathematical tools used in these existing methods, finite screw has the following advantages in describing displacements of serial robots and their joints:

(1) It has only six items, which is the simplest and nonredundant description of the special Euclidean group (SE(3)) [15].

(2) The algebraic structure of the entire set of finite screws has been proved to be a Lie group under the screw triangle product, while the entire set of instantaneous screw is its corresponding Lie algebra [16].

Considering that the instantaneous screw and its isomorphism have been successfully and widely used in mobility, velocity, dynamic, accuracy, and stiffness modelling of SRs [17,18,19,20], we intend to use the counterpart of it, i.e., finite screw, in implementing displacement modelling. Based upon Dai’s works [21,22,23], finite screw will be used in displacement modelling of SRs in this paper.

The contributions of this paper lie in the following points.

(1) Displacement modelling based upon finite screw formulates the nonredundant and most concise expression for forward and inverse kinematics of SRs [24,25].

(2) The computational cost of displacement modelling is improved by introducing finite screw and the corresponding algorithms of screw triangle product [26].

(3) The formulated displacement models of SRs are highly related with the velocity and higher kinematic models because of the differential mapping between finite and instantaneous screws, providing integrated method for kinematics and dynamics modelling [27,28].

This paper presents a new method for displacement modelling of SRs using finite screws as the mathematical tool, making it applicable to any SR. The organization of this paper is as follows. After a brief review of the state-of-the-art of the existing methods for displacement modelling and kinematics in Section 1, Section 2 presents a new method to algebraically formulate displacement equation of an SR through describing displacements generated by an SR and its joints as finite screws. Then, the procedures of forward and inverse kinematics using the formulated finite-screw-based displacement model are given in Section 3. In Section 4, two topical serial robots are taken as examples to illustrate the proposed method, which is followed by the verification of the correctness of the obtained analytical expressions in Matlab simulation in Section 5. The conclusions are drawn in Section 6.

2. Displacement Modelling Using Finite Screw

In this section, the displacement equation will be formulated through describing displacements generated by an SR and its joints by finite screws, which clearly shows the algebraic mappings between all the joint parameters and the pose of the end-effector [29].

In Chasles’ theorem, a rigid body displacement from one configuration to another configuration can always be regarded as the rigid body rotating around an axis together with translating along the axis. The axis is called Chasles’ axis. As is well known, for the Chasles’ axis, the rotational angle and translational distance about/along that axis are the basic elements of displacement [30,31]. Among all the mathematical tools used to describe displacement, the finite screw in quasi-vector form can clearly express these basic elements in the simplest and most nonredundant manner, which can be analytically composited [32,33]. Generally, a displacement of a rigid body can be expressed by a finite screw as follows:

$${\mathit{S}}_f=2tan{\displaystyle \frac{\mathit{\theta }}{2}}\left(\begin{array}{c}{\mathit{s}}_f\\ {\mathit{r}}_f\times {\mathit{s}}_f\end{array}\right)+\mathit{\tau }\left(\begin{array}{c}\mathbf{0}\\ {\mathit{s}}_f\end{array}\right),$$

(1)

where ${\mathit{s}}_f$ ($\left|{\mathit{s}}_f\right|=1$) and ${\mathit{r}}_f$ denote the unit direction vector and position vector of the Chasles’ axis, $\left({\mathit{s}}_f;{\mathit{r}}_f\times {\mathit{s}}_f\right)$ is the Plücker coordinates of the screw axis of the displacement, and $\mathit{\theta }$ and $\mathit{\tau }$ are the rotational angle and translational distance about/along that axis [34,35].

Suppose that a rigid body realizes two successive displacements expressed by

$${\mathit{S}}_{f,i}=2tan{\displaystyle \frac{{\mathit{\theta }}_i}{2}}\left(\begin{array}{c}{\mathit{s}}_{f,i}\\ {\mathit{r}}_{f,i}\times {\mathit{s}}_{f,i}\end{array}\right)+{\mathit{\tau }}_i\left(\begin{array}{c}\mathbf{0}\\ {\mathit{s}}_{f,i}\end{array}\right), i=a,b.$$

(2)

The screw triangle product [23] is the composition algorithm of finite screws. The composition of two finite screws results in the linear addition of these two screws, the screw along their common perpendicular, and their translational parts. The resultant displacement of this rigid body can be obtained by compositing the two finite screws using the screw triangle product,

$${\mathit{S}}_{f,ab}={\mathit{S}}_{f,a}\overline{\circ }{\mathit{S}}_{f,b}={\displaystyle \frac{{\mathit{S}}_{f,a}+{\mathit{S}}_{f,b}+\frac{{\mathit{S}}_{f,b}\times {\mathit{S}}_{f,a}}{2}-tan\frac{{\mathit{\theta }}_a}{2}tan\frac{{\mathit{\theta }}_b}{2}\left({\mathit{\tau }}_b\left(\begin{array}{c}\mathbf{0}\\ {\mathit{s}}_{f,a}\end{array}\right)+{\mathit{\tau }}_a\left(\begin{array}{c}\mathbf{0}\\ {\mathit{s}}_{f,b}\end{array}\right)\right)}{1-tan\frac{{\mathit{\theta }}_a}{2}tan\frac{{\mathit{\theta }}_b}{2}{\mathit{s}}_{f,a}^{\mathrm{T}}{\mathit{s}}_{f,b}}},$$

(3)

where the symbol “$\overline{\circ }$” is used to denote the screw triangle product.

By employing finite screws to describe the displacements generated by an SR and its joints, the displacement model of the robot can be directly formulated utilizing screw triangle products [36,37]. In this way, for an SR constituted by n one-DoF joints (revolute (R) and prismatic (P) joints), its displacement model can be obtained as follows using Equations (1) and (3),

$${\mathit{S}}_{f,\mathrm{SR}}={\mathit{S}}_{f,n}\overline{\circ }{\mathit{S}}_{f,n-1}\overline{\circ }\cdots \overline{\circ }{\mathit{S}}_{f,1},$$

(4)

where ${\mathit{S}}_{f,\mathrm{SR}}$ and ${\mathit{S}}_{f,k}$ ($k=1, 2,\cdots , n$) denote the displacements generated by the SR and its kth joint measured from a preset initial pose. According to Equation (1), ${\mathit{S}}_{f,k}$ can be expressed as follows:

$${\mathit{S}}_{f,k}=\left\{\begin{array}{cc}2tan{\displaystyle \frac{{\mathit{\theta }}_k}{2}}\left(\begin{array}{c}{\mathit{s}}_k\\ {\mathit{r}}_k\times {\mathit{s}}_k\end{array}\right)& \mathrm{R joint}\\ {\mathit{\tau }}_k\left(\begin{array}{c}\mathbf{0}\\ {\mathit{s}}_k\end{array}\right)& \mathrm{P joint}\end{array}\right.$$

(5)

where ${\mathit{s}}_k$ and ${\mathit{r}}_k$ are the unit direction vector and position vector of the kth joint at its initial pose; ${\mathit{\theta }}_k$ (${\mathit{\tau }}_k$) is the rotational angle (translational distance) of the kth joint with respect to its initial pose [38,39]. Conventionally, joints in the SR are numbered from the fixed base to the end-effector. Equation (4) is the finite-screw-based displacement model (displacement equation) of the SR [40,41].

3. Kinematics Using Finite Screw

Having the finite-screw-based displacement equation at hand, the kinematics of SRs can be carried out [42,43]. The procedures for forward and inverse kinematics by utilizing this displacement equation will be discussed in this section.

According to Chasles’ theorem and screw triangle product, the composition of several joint displacements can always be rewritten as a finite screw [44,45,46]. Hence, the displacement model of SR can be rewritten into the following form:

$${\mathit{S}}_{f,\mathrm{SR}}={\mathit{S}}_{f,n}\overline{\circ }{\mathit{S}}_{f,n-1}\overline{\circ }\cdots \overline{\circ }{\mathit{S}}_{f,1}=2tan{\displaystyle \frac{{\mathit{\theta }}_C}{2}}\left(\begin{array}{c}{\mathit{s}}_C\\ {\mathit{r}}_C\times {\mathit{s}}_C\end{array}\right)+{\mathit{\tau }}_C\left(\begin{array}{c}\mathbf{0}\\ {\mathit{s}}_C\end{array}\right),$$

(6)

where ${\mathit{s}}_C$ and ${\mathit{r}}_C$ are unit direction vector and position vector of the axis of the composition displacement, and ${\mathit{\theta }}_C$ and ${\mathit{\tau }}_C$ are rotational angle and translational distance about and along that axis [47,48]. It is easy to see that they are all functions of the joint parameters ${\mathit{\theta }}_k$ and ${\mathit{\tau }}_k$ ($k=1, 2,\cdots , n$) as

$${\mathit{s}}_C=f_s\left({\mathit{\theta }}_1,{\mathit{\theta }}_2,\cdots ,{\mathit{\theta }}_n\right), {\mathit{r}}_C=f_r\left({\mathit{\theta }}_1/{\mathit{\tau }}_1,{\mathit{\theta }}_2/{\mathit{\tau }}_2,\cdots ,{\mathit{\theta }}_n/{\mathit{\tau }}_n\right),$$

$$tan{\displaystyle \frac{{\mathit{\theta }}_C}{2}}=f_{\mathit{\theta }}\left({\mathit{\theta }}_1,{\mathit{\theta }}_2,\cdots ,{\mathit{\theta }}_n\right), {\mathit{\tau }}_C=f_t\left({\mathit{\theta }}_1/{\mathit{\tau }}_1,{\mathit{\theta }}_2/{\mathit{\tau }}_2,\cdots ,{\mathit{\theta }}_n/{\mathit{\tau }}_n\right).$$

(7)

We write the current pose of the SR’s end-effector as

$${\mathit{S}}_{f,\mathrm{SR}}=2tan{\displaystyle \frac{{\mathit{\theta }}_{\mathrm{SR}}}{2}}\left(\begin{array}{c}{\mathit{s}}_{\mathrm{SR}}\\ {\mathit{r}}_{\mathrm{SR}}\times {\mathit{s}}_{\mathrm{SR}}\end{array}\right)+{\mathit{\tau }}_{\mathrm{SR}}\left(\begin{array}{c}\mathbf{0}\\ {\mathit{s}}_{\mathrm{SR}}\end{array}\right).$$

(8)

Because the composition displacement generated by all the joints is equivalent to the pose of the end-effector [49,50,51], Equation (6) can be rewritten into the following three mappings between the joint parameters and the end-effector pose,

$${\mathit{s}}_{\mathrm{SR}}=f_s\left({\mathit{\theta }}_1,{\mathit{\theta }}_2,\cdots ,{\mathit{\theta }}_n\right),$$

(9)

$$tan{\displaystyle \frac{{\mathit{\theta }}_{\mathrm{SR}}}{2}}=f_{\mathit{\theta }}\left({\mathit{\theta }}_1,{\mathit{\theta }}_2,\cdots ,{\mathit{\theta }}_n\right),$$

(10)

$$\begin{array}{l}2tan{\displaystyle \frac{{\mathit{\theta }}_{\mathrm{SR}}}{2}}{\mathit{r}}_{\mathrm{SR}}\times {\mathit{s}}_{\mathrm{SR}}+{\mathit{\tau }}_{\mathrm{SR}}{\mathit{s}}_{\mathrm{SR}}\\ =2f_{\mathit{\theta }}\left({\mathit{\theta }}_1,{\mathit{\theta }}_2,\cdots ,{\mathit{\theta }}_n\right)\cdot \left(f_r\left({\mathit{\theta }}_1/{\mathit{\tau }}_1,{\mathit{\theta }}_2/{\mathit{\tau }}_2,\cdots ,{\mathit{\theta }}_n/{\mathit{\tau }}_n\right)\times f_s\left({\mathit{\theta }}_1,{\mathit{\theta }}_2,\cdots ,{\mathit{\theta }}_n\right)\right)\\ +f_t\left({\mathit{\theta }}_1/{\mathit{\tau }}_1,{\mathit{\theta }}_2/{\mathit{\tau }}_2,\cdots ,{\mathit{\theta }}_n/{\mathit{\tau }}_n\right)\cdot f_s\left({\mathit{\theta }}_1,{\mathit{\theta }}_2,\cdots ,{\mathit{\theta }}_n\right)\\ =g\left({\mathit{\theta }}_1/{\mathit{\tau }}_1,{\mathit{\theta }}_2/{\mathit{\tau }}_2,\cdots ,{\mathit{\theta }}_n/{\mathit{\tau }}_n\right)\end{array}$$

(11)

In the above three equations, $f_s$ and $f_{\mathit{\theta }}$ are the mappings between the rotational parameters of joints and the end-effector orientation, and $g$ is the mapping between the joint parameters and the end-effector position [52,53,54]. They clearly reveal the algebraic mappings between the joint parameters and the end-effector pose [55,56,57]. In these mappings, ${\mathit{s}}_k$ and ${\mathit{r}}_k$ are determined by the preset initial pose of the SR. They are known quantities, which are invariable with the pose that the end-effector moves to [58,59,60]. In forward kinematics, the n joint parameters, ${\mathit{\theta }}_1$/${\mathit{\tau }}_1$, ${\mathit{\theta }}_2$/${\mathit{\tau }}_2$, …, ${\mathit{\theta }}_n$/${\mathit{\tau }}_n$, are given, and the displacement ${\mathit{S}}_{f,\mathrm{SR}}$ generated by the SR with respect to the initial pose, i.e., the current pose of the end-effector, is what needs to be solved. In inverse kinematics, the displacement of the SR is given, and the n joint parameters are the variables needed to be solved. It is noted that the symbol “/” here means “or”, because only one of ${\mathit{\theta }}_k$ and ${\mathit{\tau }}_k$ exists [61,62,63].

From the above analysis, the procedures for forward and inverse kinematics utilizing finite screw can be concluded as follows:

(1) Formulate the displacement model of an SR using the finite screws generated by its joints [64,65].

(2) Rewrite the displacement equation into the mappings between the joint parameters of the end-effector pose [66].

(3) Given the joint parameters, the forward kinematics leads to the end-effector pose [67,68].

(4) Given the end-effector pose, the inverse kinematics results in the joint parameters [69,70].

The procedures for forward and inverse kinematics are illustrated in Figure 1.

Through vector and polynomial analysis, the analytical solutions of forward and inverse kinematics can be derived by following the procedures [71,72]. In the next section, two topical SRs will be taken as examples to show the validity of the proposed finite screw method.

4. Examples

The method and procedures proposed in this paper can be applied to displacement modelling and kinematics of SRs. In this section, P₁P₂P₃R_aR_b and R_aR_aR_aR_bR_b robots with three translations and two rotations are taken as typical examples to illustrate the method. In order to show the procedures more clearly, we firstly solve the P₁P₂P₃R_aR_b robot, and then use the result to solve R_aR_aR_aR_bR_b. Here, the subscript of each R or P joint denotes its direction.

4.1. P₁P₂P₃R_aR_b Robot

As shown in Figure 2, P₁P₂P₃R_aR_b is the three-translational and two-rotational robot with the simplest structure. It generates two rotations with fixed directions. Its displacement equation can be formulated based upon Equation (4),

$${\mathit{S}}_{f,\mathrm{SR}}=2tan{\displaystyle \frac{{\mathit{\theta }}_b}{2}}\left(\begin{array}{c}{\mathit{s}}_b\\ {\mathit{r}}_b\times {\mathit{s}}_b\end{array}\right)\overline{\circ }2tan{\displaystyle \frac{{\mathit{\theta }}_a}{2}}\left(\begin{array}{c}{\mathit{s}}_a\\ {\mathit{r}}_a\times {\mathit{s}}_a\end{array}\right)\overline{\circ }{\mathit{\tau }}_3\left(\begin{array}{c}\mathbf{0}\\ {\mathit{s}}_3\end{array}\right)\overline{\circ }{\mathit{\tau }}_2\left(\begin{array}{c}\mathbf{0}\\ {\mathit{s}}_2\end{array}\right)\overline{\circ }{\mathit{\tau }}_1\left(\begin{array}{c}\mathbf{0}\\ {\mathit{s}}_1\end{array}\right),$$

(12)

where ${\mathit{s}}_a$ and ${\mathit{s}}_b$, respectively, denote the unit direction vectors of R_a and R_b, ${\mathit{r}}_a$ and ${\mathit{r}}_b$ denote the position vectors of the two R joints, ${\mathit{\theta }}_a$ and ${\mathit{\theta }}_b$ denote their rotational angles, ${\mathit{s}}_1$, ${\mathit{s}}_2$, and ${\mathit{s}}_3$ are the unit vectors of P₁, P_2, and P₃, and ${\mathit{\tau }}_1$, ${\mathit{\tau }}_2$, and ${\mathit{\tau }}_3$ are their translational distances.

Computing the composition displacements generated by all joints in P₁P₂P₃R_aR_b, Equation (12) can be rewritten as follows by utilizing the screw triangle product,

$$2tan{\displaystyle \frac{{\mathit{\theta }}_{\mathrm{SR}}}{2}}\left(\begin{array}{c}{\mathit{s}}_{\mathrm{SR}}\\ {\mathit{r}}_{\mathrm{SR}}\times {\mathit{s}}_{\mathrm{SR}}+{\displaystyle \frac{{\mathit{t}}_{\mathrm{SR}}}{2tan\frac{{\mathit{\theta }}_{\mathrm{SR}}}{2}}}{\mathit{s}}_{\mathrm{SR}}\end{array}\right)=2tan{\displaystyle \frac{{\mathit{\theta }}_C}{2}}\left(\begin{array}{c}{\mathit{s}}_C\\ {\mathit{p}}_{ba}+{\displaystyle \frac{\mathit{t}\times {\mathit{s}}_C}{2}}+{\displaystyle \frac{\mathit{t}}{2tan\frac{{\mathit{\theta }}_C}{2}}}\end{array}\right),$$

(13)

where

$$tan{\displaystyle \frac{{\mathit{\theta }}_{\mathrm{SR}}}{2}}=tan{\displaystyle \frac{{\mathit{\theta }}_C}{2}}={\displaystyle \frac{\left|tan\frac{{\mathit{\theta }}_a}{2}{\mathit{s}}_a+tan\frac{{\mathit{\theta }}_b}{2}{\mathit{s}}_b+tan\frac{{\mathit{\theta }}_a}{2}tan\frac{{\mathit{\theta }}_b}{2}\left({\mathit{s}}_a\times {\mathit{s}}_b\right)\right|}{1-tan\frac{{\mathit{\theta }}_a}{2}tan\frac{{\mathit{\theta }}_b}{2}{\mathit{s}}_a^{\mathrm{T}}{\mathit{s}}_b}},$$

$${\mathit{s}}_{\mathrm{SR}}={\mathit{s}}_C={\displaystyle \frac{tan\frac{{\mathit{\theta }}_a}{2}{\mathit{s}}_a+tan\frac{{\mathit{\theta }}_b}{2}{\mathit{s}}_b+tan\frac{{\mathit{\theta }}_a}{2}tan\frac{{\mathit{\theta }}_b}{2}\left({\mathit{s}}_a\times {\mathit{s}}_b\right)}{\left|tan\frac{{\mathit{\theta }}_a}{2}{\mathit{s}}_a+tan\frac{{\mathit{\theta }}_b}{2}{\mathit{s}}_b+tan\frac{{\mathit{\theta }}_a}{2}tan\frac{{\mathit{\theta }}_b}{2}\left({\mathit{s}}_a\times {\mathit{s}}_b\right)\right|}},$$

$${\mathit{p}}_{ba}={\displaystyle \frac{tan\frac{{\mathit{\theta }}_a}{2}\left({\mathit{r}}_a\times {\mathit{s}}_a\right)+tan\frac{{\mathit{\theta }}_b}{2}\left({\mathit{r}}_b\times {\mathit{s}}_b\right)+tan\frac{{\mathit{\theta }}_a}{2}tan\frac{{\mathit{\theta }}_b}{2}\left({\mathit{s}}_a\times \left({\mathit{r}}_b\times {\mathit{s}}_b\right)+\left({\mathit{r}}_a\times {\mathit{s}}_a\right)\times {\mathit{s}}_b\right)}{\left|tan\frac{{\mathit{\theta }}_a}{2}{\mathit{s}}_a+tan\frac{{\mathit{\theta }}_b}{2}{\mathit{s}}_b+tan\frac{{\mathit{\theta }}_a}{2}tan\frac{{\mathit{\theta }}_b}{2}\left({\mathit{s}}_a\times {\mathit{s}}_b\right)\right|}}$$

$$\mathit{t}={\mathit{\tau }}_1{\mathit{s}}_1+{\mathit{\tau }}_2{\mathit{s}}_2+{\mathit{\tau }}_3{\mathit{s}}_3$$

Using these equations, the forward kinematics can be derived in a straightforward way, when the joint parameters are given.

When the end-effector pose is given, the two rotational parameters, ${\mathit{\theta }}_a$ and ${\mathit{\theta }}_b$, can be analytically solved through vector analysis,

$${\mathit{\theta }}_a=2\arctan \left({\displaystyle \frac{{\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}\left({\mathit{s}}_a\times {\mathit{s}}_b\right)}{{\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}{\mathit{s}}_b-{\mathit{s}}_a^{\mathrm{T}}{\mathit{s}}_b{\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}{\mathit{s}}_a}}\right), {\mathit{\theta }}_b=2\arctan \left({\displaystyle \frac{{\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}\left({\mathit{s}}_a\times {\mathit{s}}_b\right)}{{\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}{\mathit{s}}_a-{\mathit{s}}_a^{\mathrm{T}}{\mathit{s}}_b{\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}{\mathit{s}}_b}}\right).$$

(14)

In order to solve the three P joint parameters, ${\mathit{\tau }}_1$, ${\mathit{\tau }}_2$, and ${\mathit{\tau }}_3$, the mapping between these three parameters and the given position is built as

$$\mathit{t}={\left({\displaystyle \frac{{\mathit{E}}_3}{2tan\frac{{\mathit{\theta }}_C}{2}}}-{\displaystyle \frac{{\tilde{\mathit{s}}}_C}{2}}\right)}^{-1}\left({\mathit{r}}_{\mathrm{SR}}\times {\mathit{s}}_{\mathrm{SR}}+{\displaystyle \frac{{\mathit{\tau }}_{\mathrm{SR}}}{2tan\frac{{\mathit{\theta }}_{\mathrm{SR}}}{2}}}{\mathit{s}}_{\mathrm{SR}}-{\mathit{p}}_{ba}\right).$$

(15)

where ${\tilde{\mathit{s}}}_{ba}$ denotes the skew-symmetric matrix of ${\mathit{s}}_{ba}$, and ${\mathit{E}}_3$ is a unit matrix of order three.

Computing the projections of the translation vector $\mathit{t}$ on directions of ${\mathit{s}}_1$, ${\mathit{s}}_2$, and ${\mathit{s}}_3$ leads to the parameters ${\mathit{\tau }}_1$, ${\mathit{\tau }}_2$, and ${\mathit{\tau }}_3$,

$${\mathit{\tau }}_1={\displaystyle \frac{{\mathit{t}}^{\mathrm{T}}\left({\mathit{s}}_2\times {\mathit{s}}_3\right)}{{\mathit{s}}_1^{\mathrm{T}}\left({\mathit{s}}_2\times {\mathit{s}}_3\right)}}, {\mathit{\tau }}_2={\displaystyle \frac{{\mathit{t}}^{\mathrm{T}}\left({\mathit{s}}_1\times {\mathit{s}}_3\right)}{{\mathit{s}}_2^{\mathrm{T}}\left({\mathit{s}}_1\times {\mathit{s}}_3\right)}}, {\mathit{\tau }}_3={\displaystyle \frac{{\mathit{t}}^{\mathrm{T}}\left({\mathit{s}}_1\times {\mathit{s}}_2\right)}{{\mathit{s}}_3^{\mathrm{T}}\left({\mathit{s}}_1\times {\mathit{s}}_2\right)}}.$$

(16)

In fact, the displacement equation of any three-translational and two-rotational SR with two fixed rotation directions can be rewritten into a form that is similar to Equation (13). Thus, its two rotational parameters can be obtained using the mapping built in the similar way. The differences between solving different three-translational and two-rotational SRs come from the diversification mappings between the three translational parameters and the given position of end-effectors. Hence, forward and inverse kinematics of the R_aR_aR_aR_bR_b robot can be solved as follows.

4.2. R_aR_aR_aR_bR_b Robot

The R_aR_aR_aR_bR_b robot in Figure 3 consists of three R_a and two R_b joints, and its displacement equation can be formulated as

$$\begin{array}{ll}{\mathit{S}}_{f,\mathrm{SR}}& =2tan{\displaystyle \frac{{\mathit{\theta }}_{b_2}}{2}}\left(\begin{array}{c}{\mathit{s}}_b\\ {\mathit{r}}_{b_2}\times {\mathit{s}}_b\end{array}\right)\overline{\circ }2tan{\displaystyle \frac{{\mathit{\theta }}_{b_1}}{2}}\left(\begin{array}{c}{\mathit{s}}_b\\ {\mathit{r}}_{b_1}\times {\mathit{s}}_b\end{array}\right)\overline{\circ }\\ & 2tan{\displaystyle \frac{{\mathit{\theta }}_{a_3}}{2}}\left(\begin{array}{c}{\mathit{s}}_a\\ {\mathit{r}}_{a_3}\times {\mathit{s}}_a\end{array}\right)\overline{\circ }2tan{\displaystyle \frac{{\mathit{\theta }}_{a_2}}{2}}\left(\begin{array}{c}{\mathit{s}}_a\\ {\mathit{r}}_{a_2}\times {\mathit{s}}_a\end{array}\right)\overline{\circ }2tan{\displaystyle \frac{{\mathit{\theta }}_{a_1}}{2}}\left(\begin{array}{c}{\mathit{s}}_a\\ {\mathit{r}}_{a_1}\times {\mathit{s}}_a\end{array}\right)\end{array}$$

(17)

where ${\mathit{r}}_{a_1}$, ${\mathit{r}}_{a_2}$, and ${\mathit{r}}_{a_3}$ denote the position vectors of the three R_a, ${\mathit{\theta }}_{a_1}$, ${\mathit{\theta }}_{a_2}$, and ${\mathit{\theta }}_{a_3}$ denote their rotational angles, ${\mathit{r}}_{b_1}$ and ${\mathit{r}}_{b_2}$ are the position vectors of the two R_b, and ${\mathit{\theta }}_{b_1}$ and ${\mathit{\theta }}_{b_2}$ are their rotational angles. The subscripts of the position vectors and rotational angles of the R_a and R_b joints are, respectively, numbered from the joint near the fixed base to the joint near the end-effector.

Equation (17) can be rewritten as

$$2tan{\displaystyle \frac{{\mathit{\theta }}_{\mathrm{SR}}}{2}}\left(\begin{array}{c}{\mathit{s}}_{\mathrm{SR}}\\ {\mathit{r}}_{\mathrm{SR}}\times {\mathit{s}}_{\mathrm{SR}}+{\displaystyle \frac{{\mathit{t}}_{\mathrm{SR}}}{2tan\frac{{\mathit{\theta }}_{\mathrm{SR}}}{2}}}{\mathit{s}}_{\mathrm{SR}}\end{array}\right)=2tan{\displaystyle \frac{{\mathit{\theta }}_C}{2}}\left(\begin{array}{c}{\mathit{s}}_C\\ {\mathit{p}}_{ba}+{\displaystyle \frac{\mathit{t}\times {\mathit{s}}_C}{2}}+{\displaystyle \frac{\mathit{t}}{2tan\frac{{\mathit{\theta }}_C}{2}}}\end{array}\right)$$

(18)

where

$$tan{\displaystyle \frac{{\mathit{\theta }}_{\mathrm{SR}}}{2}}=tan{\displaystyle \frac{{\mathit{\theta }}_C}{2}}={\displaystyle \frac{\left|tan\frac{{\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}}{2}{\mathit{s}}_a+tan\frac{{\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}}{2}{\mathit{s}}_b+tan\frac{{\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}}{2}tan\frac{{\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}}{2}\left({\mathit{s}}_a\times {\mathit{s}}_b\right)\right|}{1-tan\frac{{\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}}{2}tan\frac{{\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}}{2}{\mathit{s}}_a^{\mathrm{T}}{\mathit{s}}_b}},$$

$$\begin{array}{ll}{\mathit{s}}_{\mathrm{SR}}& ={\mathit{s}}_C\\ & ={\displaystyle \frac{tan\frac{{\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}}{2}{\mathit{s}}_a+tan\frac{{\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}}{2}{\mathit{s}}_b+tan\frac{{\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}}{2}tan\frac{{\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}}{2}\left({\mathit{s}}_a\times {\mathit{s}}_b\right)}{\left|tan\frac{{\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}}{2}{\mathit{s}}_a+tan\frac{{\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}}{2}{\mathit{s}}_b+tan\frac{{\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}}{2}tan\frac{{\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}}{2}\left({\mathit{s}}_a\times {\mathit{s}}_b\right)\right|}}\end{array},$$

$${\mathit{p}}_{ba}={\displaystyle \frac{\begin{array}{l}tan\frac{{\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}}{2}\left({\mathit{r}}_{a_3}\times {\mathit{s}}_a\right)+tan\frac{{\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}}{2}\left({\mathit{r}}_{b_2}\times {\mathit{s}}_b\right)+\\ tan\frac{{\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}}{2}tan\frac{{\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}}{2}\left({\mathit{s}}_a\times \left({\mathit{r}}_{b_2}\times {\mathit{s}}_b\right)+\left({\mathit{r}}_{a_3}\times {\mathit{s}}_a\right)\times {\mathit{s}}_b\right)\end{array}}{\left|tan\frac{{\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}}{2}{\mathit{s}}_a+tan\frac{{\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}}{2}{\mathit{s}}_b+tan\frac{{\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}}{2}tan\frac{{\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}}{2}\left({\mathit{s}}_a\times {\mathit{s}}_b\right)\right|}}$$

$$\begin{array}{c}\mathit{t}=\left(\exp \left(\left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}\right){\tilde{\mathit{s}}}_a\right)-{\mathit{E}}_3\right)\left({\mathit{r}}_{a_3}-{\mathit{r}}_{a_2}\right)+\left(\exp \left({\mathit{\theta }}_{a_1}{\tilde{\mathit{s}}}_a\right)-{\mathit{E}}_3\right)\left({\mathit{r}}_{a_2}-{\mathit{r}}_{a_1}\right)\\ +\exp \left(\left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}\right){\tilde{\mathit{s}}}_a\right)\left(\exp \left({\mathit{\theta }}_{b_1}{\tilde{\mathit{s}}}_b\right)-{\mathit{E}}_3\right)\left({\mathit{r}}_{b_2}-{\mathit{r}}_{b_1}\right)\end{array}$$

Similarly, the forward kinematics can be derived in a straightforward way using these equations when the joint parameters are given.

When the end-effector pose is given, the inverse kinematics of this robot is solved as follows. It is easy to see from the expression of ${\mathit{s}}_C$ that the two rotations of the SR are, respectively, generated by the three R_a and the two R_b. Thus, ${\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}$ and ${\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}$ can be regraded to be the two rotational parameters. From the expression of $\mathit{t}$, it can be seen that the three translations are, respectively, generated by the first two R_a, the first R_a, and the first R_b. In this way, ${\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}$, ${\mathit{\theta }}_{a_1}$, and ${\mathit{\theta }}_{b_1}$ are the three translational parameters.

The two rotational parameters ${\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}$ and ${\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}$ can be solved in the similar way as shown in Equation (14). The solutions are

$$\begin{gathered} {\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}=2\arctan \left({\displaystyle \frac{{\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}\left({\mathit{s}}_a\times {\mathit{s}}_b\right)}{{\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}{\mathit{s}}_b-{\mathit{s}}_a^{\mathrm{T}}{\mathit{s}}_b{\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}{\mathit{s}}_a}}\right), \\ {\mathit{\theta }}_{b_1}+{\mathit{\theta }}_{b_2}=2\arctan \left({\displaystyle \frac{{\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}\left({\mathit{s}}_a\times {\mathit{s}}_b\right)}{{\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}{\mathit{s}}_a-{\mathit{s}}_a^{\mathrm{T}}{\mathit{s}}_b{\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}{\mathit{s}}_b}}\right). \end{gathered}$$

(19)

Thus, $\mathit{t}$ can be obtained like Equation (15). Using Equation (18), the mapping between $\mathit{t}$ and the three translational parameters ${\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}$, ${\mathit{\theta }}_{a_1}$, and ${\mathit{\theta }}_{b_1}$ can be obtained as

$$\begin{array}{l}\exp \left(\left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}\right){\tilde{\mathit{s}}}_a\right)\left({\mathit{r}}_{a_3}-{\mathit{r}}_{a_2}\right)+\exp \left({\mathit{\theta }}_{a_1}{\tilde{\mathit{s}}}_a\right)\left({\mathit{r}}_{a_2}-{\mathit{r}}_{a_1}\right)\\ +\exp \left(\left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}\right){\tilde{\mathit{s}}}_a\right)\exp \left({\mathit{\theta }}_{b_1}{\tilde{\mathit{s}}}_b\right)\left({\mathit{r}}_{b_2}-{\mathit{r}}_{b_1}\right)\\ =\mathit{t}+\left({\mathit{r}}_{a_3}-{\mathit{r}}_{a_1}\right)+\exp \left(\left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}\right){\tilde{\mathit{s}}}_a\right)\left({\mathit{r}}_{b_2}-{\mathit{r}}_{b_1}\right)\end{array}$$

(20)

Taking dot product of ${\mathit{s}}_a$ on both sides of Equation (20) to eliminate ${\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}$ and ${\mathit{\theta }}_{a_1}$, the equation having the only parameter ${\mathit{\theta }}_{b_1}$ can be obtained as

$$A\sin {\mathit{\theta }}_{b_1}+B\cos {\mathit{\theta }}_{b_1}=C,$$

(21)

where

$$A={\left(\exp \left(\left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}\right){\tilde{\mathit{s}}}_a\right)\left({\mathit{s}}_b\times \left({\mathit{r}}_{b_2}-{\mathit{r}}_{b_1}\right)\right)\right)}^{\mathrm{T}}{\mathit{s}}_a,$$

$$B={\left(\exp \left(\left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}\right){\tilde{\mathit{s}}}_a\right)\left({\mathit{r}}_{b_2}-{\mathit{r}}_{b_1}-{\mathit{r}}_{b_2}^{\mathrm{T}}{\mathit{s}}_b{\mathit{s}}_b+{\mathit{r}}_{b_1}^{\mathrm{T}}{\mathit{s}}_b{\mathit{s}}_b\right)\right)}^{\mathrm{T}}{\mathit{s}}_a,$$

$$C={\left(\mathit{t}+\left({\mathit{r}}_{a_3}-{\mathit{r}}_{a_1}-{\mathit{r}}_{a_3}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a+{\mathit{r}}_{a_1}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a\right)\right)}^{\mathrm{T}}{\mathit{s}}_a.$$

In Equation (21), A, B, and C are known quantities. Thus, ${\mathit{\theta }}_{b_1}$ can be solved as follows using the half-tangent formula

$${\mathit{\theta }}_{b_1}=2\arctan \left({\displaystyle \frac{A\pm \sqrt{A^2+B^2-C^2}}{B+C}}\right).$$

(22)

We substitute the obtained ${\mathit{\theta }}_{b_1}$ into Equation (20) to solve ${\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}$ and ${\mathit{\theta }}_{a_1}$,

$$\begin{array}{l}\exp \left(\left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}\right){\tilde{\mathit{s}}}_a\right)\left({\mathit{r}}_{a_3}-{\mathit{r}}_{a_2}\right)+\exp \left({\mathit{\theta }}_{a_1}{\tilde{\mathit{s}}}_a\right)\left({\mathit{r}}_{a_2}-{\mathit{r}}_{a_1}\right)\\ =\mathit{t}+\left({\mathit{r}}_{a_3}-{\mathit{r}}_{a_1}\right)-\exp \left(\left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}\right){\tilde{\mathit{s}}}_a\right)\left(\exp \left({\mathit{\theta }}_{b_1}{\tilde{\mathit{s}}}_b\right)-{\mathit{E}}_3\right)\left({\mathit{r}}_{b_2}-{\mathit{r}}_{b_1}\right)\end{array}$$

(23)

In order to solve ${\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}$ and ${\mathit{\theta }}_{a_1}$, the following two equations are obtained by using Euler formula and taking variable substitutions ${\mathit{\theta }}_1={\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}$ and ${\mathit{\theta }}_2={\mathit{\theta }}_{a_1}$,

$$A_1\sin {\mathit{\theta }}_1+B_1\cos {\mathit{\theta }}_1+C\sin {\mathit{\theta }}_2=D_1,$$

(24)

$$A_2\sin {\mathit{\theta }}_1+B_2\cos {\mathit{\theta }}_1+C\cos {\mathit{\theta }}_2=D_2,$$

(25)

where

$$A_1={\left({\mathit{s}}_a\times \left({\mathit{r}}_{a_3}-{\mathit{r}}_{a_2}\right)\right)}^{\mathrm{T}}\left({\mathit{s}}_a\times \left({\mathit{r}}_{a_2}-{\mathit{r}}_{a_1}\right)\right), B_1={\left({\mathit{r}}_{a_3}-{\mathit{r}}_{a_2}-{\mathit{r}}_{a_3}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a+{\mathit{r}}_{a_2}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a\right)}^{\mathrm{T}}\left({\mathit{s}}_a\times \left({\mathit{r}}_{a_2}-{\mathit{r}}_{a_1}\right)\right),$$

$$C={\left|{\mathit{r}}_{a_2}-{\mathit{r}}_{a_1}-{\mathit{r}}_{a_2}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a+{\mathit{r}}_{a_1}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a\right|}^2,$$

$$D_1={\left(\begin{array}{l}\mathit{t}+\left({\mathit{r}}_{a_3}-{\mathit{r}}_{a_1}-{\mathit{r}}_{a_3}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a+{\mathit{r}}_{a_1}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a\right)\\ -\exp \left(\left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}\right){\tilde{\mathit{s}}}_a\right)\left(\exp \left({\mathit{\theta }}_{b_1}{\tilde{\mathit{s}}}_b\right)-{\mathit{E}}_3\right)\left({\mathit{r}}_{b_2}-{\mathit{r}}_{b_1}\right)\end{array}\right)}^{\mathrm{T}}\left({\mathit{s}}_a\times \left({\mathit{r}}_{a_2}-{\mathit{r}}_{a_1}\right)\right),$$

$$A_2={\left({\mathit{s}}_a\times \left({\mathit{r}}_{a_3}-{\mathit{r}}_{a_2}\right)\right)}^{\mathrm{T}}\left({\mathit{r}}_{a_2}-{\mathit{r}}_{a_1}-{\mathit{r}}_{a_2}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a+{\mathit{r}}_{a_1}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a\right),$$

$$B_2={\left({\mathit{r}}_{a_3}-{\mathit{r}}_{a_2}-{\mathit{r}}_{a_3}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a+{\mathit{r}}_{a_2}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a\right)}^{\mathrm{T}}\left({\mathit{r}}_{a_2}-{\mathit{r}}_{a_1}-{\mathit{r}}_{a_2}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a+{\mathit{r}}_{a_1}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a\right)$$

$$D_2={\left(\begin{array}{l}\mathit{t}+\left({\mathit{r}}_{a_3}-{\mathit{r}}_{a_1}-{\mathit{r}}_{a_3}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a+{\mathit{r}}_{a_1}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a\right)\\ -\exp \left(\left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}+{\mathit{\theta }}_{a_3}\right){\tilde{\mathit{s}}}_a\right)\left(\exp \left({\mathit{\theta }}_{b_1}{\tilde{\mathit{s}}}_b\right)-{\mathit{E}}_3\right)\left({\mathit{r}}_{b_2}-{\mathit{r}}_{b_1}\right)\end{array}\right)}^{\mathrm{T}}\left({\mathit{r}}_{a_2}-{\mathit{r}}_{a_1}-{\mathit{r}}_{a_2}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a+{\mathit{r}}_{a_1}^{\mathrm{T}}{\mathit{s}}_a{\mathit{s}}_a\right)$$

Combining these two equations and using the half-tangent formula leads to a quartic equation of $tan{\displaystyle \frac{{\mathit{\theta }}_1}{2}}$ as follows:

$$\begin{array}{l}\left(B_1^2+B_2^2+D_1^2+D_2^2-C^2+2B_1{D}_1+2B_2{D}_2\right){tan}^4{\displaystyle \frac{{\mathit{\theta }}_1}{2}}\\ -4\left(A_1{B}_1+A_2{B}_2+A_1{D}_1+A_2{D}_2\right){tan}^3{\displaystyle \frac{{\mathit{\theta }}_1}{2}}\\ +2\left(2A_1^2+2A_2^2+D_1^2+D_2^2-C^2-B_1^2-B_2^2\right){tan}^2{\displaystyle \frac{{\mathit{\theta }}_1}{2}}\\ +4\left(A_1{B}_1+A_2{B}_2-A_1{D}_1-A_2{D}_2\right)tan{\displaystyle \frac{{\mathit{\theta }}_1}{2}}\\ +\left(B_1^2+B_2^2+D_1^2+D_2^2-C^2-2B_1{D}_1-2B_2{D}_2\right)=0\end{array}$$

(26)

As A_i, B_i, C, D_i ($i=1,2$) are known quantities, ${\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}$ can be solved using the quartic formula. Here, we do not list the detailed results due to space limitations. Substituting the analytical solution of ${\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}$ into Equations (24) and (25) leads to the solution of ${\mathit{\theta }}_{a_1}$ as follows:

$${\mathit{\theta }}_{a_1}=\arctan 2\left(\begin{array}{l}{\displaystyle \frac{D_1}{C}}-{\displaystyle \frac{A_1}{C}}\sin \left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}\right)-{\displaystyle \frac{B_1}{C}}\cos \left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}\right),\\ {\displaystyle \frac{D_2}{C}}-{\displaystyle \frac{A_2}{C}}\sin \left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}\right)-{\displaystyle \frac{B_2}{C}}\cos \left({\mathit{\theta }}_{a_1}+{\mathit{\theta }}_{a_2}\right)\end{array}\right)$$

(27)

Consequently, using the displacement model based upon finite screw and the procedures in Section 3, the forward and inverse kinematics of the two SRs are all solved, which shows the validity of the proposed new method.

5. Matlab Simulation

Using finite screw as mathematical tool, the displacement models of P₁P₂P₃R_aR_b and R_aR_aR_aR_bR_b robots are easily formulated, and the solutions for forward and inverse kinematics of these two SRs are systematically derived in an analytical manner. In this section, the obtained analytical solutions will be verified through Matlab simulation, which leads to the correctness of the proposed method.

5.1. P₁P₂P₃R_aR_b

As shown in Figure 2, P₁P₂P₃R_aR_b is the basic and simplest three-translational and two-rotational SR. Using Matlab software R2023b, the inverse kinematic solutions of the five joint parameters obtained from the derivations in Section 4.1 are programmed. We arbitrarily set the values of the unit vectors and position vectors of its joints and the pose of end-effector as follows:

$${\mathit{s}}_1^{\mathrm{T}}={\left(\begin{array}{ccc}1& 0& 0\end{array}\right)}^{\mathrm{T}}, {\mathit{s}}_2^{\mathrm{T}}={\left(\begin{array}{ccc}0& 1& 0\end{array}\right)}^{\mathrm{T}}, {\mathit{s}}_3^{\mathrm{T}}={\left(\begin{array}{ccc}0& 0& 1\end{array}\right)}^{\mathrm{T}},$$

$$\begin{gathered} {\mathit{s}}_a^{\mathrm{T}}={\left(\begin{array}{ccc}{\displaystyle \frac{\sqrt{3}}{3}}& {\displaystyle \frac{\sqrt{3}}{3}}& {\displaystyle \frac{\sqrt{3}}{3}}\end{array}\right)}^{\mathrm{T}}, {\mathit{r}}_a^{\mathrm{T}}={\left(\begin{array}{ccc}1& 1& 2\end{array}\right)}^{\mathrm{T}}, {\mathit{s}}_b^{\mathrm{T}}={\left(\begin{array}{ccc}{\displaystyle \frac{\sqrt{2}}{2}}& 0& {\displaystyle \frac{\sqrt{2}}{2}}\end{array}\right)}^{\mathrm{T}}, \\ {\mathit{r}}_b^{\mathrm{T}}={\left(\begin{array}{ccc}0& 0& 4\end{array}\right)}^{\mathrm{T}}, \end{gathered}$$

$${\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}={\left(\begin{array}{ccc}{\displaystyle \frac{\sqrt{2}}{2}}& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}}\end{array}\right)}^{\mathrm{T}}, {\mathit{r}}_{\mathrm{SR}}^{\mathrm{T}}={\left(\begin{array}{ccc}0& 0& 3\end{array}\right)}^{\mathrm{T}}, {\mathit{\theta }}_{\mathrm{SR}}=2.5717, {\mathit{\tau }}_{\mathrm{SR}}=1$$

The joint parameters can be automatically solved in Matlab as follows:

$${\mathit{\tau }}_1=-1.3719, {\mathit{\tau }}_2=0.5791, {\mathit{\tau }}_3=2.1185, {\mathit{\theta }}_a=2.0944, {\mathit{\theta }}_b=0.5698$$

Substituting these joint parameters into the forward kinematic program shows that the resultant pose is identical to the given pose (${\mathit{s}}_{\mathrm{SR}}$, ${\mathit{r}}_{\mathrm{SR}}$, ${\mathit{\theta }}_{\mathrm{SR}}$, ${\mathit{\tau }}_{\mathrm{SR}}$). Hence, the analytical solutions in Section 4.1 are verified to be correct.

5.2. R_aR_aR_aR_bR_b

R_aR_aR_aR_bR_b is a rather complicated SR, as shown in Figure 3. The inverse kinematic solutions of the five joint parameters obtained from the derivations in Section 4.2 are programmed in Matlab software. Similarly, the unit vectors and position vectors of its joints and the pose of end-effector are arbitrarily given values, as follows:

$${\mathit{s}}_a^{\mathrm{T}}={\left(\begin{array}{ccc}{\displaystyle \frac{\sqrt{3}}{3}}& {\displaystyle \frac{\sqrt{3}}{3}}& {\displaystyle \frac{\sqrt{3}}{3}}\end{array}\right)}^{\mathrm{T}}, {\mathit{r}}_{a_1}^{\mathrm{T}}={\left(\begin{array}{ccc}1& 1& 1\end{array}\right)}^{\mathrm{T}}, {\mathit{r}}_{a_2}^{\mathrm{T}}={\left(\begin{array}{ccc}1& 1& 2\end{array}\right)}^{\mathrm{T}}, {\mathit{r}}_{a_3}^{\mathrm{T}}={\left(\begin{array}{ccc}1& 1& 3\end{array}\right)}^{\mathrm{T}}$$

$${\mathit{s}}_b^{\mathrm{T}}={\left(\begin{array}{ccc}{\displaystyle \frac{\sqrt{2}}{2}}& 0& {\displaystyle \frac{\sqrt{2}}{2}}\end{array}\right)}^{\mathrm{T}}, {\mathit{r}}_{b_1}^{\mathrm{T}}={\left(\begin{array}{ccc}0& 0& 3\end{array}\right)}^{\mathrm{T}}, {\mathit{r}}_{b_2}^{\mathrm{T}}={\left(\begin{array}{ccc}0& 0& 4\end{array}\right)}^{\mathrm{T}}$$

$${\mathit{s}}_{\mathrm{SR}}^{\mathrm{T}}={\left(\begin{array}{ccc}{\displaystyle \frac{\sqrt{2}}{2}}& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}}\end{array}\right)}^{\mathrm{T}}, {\mathit{r}}_{\mathrm{SR}}^{\mathrm{T}}={\left(\begin{array}{ccc}2& 1& 2.2\end{array}\right)}^{\mathrm{T}}, {\mathit{\theta }}_{\mathrm{SR}}=2.5717, {\mathit{\tau }}_{\mathrm{SR}}=1$$

Matlab can automatically calculate the two sets of joint parameters as

$${\mathit{\theta }}_{a_1}=2.4810, {\mathit{\theta }}_{a_2}=-0.6874, {\mathit{\theta }}_{a_3}=0.3007, {\mathit{\theta }}_{b_1}=-0.6281, {\mathit{\theta }}_{b_2}=1.1980$$

and

$${\mathit{\theta }}_{a_1}=1.7936, {\mathit{\theta }}_{a_2}=0.6874, {\mathit{\theta }}_{a_3}=-0.3867, {\mathit{\theta }}_{b_1}=-0.6281, {\mathit{\theta }}_{b_2}=1.1980$$

Substituting each set of these joint parameters into the forward kinematic program can verify the correctness of the analytical solutions in Section 4.2.

In fact, numerous groups of values of the joint axes and poses of the end-effector are attempted in the inverse kinematic programs of both P₁P₂P₃R_aR_b and R_aR_aR_aR_bR_b robots. All the solved joint parameters lead to the given pose in the forward kinematic program. In this way, the correctness of the proposed finite screw method is verified through Matlab simulation. Here, we just list one group of values and solutions for each robot for an example.

6. Discussions and Conclusions

Employing finite screw as mathematical tool, this paper presents a new method for displacement modelling. Forward kinematics of the SR can be directly carried out using the formulated displacement equation. Meanwhile, the process for inverse kinematics is also given. Two SRs with three translations and two rotations are given as examples to show the validity of the proposed method. However, it should be pointed out that the detailed computations on solving the inverse kinematics of SRs with more than two rotations are still under investigations by the authors.

The following conclusions are drawn:

(1) The displacement equation of SR is algebraically formulated through describing displacements generated by SR and its joints employing finite screws;

(2) The procedures for forward and inverse kinematics by analytically solving the formulated displacement equation are discussed in detail;

(3) P₁P₂P₃R_aR_b and R_aR_aR_aR_bR_b robots are taken as examples to illustrate the proposed finite screw method. The correctness of the obtained solutions is verified by Matlab simulation.

This paper provides a new idea and method for displacement modelling and kinematics of SRs. The proposed method for displacement modelling of SR offers a novel approach, which provides a more accurate and efficient way for displacement modelling, which may have significant implications for various applications. Our study mainly focuses on theoretical analysis and simulation. Though we have demonstrated the effectiveness of our method in these aspects, further experimental validation is needed to fully assess its performance in practical applications.