A Novel Geometric Modeling and Calculation Method for Forward Displacement Analysis of 6-3 Stewart Platforms

Abstract

A novel geometric modeling and calculation method for forward displacement analysis of the 6-3 Stewart platforms is proposed by using the conformal geometric algebra (CGA) framework. Firstly, two formulas between 2-blade and 1-blade are formulated. Secondly, the expressions for two spherical joints of the moving platform are given via CGA operation. Thirdly, a coordinate-invariant geometric constraint equation is deduced. Fourthly, a 16-degree univariate polynomial equation without algebraic elimination by using the Euler angle substitution is presented. Fifthly, the coordinates of three spherical joints on the moving platform are calculated without judging the radical symbols. Finally, two numerical examples are used to verify the method. The highlight of this paper is that a new geometric modeling and calculation method without algebraic elimination is obtained by using the determinant form of the CGA inner product algorithm, which provides a new idea to solve a more complex spatial parallel mechanism in the future.

1. Introduction

The parallel mechanism (PM) [1,2,3,4] is composed of a moving platform and a fixed platform, which is connected by at least two independent kinematic chains and has two or more degrees of freedom (DOF). Compared with the traditional series mechanism, the spatial parallel mechanism has the advantages of higher rigidity, precision, and strong bearing capacity, etc. At present, the most widely studied parallel mechanism is the 6-DOF parallel mecha-nisms first proposed by Gough [1] and widely used by Stewart [2]. The classification of all self-motions of the origi-nal Stewart–Gough platform was achieved by Luces and Benhabib [3]. The redundant parallel kinematic mecha-nisms are introduced by Karger and Husty [4]. Traditionally, Eduard Study [5] put forward a novel method to de-fine a rigid body displacement in 3-dimension space, and mapped each position of a rigid body onto a point on a quadric, now called the Study quadric. In this paper, the motion of the rigid body (here, the moving platform) is described by the position coordinates of three spherical joints on the moving platform. The forward displacement analysis of the parallel mechanism is to calculate the position and attitude of the moving platform relative to the static platform via the driving rod length of the parallel mechanism. This problem is the theoretical basis of the ve-locity, acceleration, error analysis, workspace analysis, singular position analysis, dynamic analysis, and other problems of the mechanism. For the elimination of equations, the Newton–Raphson method [6] and polynomial constraints solver [7] are commonly used.

In this discourse, we will revisit the forward displacement analysis of the 6-3 Stewart platforms (6-3SPS mechanisms), which has been researched in [8,9,10,11,12,13,14]. Griffis et al. [8] managed to present a closed-form solution to the platform mechanisms of the 6-3SPS. Nanua et al. [9] published the 16th-order input-output equation. Innocenti et al. [10] revealed the solution of a system of three second-order nonlinear equations. Merlet et al. [11] put forward a method for the forward displacement analysis of 6-3 Stewart platforms by finding a polynomial in one variable whose roots determine the posture of the mobile plate. Zhang et al. [12] and Wei et al. [13] committed to an inves-tigation into forward displacement analysis of the 6-3 Stewart platforms using the resultant method and the CGA theory. Zhang et al. [14] settled the forward displacement analysis for the 6-3 Stewart platforms only using the CGA method.

Conformal geometric algebra (CGA) [14,15,16,17,18,19,20,21,22,23,24,25,26,27] is a new geometric representation and calculation framework, which can uniformly represent point, line, plane, circle, and other geometric elements. CGA is committed to present direct algebraic operations on these geometric entities. These operations usually produce formulas that are inde-pendent of the coordinate system and make complex symbolic geometry computation possible. The above features are two important intrinsic attributes of CGA. Therefore, it is useful and suitable for the geometric modeling and calculation of mechanism kinematics. In recent decades, CGA had been used to deal with the inverse kinematics problem of the serial mechanism through CGA operation of geometric entities [19,20,21]. Additionally, Song et al. [22] employed the CGA theory in topology synthesis of mechanisms. Furthermore, Hrdina et al. [23,24,25] committed to an investigation into robotic modeling and control by CGA theory. Zhang et al. [14,26] and Huang et al. [27] per-formed exploration on addressing the forward displacement analysis of the serial mechanism in virtue of CGA the-ory.

In this paper, a new geometric modeling and calculation method for closed-form solution of the 6-3 Stewart platforms is proposed by using the CGA framework which contributes to a 16-degree univariate input-output pol-ynomial equation without algebraic elimination and superfluous roots. Although the forward displacement prob-lems of the 6-3 Stewart platforms have been addressed by Zhang et al. [12,14] and Wei et al. [13] using CGA, the proposed method based on the determinant operation formula (Equation (8)) of the CGA inner product provides a new idea to solve more a complex spatial parallel mechanism in the future, which is not available in Refs. [12,13,14]. In addition, compared with [14] in the back substitution procedure, the proposed method does not need to determine the radical symbols. Moreover, two formulas between 2-blade and 1-blade are first formulated.

The remaining of the paper is formulated as follows. In Section 2, the foundations of CGA will be presented. In Section 3, the derivation of two formulas between 2-blade and 1-blade under the CGA framework is elaborated. In Section 4, the geometric modeling and calculation process of the 6-3 Stewart platforms will be revealed. In Section 5, two numerical examples of 6-3 Stewart platforms will be given. Finally, the conclusions will be built in Section 6.

2. Conformal Geometric Algebra

2.1. Foundations of CGA

The 5-dimension (5D) CGA ${\mathrm{G}}^{4,1}$ consists of 3D Euclidean space ${\mathrm{G}}^3$ and a 2D Minkowski vector space ${\mathrm{G}}^{1,1}$. CGA has five orthonormal basis vectors introduced by $\left\{{\mathit{e}}_1,{\mathit{e}}_2,{\mathit{e}}_3,{\mathit{e}}_{+},{\mathit{e}}_{-}\right\}$ with the following properties:

$$\left\{\begin{array}{l}{e_1}^2={e_2}^2={e_3}^2={e_{+}}^2=-{e_{-}}^2=1\\ e_i·e_j=0\left(i\ne j;i,j=1,2,3,+,-\right)\\ e_i\wedge e_j=-e_j\wedge e_i\left(i,j=1,2,3,+,-\right)\end{array}\right.$$

(1)

where $\left\{e_1,e_2,e_3\right\}$ are three orthonormal basis vectors in the Euclidean space and $\left\{e_{+},e_{-}\right\}$ are two orthogonal basis vectors in the Minkowski space.

In addition, two null bases can be presented by the vectors:

$$e_0=\frac{1}{2}(e_{-}-e_{+}),e_{\infty }=e_{+}+e_{-}$$

(2)

with the properties of:

$$e_0^2=e_{\infty }^2=0,e_{\infty }·e_0=-1$$

(3)

where $e_0$ is the conformal origin and $e_{\infty }$ is the conformal infinity.

Blade is the basic calculation element and basic geometric entity of geometric algebra. The 5D CGA consists of 0, 1, 2, 3, 4, and 5 blades. The blade with the largest grade, namely 5-blades, are called pseudo-scalars and defined by $I_C\left(e_{\infty 0123},I_C{}^2=-1\right)$. A linear combination of blades with different grades is called a multi-vector.

The dual ${\mathit{X}}^{\ast }$ of a multi-vector $\mathit{X}$ is denoted by:

$${\mathit{X}}^{\ast }=\mathit{X}{I}_C{}^{-1}=-\mathit{X}{I}_C$$

(4)

where $I_C{}^{-1}$ is the inverse of $I_C$.

Conformal geometric algebra has three algebraic operators: the inner product ($u·v$), the outer product ($u\wedge v$), and the geometric product ($uv=u·v+u\wedge v$). Inner product is used to solve geometric scalars. The outer product is applied to build basic geometry or the geometry intersection. The inner and outer products of two vectors $u$ and $v$ can also be built as:

$$\left\{\begin{array}{l}u·v=\frac{1}{2}\left(uv+vu\right)\\ u\wedge v=\frac{1}{2}\left(uv-vu\right)\end{array}\right.,$$

(5)

and:

$$\left\{\begin{array}{l}u·v=v·u\\ u\wedge v=-v\wedge u\end{array}\right..$$

(6)

The inner product of between an r-blade $u_1\wedge \cdots \wedge u_r$ and an s-blade $v_1\wedge \cdots \wedge v_s$ can be defined by:

$$\begin{array}{l}\left(u_1\wedge \cdots \wedge u_r\right)·\left(v_1\wedge \cdots \wedge v_s\right)\\ =\left\{\begin{array}{l}\left(\left(u_1\wedge \cdots \wedge u_r\right)·v_1\right)·\left(v_2\wedge \cdots \wedge v_s\right)\begin{array}{ccc}& & \mathrm{if}\begin{array}{c}\begin{array}{c}\end{array}r\ge s\end{array}\end{array}\\ \left(u_1\wedge \cdots \wedge u_{r-1}\right)·\left(u_r·\left(v_1\wedge \cdots \wedge v_s\right)\right)\begin{array}{ccc}& & \mathrm{if}\begin{array}{c}\begin{array}{c}\end{array}r<s\end{array}\end{array}\end{array}\right.\end{array}$$

(7)

with:

$$\left(u_1\wedge \cdots \wedge u_r\right)·v_1={\displaystyle \sum _{i=1}^r{\left(-1\right)}^{r-i}}{u}_1\wedge \cdots \wedge u_{i-1}\wedge \left(u_i·v_1\right)\wedge u_{i+1}\wedge \dots \wedge u_r,$$

$$u_r·\left(v_1\wedge \cdots \wedge v_s\right)={\displaystyle \sum _{i=1}^s{\left(-1\right)}^{i-1}}{v}_1\wedge \cdots \wedge v_{i-1}\wedge \left(u_r·v_i\right)\wedge v_{i+1}\wedge \dots \wedge v_s.$$

As a supplement, if $\begin{array}{c}r\end{array}$ is equal to $s$, the inner product of between an r-blade $u_1\wedge \cdots \wedge u_r$ and an s-blade $v_1\wedge \cdots \wedge v_s$ can also be obtained by:

$$\left(u_1\wedge \cdots \wedge u_r\right)·\left(v_1\wedge \cdots \wedge v_s\right)=\left|\begin{array}{ccc}{u}_r·v_1& \cdots & u_r·v_s\\ ⋮& \ddots & ⋮\\ u_1·v_1& \cdots & u_1·v_s\end{array}\right|.$$

(8)

For two blades ${\mathit{A}}_{\left\{r\right\}}$ and ${\mathit{B}}_{\left\{s\right\}}$ with non-zero grades $r$ and $s$, the inner and outer products can be denoted as:

$$\left\{\begin{array}{l}{\mathit{A}}_{\left\{r\right\}}·{\mathit{B}}_{\left\{s\right\}}={〈\mathit{A}\mathit{B}〉}_{\left|r-s\right|}\\ {\mathit{A}}_{\left\{r\right\}}\wedge {\mathit{B}}_{\left\{s\right\}}={〈\mathit{A}\mathit{B}〉}_{r+s}\end{array}\right..$$

(9)

2.2. Conformal Geometric Entities

CGA has a visual representation of basic geometric entities. The representations of geometric entities relative to inner product null space (IPNS) and outer product null space (OPNS) are listed in

Table 1. The geometric element of CGA.

Entity	IPNS	Grade	OPNS	Grade
Point	$P=p+\frac{1}{2}{p}^2{e}_{\infty }+e_0$	1	$P^{*}=S_1\wedge S_2\wedge S_3\wedge S_4$	4
Sphere	$S=p+\frac{1}{2}\left(p^2-{\rho }^2\right)e_{\infty }+e_0$	1	$S^{\ast }=P_1\wedge P_2\wedge P_3\wedge P_4$	4
Plane	$\mathit{\Pi }=n+d{e}_{\infty }$	1	${\mathit{\Pi }}^{\ast }=e_{\infty }\wedge P_1\wedge P_2\wedge P_3$	4
Line	$L={\mathit{\Pi }}_1\wedge {\mathit{\Pi }}_2$	2	$L^{\ast }=e_{\infty }\wedge P_1\wedge P_2$	3
Circle	$C=S_1\wedge S_2$	2	$C^{\ast }=P_1\wedge P_2\wedge P_3$	3
Point pair	$P_p=S_1\wedge S_2\wedge S_3$	3	$P_p{}^{\ast }=P_1\wedge P_2$	2

. IPNS refers to the geometric entity achieved by the intersection of geometric entities, and OPNS refers to the geometric entity described by the points belonging to the geometric entity. The two representations can be transformed by dual operators. Detailed information is given by references [15,16,17,18]. In Table 1, small italics and bold characters represent points or vectors in Euclidean space. The character $\rho$ is the radius of sphere $\mathit{S}$. The character $n$ is a normal vector of plane $\mathit{\Pi }$. The symbol $d$ represents the distance from the origin of the coordinate system to the plane $\mathit{\Pi }$.

The inner product between two points ${\mathit{P}}_1$ and ${\mathit{P}}_2$ is summarized as:

$${\mathit{P}}_1·{\mathit{P}}_2=\left(p_1+\frac{1}{2}{p}_1{}^2{e}_{\infty }+e_0\right)·\left(p_2+\frac{1}{2}{p}_2{}^2{e}_{\infty }+e_0\right)=-\frac{1}{2}{\left(p_1-p_2\right)}^2.$$

(10)

3. Two Formulas under the Framework of CGA

In this section, the derivation of two formulas between 2-blade and 1-blade under the CGA framework is elaborated. In order to facilitate the derivation of the formula, the following formula is first given using Equation (7) under the framework of CGA, i.e.,:

$${\mathit{P}}_p^{*}·{\mathit{Q}}_q^{*}=\left({\mathit{P}}_1\wedge {\mathit{P}}_2\right)·\left({\mathit{Q}}_1\wedge {\mathit{Q}}_2\right)=\left({\mathit{P}}_2·{\mathit{Q}}_1\right)\left({\mathit{P}}_1·{\mathit{Q}}_2\right)-\left({\mathit{P}}_1·{\mathit{Q}}_1\right)\left({\mathit{P}}_2·{\mathit{Q}}_2\right),$$

(11)

$$\left({\mathit{Q}}_q^{*}·{\mathit{X}}_1\right)\wedge \left({\mathit{Q}}_q^{*}·{\mathit{X}}_2\right)+\left({\mathit{Q}}_q^{*}·\left({\mathit{X}}_1\wedge {\mathit{X}}_2\right)\right){\mathit{Q}}_q^{*}=0,$$

(12)

$${\mathit{Q}}_q^{*}·\left({\mathit{X}}_1\wedge {\mathit{X}}_2\right)=\left({\mathit{Q}}_q^{*}·{\mathit{X}}_1\right)·{\mathit{X}}_2,$$

(13)

$${\mathit{Q}}_q^{*}·\left(e_{\infty }\wedge {\mathit{X}}_i\right)={\mathit{Q}}_e·{\mathit{X}}_i\left(i=1,2\right),$$

(14)

$${\mathit{Q}}_q^{*}·\left({\mathit{Q}}_e\wedge {\mathit{X}}_2\right)=\left({\mathit{Q}}_q^{*}·{\mathit{Q}}_e\right)·{\mathit{X}}_2,$$

(15)

$$\left({\mathit{Q}}_q^{*}·{\mathit{X}}_i\right)·{\mathit{Q}}_e=-\left({\mathit{Q}}_q^{*}·{\mathit{Q}}_e\right)·{\mathit{X}}_i\left(i=1,2\right),$$

(16)

where ${\mathit{P}}_p^{*}$ and ${\mathit{Q}}_q^{*}$ are both 2-blades, ${\mathit{P}}_p^{*}={\mathit{P}}_1\wedge {\mathit{P}}_2$, ${\mathit{Q}}_q^{*}={\mathit{Q}}_1\wedge {\mathit{Q}}_2$ and ${\mathit{Q}}_e={\mathit{Q}}_q^{*}·{\mathit{e}}_{\infty }$. ${\mathit{P}}_1$, ${\mathit{P}}_2$, ${\mathit{Q}}_1$, ${\mathit{Q}}_2$, ${\mathit{Q}}_e$, ${\mathit{X}}_1$ and ${\mathit{X}}_2$ are all 1-blades. The detailed derived procedure of Equation (12) is given in the Appendix A.

3.1. Derivation of the First Formula

According to Equation (12), we can obtain:

$$\left({\mathit{Q}}_q^{*}·{\mathit{e}}_{\infty }\right)\wedge \left({\mathit{Q}}_q^{*}·{\mathit{X}}_1\right)+\left({\mathit{Q}}_q^{*}·\left({\mathit{e}}_{\infty }\wedge {\mathit{X}}_1\right)\right){\mathit{Q}}_q^{*}=0.$$

(17)

By multiplying both sides of Equation (17) with $\left({\mathit{Q}}_q^{*}·{\mathit{e}}_{\infty }\right)\wedge {\mathit{X}}_2,$ and simplifying it, we obtain:

$$\left({\mathit{Q}}_e·{\mathit{X}}_2\right)\left({\mathit{Q}}_e·\left({\mathit{Q}}_q^{*}·{\mathit{X}}_1\right)\right)-\left({\mathit{Q}}_e·{\mathit{Q}}_e\right)\left({\mathit{X}}_2·\left({\mathit{Q}}_q^{*}·{\mathit{X}}_1\right)\right)+\left({\mathit{Q}}_q^{*}·\left({\mathit{Q}}_e\wedge {\mathit{X}}_2\right)\right)·\left({\mathit{Q}}_q^{*}·\left({\mathit{e}}_{\infty }\wedge {\mathit{X}}_1\right)\right)=0.$$

(18)

Using Equations (13)–(16), we obtain:

$$\left({\mathit{Q}}_e·{\mathit{X}}_1\right)·\left({\mathit{Q}}_q^{*}·\left({\mathit{Q}}_e·{\mathit{X}}_2\right)\right)-\left({\mathit{Q}}_e·{\mathit{X}}_2\right)·\left({\mathit{Q}}_q^{*}·\left({\mathit{Q}}_e·{\mathit{X}}_1\right)\right)=\left({\mathit{Q}}_e·{\mathit{Q}}_e\right)\left({\mathit{Q}}_q^{*}·\left({\mathit{X}}_1\wedge {\mathit{X}}_2\right)\right).$$

(19)

3.2. Derivation of the Second Formula

According to Equation (14), we can obtain:

$$\left({\mathit{Q}}_q^{*}·{\mathit{Q}}_q^{*}\right)\left(\left({\mathit{Q}}_e·{\mathit{X}}_1\right)·\left({\mathit{Q}}_e·{\mathit{X}}_2\right)\right)=\left(\left({\mathit{Q}}_q^{*}·\left({\mathit{e}}_{\infty }\wedge {\mathit{X}}_1\right)\right){\mathit{Q}}_q^{*}\right)·\left(\left({\mathit{Q}}_q^{*}·\left({\mathit{e}}_{\infty }\wedge {\mathit{X}}_2\right)\right){\mathit{Q}}_q^{*}\right).$$

(20)

According to Equation (12), Equation (20) is simplified as:

$$\left({\mathit{Q}}_q^{*}·{\mathit{Q}}_q^{*}\right)\left(\left({\mathit{Q}}_e·{\mathit{X}}_1\right)·\left({\mathit{Q}}_e·{\mathit{X}}_2\right)\right)=\left(-\left({\mathit{Q}}_q^{*}·{\mathit{e}}_{\infty }\right)\wedge \left({\mathit{Q}}_q^{*}·{\mathit{X}}_1\right)\right)·\left(-\left({\mathit{Q}}_q^{*}·{\mathit{e}}_{\infty }\right)\wedge \left({\mathit{Q}}_q^{*}·{\mathit{X}}_2\right)\right).$$

(21)

Expanding the right side of Equation (21) using Equation (11), we can obtain:

$$\begin{array}{l}\left({\mathit{Q}}_q^{*}·{\mathit{Q}}_q^{*}\right)\left(\left({\mathit{Q}}_e·{\mathit{X}}_1\right)·\left({\mathit{Q}}_e·{\mathit{X}}_2\right)\right)\\ =\left(\left({\mathit{Q}}_q^{*}·{\mathit{X}}_1\right)·\left({\mathit{Q}}_q^{*}·{\mathit{e}}_{\infty }\right)\right)\left(\left({\mathit{Q}}_q^{*}·{\mathit{e}}_{\infty }\right)·\left({\mathit{Q}}_q^{*}·{\mathit{X}}_1\right)\right)-\left({\mathit{Q}}_q^{*}·{\mathit{e}}_{\infty }\right)\left({\mathit{Q}}_q^{*}·{\mathit{e}}_{\infty }\right)\left(\left({\mathit{Q}}_q^{*}·{\mathit{X}}_1\right)·\left({\mathit{Q}}_q^{*}·{\mathit{X}}_2\right)\right).\end{array}$$

(22)

Using Equation (15), Equation (22) is simplified as:

$$\left({\mathit{Q}}_q^{*}·\left({\mathit{Q}}_e·{\mathit{X}}_1\right)\right)·\left({\mathit{Q}}_q^{*}·\left({\mathit{Q}}_e·{\mathit{X}}_2\right)\right)-\left({\mathit{Q}}_q^{*}·{\mathit{Q}}_q^{*}\right)\left(\left({\mathit{Q}}_e·{\mathit{X}}_1\right)·\left({\mathit{Q}}_e·{\mathit{X}}_2\right)\right)=\left({\mathit{Q}}_e·{\mathit{Q}}_e\right)\left(\left({\mathit{Q}}_q^{*}·{\mathit{X}}_1\right)·\left({\mathit{Q}}_q^{*}·{\mathit{X}}_2\right)\right).$$

(23)

Equations (19) and (23) are derived from Equation (12) and are frequently used in this paper.

4. CGA-Based Geometric Modeling and Calculation Procedure

In this section, the structure and coordinate system of the 6-3 Stewart platform is first presented. Then, the expressions for two point pairs $B_{b2}^{\ast }$ and $B_{b3}$ are given. Next, a coordinate-invariant geometric constraint equation is deduced. Finally, the coordinates of three spherical joints on the moving platform will be calculated. In this paper, three spherical joints on the moving platform are equivalent to three points.

4.1. The Structure and Coordinate System of the 6-3 Stewart Platforms

The 6-3 Stewart platforms structure is shown in Figure 1. The 6-3 Stewart platforms are connected by a moving platform $\Delta {\mathit{B}}_1{\mathit{B}}_2{\mathit{B}}_3$, a static platform $\Delta {\mathit{A}}_1{\mathit{A}}_2{\mathit{A}}_3$ and six SPS kinematic chains. The SPS kinematic chain is composed of revolute joint $R$, prismatic joint $P$, and spherical joint $S$, where prismatic joint $P$ is the actuated joint. The coordinate system $O-XYZ$ is the geodetic coordinate system. The coordinate systems $O_1-X_1{Y}_1{Z}_1$ are attached to the static platform $\Delta {\mathit{A}}_1{\mathit{A}}_2{\mathit{A}}_3$. The origin of the coordinate system $O-XYZ$ is considered to coincide with the origin of the coordinate system $O_1-X_1{Y}_1{Z}_1$. The geometric center coordinates of the spherical joints on the static platform and the spherical joints on moving platform are represented by $a_i\left(i=1,2,\cdots ,6\right)$ and $b_j\left(j=1,2,3\right)$. The distance between the three spherical joints on the moving platform is denoted by $r_j\left(j=1,2,3\right)$. $l_i\left(i=1,2,\cdots ,6\right)$ are the input of the six SPS kinematic chains and $b_j\left(j=1,2,3\right)$ are unknown.

The coordinate system $O-XYZ$ is rotated by ${\beta }_1$ about the Z₁-axis, and the new X₁-axis is rotated by ${\alpha }_1$. The unit vector of the Z₁-axis is defined as ${\left(z_{1x},z_{1y},z_{1z}\right)}^T$. As a consequence, the coordinate system $O-XYZ$ coincides with the coordinate system $O_1-X_1{Y}_1{Z}_1$, and the expressions of ${\alpha }_1$ and ${\beta }_1$ are denoted as:

$$\left\{\begin{array}{l}{\alpha }_1=\mathrm{sgn}(-z_{1y}){\cos }^{-1}(z_{1z}),{\beta }_1={\tan }^{-1}(-z_{1x}/z_{1y}), z_{1y}\ne 0\\ {\alpha }_1=\mathrm{sgn}(z_{1x}){\cos }^{-1}(z_{1z}),{\beta }_1=\pi /2, z_{1y}=0\end{array}\right..$$

(24)

In addition, the matrix ${\mathit{M}}_r$ is represented as:

$${\mathit{M}}_r=\left[\begin{array}{ccc}{c}_{{\beta }_1}& -s_{{\beta }_1}& 0\\ s_{{\beta }_1}& c_{{\beta }_1}& 0\\ 0& 0& 1\end{array}\right]·\left[\begin{array}{ccc}1& 0& 0\\ 0& c_{{\alpha }_1}& -s_{{\alpha }_1}\\ 0& s_{{\alpha }_1}& c_{{\alpha }_1}\end{array}\right],$$

(25)

where $s_{{\beta }_1}=\sin ({\beta }_1)$,$c_{{\beta }_1}=\cos ({\beta }_1)$,$s_{{\alpha }_1}=\sin ({\alpha }_1)$, and $c_{{\alpha }_1}=\cos ({\alpha }_1)$.

4.2. The Expressions for Two Point Pairs $B_{b2}^{\ast }$ and $B_{b3}$.

According to Figure 1, point $B_2$ is situated on sphere $S_{A_3{B}_2}$, whose center is point $A_3$ and radius is $l_3$, and sphere $S_{A_4{B}_2}$, whose center is point $A_4$ and radius is $l_4$. In addition, point $B_2$ is also located on sphere $S_{B_1{B}_2}$ whose center is point $B_1$ and radius is $r_3$. Considering Table 1, the point pair $B_{b2}$ can be denoted as:

$$B_{b2}=S_{B_1{B}_2}\wedge S_{A_3{B}_2}\wedge S_{A_4{B}_2},$$

(26)

where $S_{B_1{B}_2}$, $S_{A_3{B}_2}$ and $S_{A_4{B}_2}$ can be formulated as:

$$S_{B_1{B}_2}=B_1-\frac{1}{2}{r_3}^2{e}_{\infty }, S_{A_3{B}_2}=A_3-\frac{1}{2}{l_3}^2{e}_{\infty }, S_{A_4{B}_2}=A_4-\frac{1}{2}{l_4}^2{e}_{\infty }$$

Using Equation (4), the dual of point pair $B_{b2}$ can be represented as:

$$B_{b2}^{\ast }=-\left(S_{B_1{B}_2}\wedge S_{A_3{B}_2}\wedge S_{A_4{B}_2}\right)I_C.$$

(27)

Homogeneously, point $B_3$ is located on sphere $S_{A_5{B}_3}$ whose center is point $A_5$ and radius is $l_5$, sphere $S_{A_6{B}_3}$ whose center is point $A_6$ and radius is $l_6$, and sphere $S_{B_1{B}_3}$ whose center is point $B_1$ and radius is $r_2$. Hence, the point pair $B_{b3}$ can be formulated as:

$$B_{b3}=S_{B_1{B}_3}\wedge S_{A_5{B}_3}\wedge S_{A_6{B}_3}$$

(28)

where $S_{B_1{B}_3}$,$S_{A_5{B}_3}$ and $S_{A_6{B}_3}$ can be expressed as:

$$S_{B_1{B}_3}=B_1-\frac{1}{2}{r_2}^2{e}_{\infty }, S_{A_5{B}_3}=A_5-\frac{1}{2}{l_5}^2{e}_{\infty }, S_{A_6{B}_3}=A_6-\frac{1}{2}{l_6}^2{e}_{\infty }$$

4.3. The Coordinate-Invariant Geometric Constraint Equation

According to Table 1 and Equation (10), we can obtain:

$$\begin{array}{l}{\mathit{B}}_3^{\ast }·{\mathit{B}}_3^{\ast }=\left(S_{B_1{B}_3}\wedge S_{A_5{B}_3}\wedge S_{A_6{B}_3}\wedge S_{B_2{B}_3}\right)·\left(S_{B_1{B}_3}\wedge S_{A_5{B}_3}\wedge S_{A_6{B}_3}\wedge S_{B_2{B}_3}\right)=0\\ \iff \left(B_{b3}\wedge S_{B_2{B}_3}\right)·\left(B_{b3}\wedge S_{B_2{B}_3}\right)=0\end{array}$$

(29)

According to the formula to dissect a point from the point pair [14,27], we have:

$$B_2=\pm \sqrt{{\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }}{\mathit{B}}_{2e}^{-1}+{\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{2e}^{-1},$$

(30)

where $B_{2e}=B_{b2}^{\ast }·e_{\infty }$, and $B_{2e}^{-1}={(B_{b2}^{\ast }·e_{\infty })}^{-1}=\frac{B_{2e}}{B_{2e}·B_{2e}}$, $B_{2e}$ and $B_{2e}^{-1}$ are both 4-blade.

According to Table 1, sphere $S_{B_2{B}_3}$ is represented as

$$S_{B_2{B}_3}=B_2-R{e}_{\infty },$$

(31)

where $R=\frac{1}{2}{r}_1^2$.

By substituting Equation (30) into Equation (31), we obtain:

$$S_{B_2{B}_3}=\pm \sqrt{{\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }}{\mathit{B}}_{2e}^{-1}+{\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{2e}^{-1}-R{\mathit{e}}_{\infty }.$$

(32)

By substituting Equation (32) into Equation (29) and simplifying it, we have:

$$H_2+H_1+H_0=0,$$

(33)

where:

$$\begin{gathered} H_2=\left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }\right)\left(\left({\mathit{B}}_{b3}\wedge {\mathit{B}}_{2e}^{-1}\right)·\left({\mathit{B}}_{b3}\wedge {\mathit{B}}_{2e}^{-1}\right)\right), H_1=\pm 2\sqrt{{\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }}\left(B_{b3}\wedge {\mathit{B}}_{2e}^{-1}\right)·\left({\mathit{B}}_{b3}\wedge \left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{2e}^{-1}\right)-R{\mathit{B}}_{3\mathit{e}}\right), \\ {H}_0=\left({\mathit{B}}_{b3}\wedge \left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{2e}^{-1}\right)-R{\mathit{B}}_{3e}\right)·\left({\mathit{B}}_{b3}\wedge \left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{2e}^{-1}\right)-R{\mathit{B}}_{3e}\right) \end{gathered}$$

${\mathit{B}}_{3e}={\mathit{B}}_{b3}\wedge {\mathit{e}}_{\infty }$, ${\mathit{B}}_{3e}$ is 4-blade.

According to Equation (33), we derive:

$$H_1^2-{\left(H_2+H_0\right)}^2=0,$$

(34)

and expanding Equation (34), we have:

$$C_4{R}^4+C_3{R}^3+C_2{R}^2+C_{1}R+C_0=0,$$

(35)

where:

$$C_4=-{\left({\mathit{B}}_{3e}·{\mathit{B}}_{3e}\right)}^2,$$

$$C_3=4\left({\mathit{B}}_{3e}·{\mathit{U}}_2\right)\left({\mathit{B}}_{3e}·{\mathit{B}}_{3e}\right),$$

$$\begin{array}{c}{C}_2=4\left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }\right){\left({\mathit{B}}_{3e}·{\mathit{U}}_1\right)}^2-4{\left({\mathit{B}}_{3e}·{\mathit{U}}_2\right)}^2-2\left({\mathit{B}}_{3e}·{\mathit{B}}_{3e}\right)\left(\left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }\right)\left({\mathit{U}}_1·{\mathit{U}}_1\right)+\left({\mathit{U}}_2·{\mathit{U}}_2\right)\right),\hfill \\ C_1=4\left({\mathit{B}}_{3e}·{\mathit{U}}_2\right)\left(\left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }\right)\left({\mathit{U}}_1·{\mathit{U}}_1\right)+\left({\mathit{U}}_2·{\mathit{U}}_2\right)\right)-8\left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }\right)\left({\mathit{B}}_{3e}·{\mathit{U}}_1\right)\left({\mathit{U}}_1·{\mathit{U}}_2\right),\hfill \\ C_0=4\left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }\right){\left({\mathit{U}}_1·{\mathit{U}}_2\right)}^2-{\left(\left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }\right)\left({\mathit{U}}_1·{\mathit{U}}_1\right)+{\mathit{U}}_2·{\mathit{U}}_2\right)}^2,\hfill \end{array}$$

${\mathit{U}}_1=\left({\mathit{B}}_{b3}\wedge {\mathit{B}}_{2e}^{-1}\right)$, ${\mathit{U}}_2=\left({\mathit{B}}_{b3}\wedge \left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{2e}^{-1}\right)\right)$. $C_i\left(i=0,1,\cdots ,4\right)$ are all scalars. ${\mathit{U}}_1$ and ${\mathit{U}}_2$ are both 4-blade.

By multiplying both sides of Equation (35) with $\left({\mathit{B}}_{2e}·{\mathit{B}}_{2e}\right)\left({\mathit{B}}_{2e}·{\mathit{B}}_{2e}\right)$ and simplifying it, we have:

$$D_4{R}^4+D_3{R}^3+D_2{R}^2+D_{1}R+D_0=0,$$

(36)

where $D_i=\left({\mathit{B}}_{2e}·{\mathit{B}}_{2e}\right)\left({\mathit{B}}_{2e}·{\mathit{B}}_{2e}\right)C_i(i=0,1,\cdots ,4)$.

Simplifying the coefficients $D_4$,$D_3$,$D_2$,$D_1$, and $D_0$ by using Equations (1)–(10), we have:

$$D_4=-{\left({\mathit{B}}_{2e}·{\mathit{B}}_{2e}\right)}^2{\left({\mathit{B}}_{3e}·{\mathit{B}}_{3e}\right)}^2$$

(37)

$$D_3=4\left({\mathit{B}}_{2e}·{\mathit{B}}_{2e}\right)\left({\mathit{B}}_{3e}·{\mathit{B}}_{3e}\right)\left({\mathit{B}}_{3e}·{\mathit{U}}_4\right),$$

(38)

$$D_2=4\left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }\right){\left({\mathit{B}}_{3e}·{\mathit{U}}_3\right)}^2-4{\left({\mathit{B}}_{3e}·{\mathit{U}}_4\right)}^2-2\left({\mathit{B}}_{3e}·{\mathit{B}}_{3e}\right)\left(\left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }\right)\left({\mathit{U}}_3·{\mathit{U}}_3\right)+\left({\mathit{U}}_4·{\mathit{U}}_4\right)\right),$$

(39)

$$\begin{array}{c}{D}_1=4\left(-V_7{M}_1{M}_8+V_9{M}_1{M}_9+V_9{M}_2{M}_8-V_8{M}_2{M}_9\right)\hfill \\ +4\left(V_5\left(V_1{M}_8\left(V_6{T}_1+V_1{R}_{r1}{T}_3\right)+V_1{M}_9\left(-V_6{T}_2-V_2{R}_{r1}{T}_3\right)+V_2{M}_9\left(V_6{T}_1+V_2{R}_{r1}{T}_6\right)+V_2{M}_8\left(-V_6{T}_4-V_1{R}_{r1}{T}_6\right)\right)\right)\hfill \\ +4\left(V_5\left(-V_1{M}_3+V_2{M}_4\right)+\left(-R_{r1}{V}_0{M}_7+V_0{M}_5+V_3{M}_8-V_4{M}_9\right)R_{r1}{M}_6+\left(-V_3{M}_1+V_4{M}_2\right)\left(-R_{r1}{M}_7+M_5\right)\right)\hfill \end{array}$$

(40)

$$\begin{array}{c}{D}_0=4V_7{V}_9{T}_1{T}_2-V_7^2{T}_2^2+4V_3{V}_7{R}_{r1}{T}_2{T}_3-4V_0{V}_7{R}_{r1}^2{T}_3^2+4V_8{V}_9{T}_1{T}_4-V_8^2{T}_4^2+8V_0{V}_9{R}_{r1}^2{T}_3{T}_6-4V_4{V}_8{R}_{r1}{T}_4{T}_6,\hfill \\ -4V_0{V}_8{R}_{r1}^2{T}_6^2+4V_0{V}_3{R}_{r1}^3{T}_3{T}_9-4V_0{V}_4{R}_{r1}^3{T}_6{T}_9-V_0^2{R}_{r1}^4{T}_9^2-2\left(V_9^2+V_7{V}_8-V_5{V}_6^2\right)T_1^2-2\left(V_0^2+V_5{V}_6^2\right)T_2{T}_4\hfill \\ -2\left(V_4{V}_7{R}_{r1}+V_3{V}_9{R}_{r1}+3V_1{V}_5{V}_6{R}_{r1}\right)T_2{T}_6-2\left(V_0{V}_7{R}_{r1}^2+2V_5{V}_1^2{R}_{r1}^2\right)T_2{T}_9\hfill \\ -2\left(3V_4{V}_7{R}_{r1}+V_3{V}_9{R}_{r1}-V_1{V}_5{V}_6{R}_{r1}\right)T_1{T}_3+2\left(V_4{V}_9{R}_{r1}+V_3{V}_8{R}_{r1}-3V_2{V}_5{V}_6{R}_{r1}\right)T_3{T}_4\hfill \\ +2\left(V_4{V}_9{R}_{r1}+3V_3{V}_8{R}_{r1}+V_2{V}_5{V}_6{R}_{r1}\right)T_1{T}_6+4\left(V_0{V}_9{R}_{r1}^2+2V_1{V}_2{V}_5{R}_{r1}^2\right)T_1{T}_9-2\left(V_0{V}_8{R}_{r1}^2+2V_5{V}_2^2{R}_{r1}^2\right)T_4{T}_9\hfill \end{array},$$

(41)

where ${\mathit{U}}_3=\left({\mathit{B}}_{b3}\wedge {\mathit{B}}_{2e}\right)$,${\mathit{U}}_4=\left({\mathit{B}}_{b3}\wedge \left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{2e}\right)\right)$, ${\mathit{U}}_3$ and ${\mathit{U}}_4$ are both 4-blade:

$$\begin{gathered} M_1=V_4{T}_1-V_3{T}_2+V_0{R}_{r1}{T}_3,M_2=V_4{T}_4-V_3{T}_5+V_0{R}_{r1}{T}_6,M_3=V_2{T}_1-V_1{T}_2,M_4=V_2{T}_4-V_1{T}_5,M_5=T_{10}, \\ {M}_6=V_4{T}_7-V_3{T}_8+V_0{R}_{r1}{T}_3,M_7=T_7-T_8+T_9,M_8=T_1-T_2+T_3,M_9=T_4-T_5+T_6,R_{r1}=\left(r_2^2-r_3^2\right)/2, \\ {V}_0=\left({\mathit{B}}_{2e}·{\mathit{B}}_{2e}\right),V_1=\left({\mathit{B}}_{2e}·{\mathit{S}}_{A_5{B}_3}\right),V_2=\left({\mathit{B}}_{2e}·{\mathit{S}}_{A_6{B}_3}\right),V_3=\left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{2e}\right)·{\mathit{S}}_{A_5{B}_3},V_4=\left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{2e}\right)·{\mathit{S}}_{A_6{B}_3},V_5=\left({\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }\right), \\ {V}_6=\left({\mathit{B}}_{b2}^{\ast }·\left({\mathit{S}}_{A_5{B}_3}\wedge {\mathit{S}}_{A_6{B}_3}\right)\right),V_7=\left({\mathit{B}}_{b2}^{\ast }·{\mathit{S}}_{A_5{B}_3}\right)·\left({\mathit{B}}_{b2}^{\ast }·{\mathit{S}}_{A_5{B}_3}\right),V_8=\left({\mathit{B}}_{b2}^{\ast }·{\mathit{S}}_{A_6{B}_3}\right)·\left({\mathit{B}}_{b2}^{\ast }·{\mathit{S}}_{A_6{B}_3}\right),V_9=\left({\mathit{B}}_{b2}^{\ast }·{\mathit{S}}_{A_5{B}_3}\right)·\left({\mathit{B}}_{b2}^{\ast }·{\mathit{S}}_{A_6{B}_3}\right). \end{gathered}$$

$$\begin{array}{c}{T}_1=\left|\begin{array}{cc}{S}_{A_5{B}_3}·S_{B_1{B}_3}& S_{A_5{B}_3}·S_{A_6{B}_3}\\ S_{B_1{B}_3}·S_{B_1{B}_3}& S_{B_1{B}_3}·S_{A_6{B}_3}\end{array}\right|,T_2=\left|\begin{array}{cc}{S}_{A_6{B}_3}·S_{B_1{B}_3}& S_{A_6{B}_3}·S_{A_6{B}_3}\\ S_{B_1{B}_3}·S_{B_1{B}_3}& S_{B_1{B}_3}·S_{A_6{B}_3}\end{array}\right|,T_3=\left|\begin{array}{cc}{S}_{A_6{B}_3}·S_{B_1{B}_3}& S_{A_6{B}_3}·S_{A_6{B}_3}\\ S_{A_5{B}_3}·S_{B_1{B}_3}& S_{A_5{B}_3}·S_{A_6{B}_3}\end{array}\right|,\hfill \\ T_4=\left|\begin{array}{cc}{S}_{A_5{B}_3}·S_{B_1{B}_3}& S_{A_5{B}_3}·S_{A_5{B}_3}\\ S_{B_1{B}_3}·S_{B_1{B}_3}& S_{B_1{B}_3}·S_{A_5{B}_3}\end{array}\right|,T_5=\left|\begin{array}{cc}{S}_{A_6{B}_3}·S_{B_1{B}_3}& S_{A_6{B}_3}·S_{A_5{B}_3}\\ S_{B_1{B}_3}·S_{B_1{B}_3}& S_{B_1{B}_3}·S_{A_5{B}_3}\end{array}\right|,T_6=\left|\begin{array}{cc}{S}_{A_6{B}_3}·S_{B_1{B}_3}& S_{A_6{B}_3}·S_{A_5{B}_3}\\ S_{A_5{B}_3}·S_{B_1{B}_3}& S_{A_5{B}_3}·S_{A_5{B}_3}\end{array}\right|,\hfill \\ T_7=\left|\begin{array}{cc}{S}_{A_5{B}_3}·S_{A_5{B}_3}& S_{A_5{B}_3}·S_{A_6{B}_3}\\ S_{B_1{B}_3}·S_{A_5{B}_3}& S_{B_1{B}_3}·S_{A_6{B}_3}\end{array}\right|,T_8=\left|\begin{array}{cc}{S}_{A_6{B}_3}·S_{A_5{B}_3}& S_{A_6{B}_3}·S_{A_6{B}_3}\\ S_{B_1{B}_3}·S_{A_5{B}_3}& S_{B_1{B}_3}·S_{A_6{B}_3}\end{array}\right|,T_9=\left|\begin{array}{cc}{S}_{A_6{B}_3}·S_{A_5{B}_3}& S_{A_6{B}_3}·S_{A_6{B}_3}\\ S_{A_5{B}_3}·S_{A_5{B}_3}& S_{A_5{B}_3}·S_{A_6{B}_3}\end{array}\right|,\hfill \\ T_{10}=\left|\begin{array}{ccc}{S}_{A_6{B}_3}·S_{B_1{B}_3}& S_{A_6{B}_3}·S_{A_5{B}_3}& S_{A_6{B}_3}·S_{A_6{B}_3}\\ S_{A_5{B}_3}·S_{B_1{B}_3}& S_{A_5{B}_3}·S_{A_5{B}_3}& S_{A_5{B}_3}·S_{A_6{B}_3}\\ S_{B_1{B}_3}·S_{B_1{B}_3}& S_{B_1{B}_3}·S_{A_5{B}_3}& S_{B_1{B}_3}·S_{A_6{B}_3}\end{array}\right|,T_5=T_1,T_7=T_6,T_8=T_3.\hfill \end{array}$$

$D_i(i=0,1,\cdots ,4)$,$M_i\left(i=1,2,\cdots ,9\right)$,$T_i\left(i=1,2,\cdots ,10\right)$, and $V_i\left(i=0,1,\cdots ,9\right)$ are all scalars. In addition, the detailed simplification procedure of $D_i(i=0,1,\cdots ,4)$ is given in the Appendix B.

For the 6-3 Stewart platforms, the geometric constraint equation Equation (36) depends on only the design parameters, the inputs, and the coordinates of point $B_1$.

4.4. The Univariate Polynomial Equation for Forward Displacement Analysis

According to Table 1, point $B_1$ can be denoted in CGA as:

$${\mathit{B}}_1={\mathit{b}}_1+\frac{1}{2}{{\mathit{b}}_1}^2{e}_{\infty }+e_0,$$

(42)

where the coordinates of point $B_1$ in coordinate system O-XYZ can be expressed as:

$$b_1=a_1+{\mathit{M}}_r·\left({\left(0,l_1\sin ({\theta }_1),0\right)}^{\mathrm{T}}+{\left(l_1\cos ({\theta }_1),0,0\right)}^{\mathrm{T}}\right).$$

(43)

According to Equations (36) and (42), a 16th-degree polynomial equation about $x$ is formulated by using the Euler angle substitution ($\cos {\theta }_1=\left(\mathrm{x}+1/x\right)/2$, $\sin {\theta }_1=\left(\mathrm{x}-1/x\right)/(2i)$, $\mathrm{i}=\sqrt{-1}$ and $x=e^{i{\theta }_1}$).

4.5. Back Substitution

After the values of $x$ are given, the coordinates of point $B_1$ can be solved by Equation (42). Two point pairs $B_{b2}^{\ast }$ and $B_{b3}$ can be calculated using Equations (27) and (28). According to Table 1, we can obtain:

$${\mathit{B}}_3^{\ast }=S_{B_1{B}_3}\wedge S_{A_5{B}_3}\wedge S_{A_6{B}_3}\wedge S_{B_2{B}_3}=B_{b3}\wedge S_{B_2{B}_3}$$

(44)

where the coordinates of point $B_2$ are unknown.

Since two values for point $B_2$ can be solved by Equation (30), aiming to receive the unique value of point $B_2$, a novel objective function ${\mathit{g}}_p$ for point $B_2$ is formulated as:

$${\mathit{g}}_p=\left(\left(B_{b3}\wedge {\mathit{S}}_{p1}\right)·\left(B_{b3}\wedge {\mathit{S}}_{p1}\right)\right){\mathit{B}}_{22}+\left(\left(B_{b3}\wedge {\mathit{S}}_{p2}\right)·\left(B_{b3}\wedge {\mathit{S}}_{p2}\right)\right){\mathit{B}}_{21},$$

(45)

where ${\mathit{S}}_{p1}={\mathit{B}}_{21}-R{\mathit{e}}_{\infty }$, ${\mathit{S}}_{p2}={\mathit{B}}_{22}-R{\mathit{e}}_{\infty }$, ${\mathit{B}}_{21}=\sqrt{{\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }}{\mathit{B}}_{2e}^{-1}+{\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{2e}^{-1}$, ${\mathit{B}}_{22}=-\sqrt{{\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{b2}^{\ast }}{\mathit{B}}_{2e}^{-1}+{\mathit{B}}_{b2}^{\ast }·{\mathit{B}}_{2e}^{-1}$, ${\mathit{B}}_{21}\wedge {\mathit{B}}_{22}={\mathit{B}}_{b2}^{\ast }$.

By expanding Equation (45), we readily obtain:

$${\mathit{g}}_p={\mathit{F}}_2{R}^2+{\mathit{F}}_{1}R+{\mathit{F}}_0,$$

(46)

where ${\mathit{F}}_2=\left({\mathit{B}}_{3e}·{\mathit{B}}_{3e}\right)\left({\mathit{B}}_{21}+{\mathit{B}}_{22}\right)$, ${\mathit{F}}_1=-2\left(\left({\mathit{B}}_{3e}·\left(B_{b3}\wedge {\mathit{B}}_{21}\right)\right){\mathit{B}}_{22}+\left({\mathit{B}}_{3e}·\left(B_{b3}\wedge {\mathit{B}}_{22}\right)\right){\mathit{B}}_{21}\right)$,

${\mathit{F}}_0=\left(\left(B_{b3}\wedge {\mathit{B}}_{21}\right)·\left(B_{b3}\wedge {\mathit{B}}_{21}\right)\right){\mathit{B}}_{22}+\left(\left(B_{b3}\wedge {\mathit{B}}_{22}\right)·\left(B_{b3}\wedge {\mathit{B}}_{22}\right)\right){\mathit{B}}_{21}$. ${\mathit{F}}_0$, ${\mathit{F}}_1$, and ${\mathit{F}}_2$ are all 1-blade.

By simplifying the 1-blades ${\mathit{F}}_2$, ${\mathit{F}}_1$ and ${\mathit{F}}_0$, we obtain:

$$\begin{gathered} {\mathit{F}}_2=2\left({\mathit{B}}_3{}_e·{\mathit{B}}_3{}_e\right)\left(B_{b2}^{\ast }·{\mathit{B}}_{2e}\right)/V_0, \\ {\mathit{F}}_1=-4\left(-V_5\left({\mathit{B}}_{3e}·\left(B_{b3}\wedge {\mathit{B}}_{2e}\right)\right){\mathit{B}}_{2e}+\left({\mathit{B}}_{3e}·\left(B_{b3}\wedge \left(B_{b2}^{\ast }·{\mathit{B}}_{2e}\right)\right)\right)\left(B_{b2}^{\ast }·{\mathit{B}}_{2e}\right)\right)/V_0^2, \\ \begin{array}{l}{\mathit{F}}_0=2\left(V_5{\left(B_{b3}\wedge {\mathit{B}}_{2e}\right)}^2+{\left(B_{b3}\wedge \left(B_{b2}^{\ast }·{\mathit{B}}_{2e}\right)\right)}^2\right)\left(B_{b2}^{\ast }·{\mathit{B}}_{2e}\right)/V_0^3\\ -4V_5\left(\left(B_{b3}\wedge {\mathit{B}}_{2e}\right)·\left(B_{b3}\wedge \left(B_{b2}^{\ast }·{\mathit{B}}_{2e}\right)\right)\right){\mathit{B}}_{2e}/V_0^3.\end{array} \end{gathered}$$

The expression of point ${\mathit{B}}_2$ is the expression for a point (${\mathit{B}}_{21}$ or ${\mathit{B}}_{22}$) of point pair $B_{b2}^{\ast }$. Analyzing carefully Equation (45), on the one hand, when $\left(B_{b3}\wedge {\mathit{S}}_{p1}\right)·\left(B_{b3}\wedge {\mathit{S}}_{p1}\right)$ is equal to zero, we can have ${\mathit{B}}_{21}={\mathit{g}}_p/\left(\left(B_{b3}\wedge {\mathit{S}}_{p2}\right)·\left(B_{b3}\wedge {\mathit{S}}_{p2}\right)\right)$. On the other hand, when $\left(B_{b3}\wedge {\mathit{S}}_{p2}\right)·\left(B_{b3}\wedge {\mathit{S}}_{p2}\right)$ is equal to zero, we can also acquire ${\mathit{B}}_{22}={\mathit{g}}_p/\left(\left(B_{b3}\wedge {\mathit{S}}_{p1}\right)·\left(B_{b3}\wedge {\mathit{S}}_{p1}\right)\right)$. $\left(B_{b3}\wedge {\mathit{S}}_{p2}\right)·\left(B_{b3}\wedge {\mathit{S}}_{p2}\right)$ and $\left(B_{b3}\wedge {\mathit{S}}_{p1}\right)·\left(B_{b3}\wedge {\mathit{S}}_{p1}\right)$ are both 4-blade. Therefore, we can obtain the coordinates of point ${\mathit{B}}_2$ by the normalization of a point in CGA, i.e.,:

$${\mathit{B}}_2=-{\mathit{g}}_p/\left({\mathit{g}}_p·{\mathit{e}}_{\infty }\right).$$

(47)

By substituting Equation (47) into Equation (44), the coordinates of point ${\mathit{B}}_3$ are equal to the normalization of 4-blade ${\mathit{B}}_3^{\ast }$, i.e.,:

$${\mathit{B}}_3=-{\mathit{B}}_3^{\ast }/\left({\mathit{B}}_3^{\ast }\wedge {\mathit{e}}_{\infty }\right).$$

(48)

4.6. Two Comparisons

Compared with the traditional algebra method for forward displacement analysis of 6-3 Stewart platforms, the method proposed in this paper has the following characteristics:

(1) The derivation process of the proposed method is geometrically intuitive due to the intuitiveness of CGA.

(2) The univariate symbolic polynomial equation (Equation (36)) is derived directly by CGA operation without algebraic elimination.

Compared with the existing geometric algebra method to solve 6-3 Stewart platforms, the characteristics of this method proposed in this paper are as follows:

(1) In the back substitution procedure, the proposed method does not need to determine the radical symbols.

(2) Two formulas between 2-blade and 1-blade are first formulated.

(3) The proposed method based on the determinant operation formula (Equation (8)) of the CGA inner product provides a new idea to solve a more complex spatial parallel mechanism in the future, which is not available in Refs. [12,13,14].

5. Numerical Example

In order to verify the proposed modeling and calculation method using the CGA framework, two numerical examples of the 6-3 Stewart platforms are given.

5.1. Example 1

The structural parameters and inputs of the 6-3 Stewart platforms are as follows:

$$\begin{gathered} {\mathit{a}}_1={\left(50,0,100\right)}^{\mathrm{T}},{\mathit{a}}_2={\left(-25,-2,40\right)}^{\mathrm{T}},{\mathit{a}}_3={\left(80,20,50\right)}^{\mathrm{T}},{\mathit{a}}_4={\left(-50,-20,70\right)}^{\mathrm{T}},{\mathit{a}}_5={\left(48,15,68\right)}^{\mathrm{T}},{\mathit{a}}_6={\left(36,31,-93\right)}^{\mathrm{T}}, \\ {l}_1=76,l_2=160,l_3=139,l_4=55,l_5=128,l_6=217,r_1=135,r_2=190,r_3=141. \end{gathered}$$

Substituting the link parameters and inputs into Equation (36), the polynomial of degree sixteen is formulated as:

$$\begin{array}{c}\left(0.349811-0.93682i\right)-\left(114.971-57.843i\right)x+\left(2631.72+1824.52i\right)x^2+\left(31170.7-13317.2i\right)x^3\hfill \\ +\left(78681.7-165524i\right)x^4-\left(147210+486737i\right)x^5-\left(347339+300463i\right)x^6\hfill \\ +\left(398116-230184i\right)x^7+\left(994070-689922i\right)x^8+\left(354906-292442i\right)x^9\hfill \\ +\left(159977+430499i\right)x^{10}+\left(404489+308175i\right)x^{11}+\left(182590-15808.4i\right)x^{12}\hfill \\ +\left(23379.7-24542.8i\right)x^{13}-\left(788.664+3103.68i\right)x^{14}-\left(94.4066-87.4728i\right)x^{15}+x^{16}=0.\hfill \end{array}$$

(49)

The sixteen sets of solutions received by solving the polynomial equation Equation (49) agree with those given in [13,14]. The four sets of real solutions are shown in

Table 2. Four sets of real solutions of point $B_i(i=1,2,3)$.

	$x$	Coordinates	B1	B2	B3
1	0.8920 + 0.4520i	$X$	79.5345	−26.0943	−70.9222
		$Y$	−45.8809	−68.9458	54.3105
		$Z$	152.9020	62.3955	94.3853
2	−0.9997 + 0.0231i	$X$	90.9017	−40.7764	14.0676
		$Y$	53.3945	6.28516	−108.359
		$Z$	135.3850	117.424	71.8847
3	−0.9597 + 0.2809i	$X$	82.5389	−48.8262	−6.78228
		$Y$	51.0783	24.7277	−100.5170
		$Z$	145.9150	101.985	74.2181
4	0.6405 + 0.7680i	$X$	68.8676	−47.0212	21.2493
		$Y$	−33.0062	21.0887	137.061
		$Z$	165.8070	106.44	95.7391

and Figure 2.

5.2. Example 2

The link parameters and inputs of the planar 6-3 Stewart platforms are shown in

Table 3. Link parameters and inputs for example 2.

O-XYZ	A1	A2	A3	A4	A5	A6
aix	0	0	4.0803	1.4802	5.7676	6.2278
aiy	−0.6749	3.2366	−2.3929	−3.5971	3.8962	0.7516
aiz	0	0	0	0	0	0
$r_1=2.0,r_2=2.0,r_3=3.0$
$l_1=5.0,l_2=4.5,l_3=5.0,l_4=5.5,l_5=5.5,l_6=5.7$

. According to the steps mentioned above, substituting the link parameters and inputs into Equation (36), we can obtain sixteen solutions. The sixteen sets of solutions are consistent with those given in [14]. The four sets of real solutions are shown in

Table 4. Four sets of real solutions of point $B_i(i=1,2,3)$.

	$x$	Coordinates	B1	B2	B3
1	0.0366 − 0.9993i	$X$	0.1570	1.6474	2.0248
		$Y$	1.8880	1.6311	2.0987
		$Z$	4.2903	1.6994	3.6070
2	0.0366 + 0.9993i	$X$	0.1570	1.6474	2.0248
		$Y$	1.8880	1.6311	2.0987
		$Z$	−4.2903	−1.6994	−3.6070
3	0.7239 − 0.6899i	$X$	3.1079	1.5287	1.1961
		$Y$	1.8880	1.8873	1.9774
		$Z$	2.9618	0.4111	2.3812
4	0.7239 + 0.6899i	$X$	3.1079	1.5287	1.1961
		$Y$	1.8880	1.8873	1.9774
		$Z$	−2.9618	−0.4111	−2.3812

and Figure 3.

6. Conclusions

In this paper, a novel geometric modeling and calculation method for closed-form solution of 6-3 Stewart platforms based on the framework of conformal geometric algebra is presented. Two formulas between 2-blade and 1-blade are first obtained. A 16-degree and coordinate-invariant polynomial equation by using the Euler angle substitution was derived without algebraic elimination. The final numerical results show that the proposed geometric modeling and calculation method is effective. Since the result obtained by the proposed method is a closed symbolic solution, the method is suitable to write the computer program. Compared with the existing methods for the forward displacement analysis of the 6-3 Stewart platforms, the highlight of this paper is that a new geometric modeling and calculation method without algebraic elimination is obtained by using the determinant form of CGA inner product algorithm (Equation (8)), which can be used for the forward displacement analysis of other complex parallel mechanisms. In addition, compared with [14] in the back substitution procedure, the proposed method does not need to determine the radical symbols.