3.1. Inverse solution of pose

The inverse solution of pose establishes the mapping relationship between the posture parameters of moving platform and the displacement of prismatic pair in the local coordinate system. The posture transformation matrix of the parallel posture alignment mechanism is:

(3) \begin{equation} \textbf{R}_{t}^{b}=\textbf{R}\!\left ({{y}_{b}},\beta _{t}^{b} \right )\textbf{R}\!\left ({{x}_{b}},\alpha _{t}^{b} \right ) = {{\left. \left [ \begin{matrix} \cos \beta _{t}^{b} & \quad \sin \beta _{t}^{b}\sin \alpha _{t}^{b} & \quad \sin \beta _{t}^{b}\cos \alpha _{t}^{b} \\[5pt] 0 & \quad \cos \alpha _{t}^{b} & \quad -\!\sin \alpha _{t}^{b} \\[5pt] {-\!\sin} \beta _{t}^{b} & \quad \cos \beta _{t}^{b}\sin \alpha _{t}^{b} & \quad \cos \beta _{t}^{b}\cos \alpha _{t}^{b} \\[5pt] \end{matrix} \right ] \right |}_{\gamma _{t}^{b}=0}} \end{equation}

where the posture alignment parallel mechanism has no degree of freedom of rotation about the $z_b$ axis of the global coordinate system. As shown in Eq. (4), the displacement of each prismatic pair can be obtained by using space vector chain method:

(4) \begin{align} & d_{i}^{x}={{\left ( \textbf{R}_{t}^{b}{\lambda }_{si}^{t}+\textbf{P}_{t}^{b}-{\lambda }_{oi}^{b} \right )}^{\text{T}}}{{\textbf{n}}^{x}} \nonumber\\[3pt] & d_{i}^{y}={{\left ( \textbf{R}_{t}^{b}{\lambda }_{si}^{t}+\textbf{P}_{t}^{b}-{\lambda }_{oi}^{b} \right )}^{\text{T}}}{{\textbf{n}}^{y}} \\[3pt] & d_{i}^{z}={{\left ( \textbf{R}_{t}^{b}{\lambda }_{si}^{t}+\textbf{P}_{t}^{b}-{\lambda }_{oi}^{b} \right )}^{\text{T}}}{{\textbf{n}}^{z}} \nonumber \end{align}

where ${\textbf{n}}_x$ = [1, 0, 0] $^T$ , ${\textbf{n}}_y$ = [0, 1, 0] $^T$ and ${\textbf{n}}_z$ = [0, 0, 1] $^T$ . ${\textbf{n}}_x$ , ${\textbf{n}}_y$ and ${\textbf{n}}_z$ denote the direction vector. ${\textbf{P}}^b_t$ = [ $x^b_t$ , $y^b_t$ , $z^b_t$ ] $^T$ denotes the vector of moving platform coordinate system origin $O_t$ relative to global coordinate system origin $O_b$ . ${\lambda }^t_{si}$ represents the vector of $S_i$ relative to $O_t$ . It can be calculated from Eq. (5) that the displacement of each prismatic pair of the positioner driven by servo motor is:

(5) \begin{equation} \begin{aligned} & d_{1}^{z}={-l\sin \beta _{t}^{b}}/{2-{w\cos \beta _{t}^{b}\sin \alpha _{t}^{b}}/{2-h\cos \beta _{t}^{b}\cos \alpha _{t}^{b}+z_{t}^{b}}\;}\; \\[3pt] & d_{2}^{x}=l-l\cos \beta _{t}^{b} \\[3pt] & d_{2}^{z}={l\sin \beta _{t}^{b}}/{2-{w\cos \beta _{t}^{b}\sin \alpha _{t}^{b}}/{2-h\cos \beta _{t}^{b}\cos \alpha _{t}^{b}+z_{t}^{b}}\;}\; \\[3pt] & d_{3}^{z}=l{\sin \beta _{t}^{b}}/{2+{w\cos \beta _{t}^{b}\sin \alpha _{t}^{b}}/{2-h\cos \beta _{t}^{b}\cos \alpha _{t}^{b}+z_{t}^{b}}\;}\; \\[3pt] & d_{4}^{z}={-l\sin \beta _{t}^{b}}/{2+{w\cos \beta _{t}^{b}\sin \alpha _{t}^{b}}/{2-h\cos \beta _{t}^{b}\cos \alpha _{t}^{b}+z_{t}^{b}}\;}\; \end{aligned} \end{equation}

where $d^z_i$ and $d^x_i$ represent the relative displacement of prismatic pair in z and $x$ direction, respectively.

3.2. Inverse solution of velocity

The inverse solution of pose establishes the mapping relationship between the generalized velocity of moving platform and the velocity of prismatic pair in the local coordinate system. Relative velocity of prismatic pair can be calculated as follow:

(6) \begin{equation} {{v}_{i}^j}=\left [ \textbf{v}_{t}^{b}+{\omega }_{t}^{b}\times \left ( \textbf{R}_{t}^{b}{\lambda }_{si}^{t} \right ) \right ] \centerdot \textbf{n}_{i}^{j}=\left [ \begin{matrix}{{\left ( \textbf{n}_{i}^{j} \right )}^{\text{T}}} &{{\left ( \textbf{R}_{t}^{b}{\lambda }_{si}^{t}\times \textbf{n}_{i}^{j} \right )}^{\text{T}}} \\ \end{matrix} \right ]\left [ \begin{matrix} \textbf{v}_{t}^{b} \\[4pt] {\omega }_{t}^{b} \\ \end{matrix} \right ] \end{equation}

where v $^b_t$ and ${{\omega }}^b_t$ denote velocity and angular velocity of moving platform in global coordinate system, respectively. The velocity v $^b_t$ and angular velocity ${\omega }^b_t$ of moving platform can be expressed as:

(7) \begin{equation} {\omega }_{t}^{b}=\textbf{R}\!\left ({{y}_{b}},\beta _{t}^{b} \right )\left [ \begin{matrix} 1 \\[4pt] 0 \\[4pt] 0 \\[4pt] \end{matrix} \right ]\dot{\alpha }_{t}^{b}+\left [ \begin{matrix} 0 \\[4pt] 1 \\[4pt] 0 \\[4pt] \end{matrix} \right ]\dot{\beta }_{t}^{b}=\left [ \begin{matrix} 0 & \quad \cos \beta _{t}^{b} & \quad 0 \\[4pt] 0 & \quad 0 & \quad 1 \\[4pt] 0 & \quad -\!\sin \beta _{t}^{b} & \quad 0 \\[4pt] \end{matrix} \right ]\left [ \begin{matrix} \dot{z}_{t}^{b} \\[4pt] \dot{\alpha }_{t}^{b} \\[4pt] \dot{\beta }_{t}^{b} \\[4pt] \end{matrix} \right ] \end{equation}

(8) \begin{equation} \textbf{v}_{t}^{b}=\left [ \begin{matrix} 0 & \quad {\partial x_{t}^{b}}/{\partial \alpha _{t}^{b}}\; & \quad {\partial x_{t}^{b}}/{\partial \beta _{t}^{b}}\; \\[4pt] 0 & \quad {\partial y_{t}^{b}}/{\partial \alpha _{t}^{b}}\; & \quad 0 \\[4pt] 1 & \quad 0 & \quad 0 \\[4pt] \end{matrix} \right ]\left [ \begin{matrix} \dot{z}_{t}^{b} \\[4pt] \dot{\alpha }_{t}^{b} \\[4pt] \dot{\beta }_{t}^{b} \\[4pt] \end{matrix} \right ] \end{equation}

Therefore, the relationship between the speed of each prismatic pair and the generalized speed of moving platform can be expressed as:

(9) \begin{equation} \left [ \begin{matrix} v_{1}^{z} \\[4pt] v_{2}^{x} \\[4pt] v_{2}^{z} \\[4pt] v_{3}^{x} \\[4pt] v_{3}^{y} \\[4pt] v_{3}^{z} \\[4pt] v_{4}^{x} \\[4pt] v_{4}^{y} \\[4pt] v_{4}^{z} \\[4pt] \end{matrix} \right ]=\left [ \begin{matrix} \textbf{n}_{z}^{\text{T}} & \quad {{\left ( {\lambda }_{s1}^{b}\times{{\textbf{n}}_{z}} \right )}^{\text{T}}} \\[4pt] \textbf{n}_{x}^{\text{T}} & \quad {{\left ( {\lambda }_{s2}^{b}\times{{\textbf{n}}_{x}} \right )}^{\operatorname{T}}} \\[4pt] \textbf{n}_{z}^{\text{T}} & \quad {{\left ( {\lambda }_{s2}^{b}\times{{\textbf{n}}_{z}} \right )}^{\text{T}}} \\[4pt] \textbf{n}_{x}^{\text{T}} & \quad {{\left ( {\lambda }_{s3}^{b}\times{{\textbf{n}}_{x}} \right )}^{\text{T}}} \\[4pt] \textbf{n}_{y}^{\text{T}} & \quad {{\left ( {\lambda }_{s3}^{b}\times{{\textbf{n}}_{y}} \right )}^{\text{T}}} \\[4pt] \textbf{n}_{z}^{\text{T}} & \quad {{\left ( {\lambda }_{s3}^{b}\times{{\textbf{n}}_{z}} \right )}^{\text{T}}} \\[4pt] \textbf{n}_{x}^{\text{T}} & \quad {{\left ( {\lambda }_{s4}^{b}\times{{\textbf{n}}_{x}} \right )}^{\text{T}}} \\[4pt] \textbf{n}_{y}^{\text{T}} & \quad {{\left ( {\lambda }_{s4}^{b}\times{{\textbf{n}}_{y}} \right )}^{\text{T}}} \\[4pt] \textbf{n}_{z}^{\text{T}} & \quad {{\left ( {\lambda }_{s4}^{b}\times{{\textbf{n}}_{z}} \right )}^{\text{T}}} \\[4pt] \end{matrix} \right ]\left [ \begin{matrix} 0 & \quad \dfrac{\partial x_{t}^{b}}{\partial \alpha _{t}^{b}} & \quad \dfrac{\partial x_{t}^{b}}{\partial \beta _{t}^{b}} \\[12pt] 0 & \quad \dfrac{\partial y_{t}^{b}}{\partial \alpha _{t}^{b}} & \quad 0 \\[8pt] 1 & \quad 0 & \quad 0 \\[4pt] 0 & \quad \cos \beta _{t}^{b} & \quad 0 \\[4pt] 0 & \quad 0 & \quad 1 \\[4pt] 0 & \quad -\!\sin \beta _{t}^{b} & \quad 0 \\[4pt] \end{matrix} \right ]\left [ \begin{matrix} \dot{z}_{t}^{b} \\[4pt] \dot{\alpha }_{t}^{b} \\[4pt] \dot{\beta }_{t}^{b} \\[4pt] \end{matrix} \right ] \end{equation}

where $v^x_i$ , $v^y_i$ and $v^z_i$ represent the relative velocity of prismatic pair in x, y and z direction, respectively. ${\lambda }^b_{si}$ represents the vector of $S_i$ relative to $O_b$ which can be expressed as ${\lambda }^b_{si}$ = R $^b_{t}$ ${\lambda }^t_{si}$ .

3.3. Inverse solution of acceleration

The inverse solution of pose establishes the mapping relationship between the generalized acceleration of moving platform and the acceleration of prismatic pair in the local coordinate system. Relative acceleration of prismatic pair can be calculated according to Eq. (10):

(10) \begin{equation} a_{i}^{j}={{\left [{\dot{v}}_{t}^{b}+{\dot{\omega }}_{t}^{b}\times {\lambda }_{si}^{b}+{\omega }_{t}^{b}\times \left ( {\omega }_{t}^{b}\times {\lambda }_{si}^{b} \right ) \right ]}^{\text{T}}}{{\textbf{n}}_{j}}=\left [ \begin{matrix} \textbf{n}_{j}^{\text{T}}{{\left ( {\lambda }_{si}^{b}\times{{\textbf{n}}_{j}} \right )}^{\text{T}}} \\ \end{matrix} \right ]\left [ \begin{matrix}{\dot{v}}_{t}^{b} \\[5pt] {\dot{\omega }}_{t}^{b} \\ \end{matrix} \right ]+{{\left [ {\omega }_{t}^{b}\times \left ( {\omega }_{t}^{b}\times {\lambda }_{si}^{b} \right ) \right ]}^{\text{T}}}{{\textbf{n}}_{j}} \end{equation}

where j represents the direction of prismatic pair and j = (x,y,z).

(11) \begin{equation} \left [ \begin{matrix}{\dot{v}}\it _{t}^{b}\rm \\[5pt] {\dot{\omega }}_{t}^{b} \\ \end{matrix} \right ]={{\textbf{J}}_{s}}{{\textbf{A}}_{s}}+{{\left ({{\textbf{V}}_{s}} \right )}^{\text{T}}}{{\textbf{H}}_{r}}{{\textbf{V}}_{s}} \end{equation}

where A $_s$ = [ ${\ddot z}^b_t$ , ${\ddot \alpha }^b_t$ , ${\ddot \beta }^b_t$ ] $^T$ . Hr is a $6\times 3\times 3$ Hessian matrix and H $_r$ = [h $_1$ h $_2$ h $_3$ h $_4$ h $_5$ h $_6$ ] $^T$ . h $_3$ and h $_5$ are 3-order matrix whose element is zero.

(12) \begin{equation} {{\textbf{h}}_{1}}=\left [ \begin{matrix} 0 & \quad 0 & \quad 0 \\[5pt] 0 & \quad \dfrac{{{\partial }^{2}}x_{t}^{b}}{\partial \alpha _{t}^{b}\partial \alpha _{t}^{b}} & \quad \dfrac{{{\partial }^{2}}x_{t}^{b}}{\partial \beta \partial \alpha _{t}^{b}} \\[12pt] 0 & \quad \dfrac{{{\partial }^{2}}x_{t}^{b}}{\partial \alpha _{t}^{b}\partial \beta _{t}^{b}} & \quad \dfrac{{{\partial }^{2}}x_{t}^{b}}{\partial \beta _{t}^{b}\partial \beta _{t}^{b}} \\[5pt] \end{matrix} \right ]\text{ }{{\textbf{h}}_{2}}=\left [ \begin{matrix} 0 & \quad 0 & \quad 0 \\[5pt] 0 & \quad \dfrac{{{\partial }^{2}}y_{t}^{b}}{\partial \alpha _{t}^{b}\partial \alpha _{t}^{b}} & \quad 0 \\[12pt] 0 & \quad 0 & \quad 0 \\[5pt] \end{matrix} \right ]\text{ } \end{equation}

(13) \begin{equation} \text{ }{{\textbf{h}}_{4}}=\left [ \begin{matrix} 0 & \quad 0 & \quad 0 \\[5pt] 0 & \quad 0 & \quad 0 \\[5pt] 0 & \quad -\!\sin \beta _{t}^{b} & \quad 0 \\[5pt] \end{matrix} \right ]\text{}{{\textbf{h}}_{6}}=\left [ \begin{matrix} 0 & \quad 0 & \quad 0 \\[5pt] 0 & \quad 0 & \quad 0 \\[5pt] 0 & \quad -\!\cos \beta _{t}^{b} & \quad 0 \\[5pt] \end{matrix} \right ] \end{equation}

4.1. Establishment of dynamic equation

During dynamic modeling, force analysis of each component is required. The force analysis of the moving platform is shown in Fig. 2. $^i{\textbf{F}}^p_t$ denotes the constraint force acting on the spherical hinge relative to the global coordinate system. $^o{\textbf{F}}^p_t$ and $^o{\textbf{M}}^p_t$ represent the equivalent external force and moment acting on the moving platform relative to the global coordinate system, respectively. $m_t$ denotes the mass of the moving platform. $g$ represents the gravity acceleration relative to the global coordinate system and $\textbf{g}$ = [0, 0, −9.8 m/s $^2$ ] $^T$ . When dynamic modeling is performed, kinematic chain and joint deformation are ignored.

Figure 2. Force and moment analysis of moving platform.

Newton equation and Euler equation of moving platform are established relative to the global coordinate system.

(14) \begin{equation} \begin{array}{l} \sum \limits _{i = 1}^4{{}^i{\textbf{F}}_t^p +{m_t}{\textbf{g}} +{}^o{\textbf{F}}_t^b} ={m_t}{{\textbf{A}}_s}\\[4pt] \sum \limits _{i = 1}^4{\left ({{\textbf{R}}_t^b{{\lambda }}_{si}^t} \right ) \times{}^i{\textbf{F}}_t^p} +{}^o{\textbf{M}}_t^b ={{\textbf{I}}_b}{{\dot{\omega } }}_t^b +{{\omega }}_t^b \times \left ({{{\textbf{I}}_b}{{\omega }}_t^b} \right ) \end{array} \end{equation}

where ${\textbf{R}}^b_t$ and ${\textbf{I}}_b$ represent the posture transform matrix and inertia matrix of the moving platform relative to global coordinate system [Reference Peng, Xu, Hu, Liang and Wu22], respectively. ${\textbf{I}}_t$ denotes inertia matrix of moving platform relative to moving platform coordinate system and ${\textbf{I}}_b$ = ${\textbf{R}}^b_t$ ${\textbf{I}}_t$ $\left({\textbf{R}}^b_t\right)$ $^T$ .

The component analyzed in Figs. 3(a)–(d) is defined as component 3x, component 3y, component 3z and component 3r, respectively. $l^{Fu}_{3z}$ and $l^{Fd}_{3z}$ denote the arm of force and in Fig. 3(a), respectively. $m^z_3$ denote the mass of components 3z. Newton equation and Euler equation of each analysis object of positioner are established in the global coordinate system. Take positioner 3 as an example to analyze the force of each component, as shown in Fig. 3. Coulomb friction and elastoplastic friction between different parts are considered. According to the mass and installation position of each part, the centroid position of each part of positioner can be calculated. The dynamic equations of each component of positioner 3 are established as follows.

Figure 3. Force analysis diagram of positioner 3. (a) Force analysis of component 3z. (b) Force analysis of component 3y. (c) Force analysis of component 3z. (d) Force analysis of component 3r.

See Fig. 3(a) for the stress analysis of telescopic rod of positioner 3 and define ${\textbf{a}}^z_3$ = [0, 0, $a^z_3$ ] $^T$ . Then, the Newton–Euler equation of component 3z are written as:

(15) \begin{equation} {}^{e}\textbf{F}_{3x}^{3z}+{}^{n}\textbf{F}_{3x}^{3z}+{}^{d}\textbf{F}_{3}^{z}+m_{3}^{z}\textbf{g}-{}^{3}\textbf{F}_{t}^{p}=m_{3}^{z}\textbf{a}_{3}^{z} \end{equation}

(16) \begin{equation} l_{3z}^{Fd}\!\left ( -\textbf{n}_{3}^{z}\times{}^{n}\textbf{F}_{3x}^{3z} \right )+l_{3z}^{Fu}\!\left [ \textbf{n}_{3}^{z}\times \left ( -{}^{3}\textbf{F}_{t}^{p} \right ) \right ]+{}^{s}\textbf{M}_{3x}^{3z}={{\textbf{E}}_{0}} \end{equation}

where F $^{3z}_{3x}$ and $^s$ M $^{3z}_{3x}$ represent the force and moment of component $3x$ applied on component $3z$ , respectively. $^n$ F $^{3z}_{3x}$ and $^e$ F $^{3z}_{3x}$ represent the perpendicular and parallel components of the spatial force to the z-axis, respectively. $^d$ F $^{z}_{3}$ indicates the driving force acting on the telescopic rod, and the definitions of other driving force symbols are defined similarly. n $^{z}_{3}$ represents the direction vector of the z-direction prismatic pair of positioner 3, and the other direction vectors are defined similarly. E $_{0}$ = [0, 0, 0] $^T$ .

The force analysis of components $3x$ , $3y$ and $3r$ is shown in Figs. 3(b)–(d), whose Newton–Euler equation is given as:

(17) \begin{equation} m_{3}^{x}\textbf{g}-{}^{d}\textbf{F}_{3}^{z}-{}^{e}\textbf{F}_{3x}^{3z}-{}^{n}\textbf{F}_{3x}^{3z}+{}^{e}\textbf{F}_{3y}^{3x}+{}^{n}\textbf{F}_{3y}^{3x}=m_{3}^{x}\textbf{a}_{3}^{x} \end{equation}

(18) \begin{equation} l_{3x}^{Fd}\!\left [ \textbf{n}_{3}^{z}\times \left ({}^{e}\textbf{F}_{3y}^{3x}+{}^{n}\textbf{F}_{3y}^{3x} \right ) \right ]+l_{3x}^{Fu}\!\left [ \textbf{n}_{3}^{z}\times \left ( -{}^{n}\textbf{F}_{3x}^{3z} \right ) \right ]-{}^{s}\textbf{M}_{3x}^{3z}+{}^{s}\textbf{M}_{3y}^{3x}={{\textbf{E}}_{0}} \end{equation}

(19) \begin{equation} {}^{e}\textbf{F}_{3r}^{3y}+{}^{n}\textbf{F}_{3r}^{3y}-{}^{e}\textbf{F}_{3y}^{3x}-{}^{n}\textbf{F}_{3y}^{3x}+m_{3}^{y}\textbf{g}=m_{3}^{y}\textbf{a}_{3}^{y} \end{equation}

(20) \begin{equation} \left ( l_{3y}^{Fu}\textbf{n}_{3}^{z}+d_{3}^{x}\textbf{n}_{3}^{x} \right )\times \left ( -{}^{e}\textbf{F}_{3y}^{3x}-{}^{n}\textbf{F}_{3y}^{3x} \right )+l_{3y}^{Fd}\!\left [ -\textbf{n}_{3}^{z}\times \left ({}^{e}\textbf{F}_{3r}^{3y}+{}^{n}\textbf{F}_{3r}^{3y} \right ) \right ]+{}^{s}\textbf{M}_{3r}^{3y}-{}^{s}\textbf{M}_{3y}^{3x}={{\textbf{E}}_{0}} \end{equation}

(21) \begin{equation} \textbf{F}_{gd}^{3r}-{}^{e}\textbf{F}_{3r}^{3y}-{}^{n}\textbf{F}_{3r}^{3y}+m_{3}^{r}\textbf{g}={{\textbf{E}}_{0}} \end{equation}

(22) \begin{equation} {}^{s}\textbf{M}_{gd}^{3r}+l_{3r}^{Fd}\!\left ( -\textbf{n}_{3}^{z}\times \textbf{F}_{gd}^{3r} \right )+\left ( l_{3r}^{Fu}\textbf{n}_{3}^{z}+d_{3}^{y}\textbf{n}_{3}^{y} \right )\times \left ( -{}^{e}\textbf{F}_{3r}^{3y}-{}^{n}\textbf{F}_{3r}^{3y} \right )-{}^{s}\textbf{M}_{3r}^{3y}={{\textbf{E}}_{0}} \end{equation}

where the variables in Eqs. (17)–(22) are similar to those in the previous equation.

Elastoplastic friction model [Reference Liu, Li, Zhang, Hu and Zhang23] considers the influence of elastic–plastic deformation of parts on friction, which is suitable for the calculation of friction when considering the elastic deformation of telescopic rod. On the premise that the friction calculation accuracy meets the requirements, to improve the efficiency of the algorithm, the Coulomb friction model is used to calculate the friction of telescopic rod. When the friction between prismatic pairs is taken into consideration, the friction is calculated as follows:

(23) \begin{equation} {}^e{\textbf{F}}_{3x}^{3z} = -{\mu _c}\!\left |{{}^n{\textbf{F}}_{3x}^{3z}} \right |{\textrm{sgn}} \!\left ({v_3^z} \right ){\textbf{n}}_3^z \end{equation}

(24) \begin{equation} \begin{aligned} &{}^{e}\textbf{F}_{3y}^{3x}= \\ &{{\sigma }_{0}}z_{3}^{x}+{{\sigma }_{1}}v_{3}^{x}\!\left (\! 1-\frac{\sigma \!\left ( z_{3}^{x},v_{3}^{x} \right )z_{3}^{x}{{\sigma }_{0}}}{\mu _c \!\left ( \left |{}^{n}\textbf{F}_{3y}^{3x}\text{ }\centerdot \text{ }\textbf{n}_{3}^{z} \right |+\left |{}^{n}\textbf{F}_{3y}^{3x}\text{ }\centerdot \text{ }\textbf{n}_{3}^{y} \right | \right )+\left [{{F}_{s}}-\mu _c \!\left ( \left |{}^{n}\textbf{F}_{3y}^{3x}\text{ }\centerdot \text{ }\textbf{n}_{3}^{z} \right |+\left |{}^{n}\textbf{F}_{3y}^{3x}\text{ }\centerdot \text{ }\textbf{n}_{3}^{y} \right | \right ) \right ]{{e}^{-{{\left (| \frac{v_{3}^{x}}{{{v}_{s}}} |\right )}^{j_s}}}}} \!\right )\!+{{\sigma }_{2}}v_{3}^{x} \\ \end{aligned} \end{equation}

(25) \begin{equation} \begin{aligned} &{}^{e}\textbf{F}_{3r}^{3y}= \\ &{{\sigma }_{0}}z_{3}^{y}+{{\sigma }_{1}}v_{3}^{y}\!\left (\! 1-\frac{\sigma \!\left ( z_{3}^{y},v_{3}^{y} \right )z_{3}^{y}{{\sigma }_{0}}}{\mu _c \!\left ( \left |{}^{n}\textbf{F}_{3r}^{3y}\text{ }\centerdot \text{ }\textbf{n}_{3}^{z} \right |+\left |{}^{n}\textbf{F}_{3r}^{3y}\text{ }\centerdot \text{ }\textbf{n}_{3}^{x} \right | \right )+\left [{{F}_{s}}-\mu _c \!\left ( \left |{}^{n}\textbf{F}_{3r}^{3y}\text{ }\centerdot \text{ }\textbf{n}_{3}^{z} \right |+\left |{}^{n}\textbf{F}_{3r}^{3y}\text{ }\centerdot \text{ }\textbf{n}_{3}^{x} \right | \right ) \right ]{{e}^{-{{\left (| \frac{v_{3}^{y}}{{{v}_{s}}} |\right )}^{j_s}}}}} \!\right )\!+{{\sigma }_{2}}v_{3}^{y} \\ \end{aligned} \end{equation}

where $\mu _c$ represents Coulomb friction coefficient and sgn represents step function. $F_s$ denotes static friction force. $\sigma _0$ represents the stiffness coefficient. $z^x_3$ and $z^y_3$ stand for average deflection of the contacting asperities, $\sigma _1$ denotes the damping coefficient of the bristle, $\sigma _2$ is the viscous friction coefficient. $\sigma (z^x_3,v^x_3)$ and $\sigma (z^y_3,v^y_3)$ stand for the zones of the elastic and plastic deformation of asperities. $v_s$ represents the Stribeck velocity. $j_s$ denotes the Stribeck shape factor.

Newton equations and Euler equations of the other positioners can be established using similar methods. Since the number of unknowns 95 is more than the number of equations 93, there are multiple sets of solutions to the driving force, so it is difficult to accurately calculate. Therefore, it is necessary to supplement the constraint equations.

4.1.1. Mapping relationship between position uncertainty of follow-up prismatic pair and pose error of moving platform

For parallel mechanism with actuation redundancy, there are many ways to increase the number of constraint equations, for example, minimizing the position errors [Reference Jiang, Li and Wang10] and internal force regulation [Reference Liu, Yao, Li and Zhao24]. Based on the deformation compatibility equation, the paper increases the number of constraint equations. Taking the third branch chain as an example, the establishment process of the mapping relationship between the telescopic rod deformation and the pose error of the moving platform is introduced in detail.

The telescopic rod is regarded as a cantilever beam. Under the action of spatial force, the position error of $S_i$ relative to $O_t$ . $\delta {\lambda }^t_{si}$ can be expressed as [Reference Hibbeler25]:

(26) \begin{equation} \delta {\lambda }_{si}^{t} = {{\left [ \begin{matrix} \dfrac{-{}_{x}^{i}F_{t}^{p}L_{ii}^{3}}{3EI} & \quad \dfrac{-{}_{y}^{i}F_{t}^{p}L_{ii}^{3}}{3EI} & \quad \dfrac{-{}_{z}^{i}F_{t}^{p}{{L}_{iz}}}{EA} \\ \end{matrix} \right ]}^{\text{T}}} \end{equation}

where $L_{ii}$ denotes the length of component iz extending from component ix. $L_{iz}$ stands for the length of component iz. A, E and I represent the cross-section area, elastic modulus and polar moment of inertia of component iz, respectively.

n $^x_i$ , n $^y_i$ and n $^z_i$ represent the direction vectors of each prismatic pair, respectively. ${\lambda }^b_{\textit{oi}}$ denotes the origin of positioner i coordinate system in the global coordinate system. $d^x_i$ , $d^y_i$ and $d^z_i$ denote the displacement of each prismatic pair of the positioner. Eq. (27) can be obtained by the space vector chain method:

(27) \begin{equation} {{\lambda }}_{oi}^b + d_i^x{\textbf{n}}_i^x + d_i^y{\textbf{n}}_i^y + d_i^z{\textbf{n}}_i^z ={\textbf{P}}_t^b +{\textbf{R}}_t^b{{\lambda }}_{si}^t \end{equation}

Considering the axial deformation and bending deformation of the telescopic rod under the action of external force and torque, Eq. (28) is established as follows:

(28) \begin{equation} {\lambda }_{oi}^{b}+\left ( d_{i}^{x}+\delta d_{i}^{x} \right )\textbf{n}_{i}^{x}+\left ( d_{i}^{y}+\delta d_{i}^{y} \right )\textbf{n}_{i}^{y}+d_{i}^{z}\textbf{n}_{i}^{z}+\delta{{\textbf{S}}_{i}}=\textbf{P}_{t}^{b}+\delta \textbf{P}_{t}^{b}+\left ( \delta \textbf{R}_{t}^{b}+\textbf{R}_{t}^{b} \right ){\lambda }_{si}^{t} \end{equation}

where the elastic deformation of moving platform, x and y prismatic pair is ignored, so there is no parameter term related to $\delta$ n $^x_i$ , $\delta$ n $^y_i$ , $\delta$ n $^z_i$ , $\delta {\lambda }^t_{si}$ and $\delta {\lambda }^b_{oi}$ . z-direction prismatic pair is driven by servo motor and ball screw, so the position error is very small and can be ignored. $\delta {S}_i$ denotes the spatial variation of point $S_i$ caused by the deformation of the telescopic rod, and $\delta S_i$ = [ $\delta S^x_i$ , $\delta S^y_i$ , $\delta S^z_i$ ] $^T$ . $\delta d^x_i$ and $\delta d^y_i$ represent the positional errors of the follow-up prismatic pair of positioner i in the x direction and y direction, respectively. $\delta$ P $^b_t$ and $\delta$ R $^b_t$ represent the position error and the differential of rotation matrix of moving platform [Reference Niku26], respectively.

(29) \begin{equation} \delta \textbf{P}_{t}^{b}={{\left [ \begin{matrix} \delta x_{t}^{b} & \quad \delta y_{t}^{b} & \quad \delta z_{t}^{b} \\[5pt] \end{matrix} \right ]}^{\text{T}}}\qquad\qquad\qquad\qquad \end{equation}

(30) \begin{equation} \delta \textbf{R}_{t}^{b}=\left [ \begin{matrix} 0 & \quad -\delta \gamma _{t}^{b} & \quad \delta \beta _{t}^{b} \\[5pt] \delta \gamma _{t}^{b} & \quad 0 & \quad -\delta \alpha _{t}^{b} \\[5pt] -\delta \beta _{t}^{b} & \quad \delta \alpha _{t}^{b} & \quad 0 \\[5pt] \end{matrix} \right ]\textbf{R}_{t}^{b}={{{\Omega }}_{3\times 3}}\textbf{R}_{t}^{b} \end{equation}

where $\delta x^b_t$ , $\delta y^b_t$ , $\delta z^b_t$ , $\delta \alpha ^b_t$ , $\delta \beta ^b_t$ and $\delta \gamma ^b_t$ represent pose error of moving platform. It can be obtained from Eq. (28) that:

(31) \begin{equation} {\lambda }_{oi}^{b}+d_{i}^{x}\textbf{n}_{i}^{x}+\delta d_{i}^{x}\textbf{n}_{i}^{x}+d_{i}^{y}\textbf{n}_{i}^{y}+\delta d_{i}^{y}\textbf{n}_{i}^{y}+d_{i}^{z}\textbf{n}_{i}^{z}+\delta{{\textbf{S}}_{i}}=\textbf{P}_{t}^{b}+\delta \textbf{P}_{t}^{b}+\delta \textbf{R}_{t}^{b}{\lambda }_{si}^{t}+\textbf{R}_{t}^{b}{\lambda }_{si}^{t} \end{equation}

Eq. (32) can be built by subtracting Eq. (27) from Eq. (31):

(32) \begin{equation} \delta d_{i}^{x}\textbf{n}_{i}^{x}+\delta d_{i}^{y}\textbf{n}_{i}^{y}+\delta{{\textbf{S}}_{i}}=\delta \textbf{P}_{t}^{b}+\delta \textbf{R}_{t}^{b}{\lambda }_{si}^{t} \end{equation}

Eq. (33) is established by matrix vector multiplication:

(33) \begin{equation} \delta d_{i}^{x}+\delta{{\textbf{S}}_{i}}\text{ }\centerdot \text{ }\textbf{n}_{i}^{x}=\left ( \delta \textbf{P}_{t}^{b}+\delta \textbf{R}_{t}^{b}{\lambda }_{si}^{t} \right )\text{ }\centerdot \text{ }\textbf{n}_{i}^{x}=\left ( \delta \textbf{P}_{t}^{b}+{{{\Omega }}_{3\times 3}}\textbf{R}_{t}^{b}{\lambda }_{si}^{t} \right )\text{ }\centerdot \text{ }\textbf{n}_{i}^{x} \end{equation}

There is the following definition,

(34) \begin{equation} \textbf{R}_{t}^{b}{\lambda }_{si}^{t}={{{\zeta }}_{i}} \end{equation}

where $\zeta_i$ = [ $\zeta ^x_i$ , $\zeta ^y_i$ , $\zeta ^z_i$ ] $^T$ ,which is a known quantity. Then, Eq. (32) can be sorted as:

(35) \begin{equation} \delta d_{i}^{x}+\delta{{\textbf{S}}_{i}}\text{ }\centerdot \text{ }\textbf{n}_{i}^{x}=\delta d_{i}^{x}+\delta \textbf{S}_{i}^{\text{T}}\textbf{n}_{i}^{x}=\left ( \delta \textbf{P}_{t}^{b}+{{{\Omega }}_{3\times 3}}{{{\zeta }}_{i}} \right )\text{ }\centerdot \text{ }\textbf{n}_{i}^{x}={{\textbf{Q}}_{1}}\delta \textbf{M} \end{equation}

where

(36) \begin{equation} {{\textbf{Q}}_{1}}=\left [ \begin{matrix} 1 & \quad 0 & \quad 0 & \quad 0 & \quad \zeta _{i}^{z} & \quad -\zeta _{i}^{y} \\ \end{matrix} \right ] \end{equation}

(37) \begin{equation} \delta \textbf{M}={{\left [ \begin{matrix} \delta x_{t}^{b} & \quad \delta y_{t}^{b} & \quad \delta z_{t}^{b} & \quad \delta \alpha _{t}^{b} & \quad \delta \beta _{t}^{b} & \quad \delta \gamma _{t}^{b} \\ \end{matrix} \right ]}^{\text{T}}} \end{equation}

Evidenced by the same token,

(38) \begin{equation} \delta d_{i}^{y}+\delta{{\textbf{S}}_{i}}\text{ }\centerdot \text{ }\textbf{n}_{i}^{y}=\delta d_{i}^{y}+\delta \textbf{S}_{i}^{\text{T}}\textbf{n}_{i}^{y}=\left [ \begin{matrix} 0 & \quad 1 & \quad 0 & \quad -\zeta _{i}^{z} & \quad 0 & \quad \zeta _{i}^{x} \\ \end{matrix} \right ]\delta \textbf{M} \end{equation}

(39) \begin{equation} \delta \textbf{S}_{i}^{\text{T}}\textbf{n}_{i}^{z}=\left [ \begin{matrix} 0 & \quad 0 & \quad 1 & \quad \zeta _{i}^{y} & \quad -\zeta _{i}^{x} & \quad 0 \\ \end{matrix} \right ]\delta \textbf{M} \end{equation}

Eq. (40) can be obtained by further analysis,

(40) \begin{equation} \vartriangle\! \textbf{d}+\Delta \textbf{SN}=\textbf{Q}\delta \textbf{M} \end{equation}

where

(41) \begin{equation} \vartriangle\! \textbf{d}=\left [ \begin{matrix} 0 & \quad \cdots & \quad \delta d_{3}^{x} & \quad \cdots & \quad \delta d_{4}^{y} & \quad 0 \\[5pt] \end{matrix} \right ]_{12\times 1}^{\text{T}} \end{equation}

(42) \begin{equation} \textbf{N}=\left [ \begin{matrix}{{\left ( \textbf{n}_{1}^{x} \right )}^{\text{T}}} & \quad {{\left ( \textbf{n}_{1}^{y} \right )}^{\text{T}}} & \quad \cdots & \quad {{\left ( \textbf{n}_{4}^{z} \right )}^{\text{T}}} \\[5pt] \end{matrix} \right ]_{36\times 1}^{\text{T}} \end{equation}

(43) \begin{equation} \vartriangle \textbf{S}={{\left [ \begin{matrix} \delta \textbf{S}_{1}^{\text{T}} & \quad \textbf{E}_{0}^{\text{T}} & \quad \cdots & \quad \textbf{E}_{0}^{\text{T}} & \quad \textbf{E}_{0}^{\text{T}} \\[5pt] \textbf{E}_{0}^{\text{T}} & \quad \delta \textbf{S}_{1}^{\text{T}} & \quad \cdots & \quad \textbf{E}_{0}^{\text{T}} & \quad \textbf{E}_{0}^{\text{T}} \\[5pt] \vdots & \quad \vdots & \quad \ddots & \quad \vdots & \quad \vdots \\[5pt] \textbf{E}_{0}^{\text{T}} & \quad \textbf{E}_{0}^{\text{T}} & \quad \cdots & \quad \textbf{E}_{0}^{\text{T}} & \quad \delta \textbf{S}_{4}^{\text{T}} \\[5pt] \end{matrix} \right ]}_{12\times 36}} \end{equation}

(44) \begin{equation} \textbf{Q}={{\left [ \begin{matrix} 1 & \quad 0 & \quad 0 & \quad 0 & \quad \zeta _{1}^{z} & \quad -\zeta _{1}^{y} \\[5pt] 0 & \quad 1 & \quad 0 & \quad -\zeta _{1}^{z} & \quad 0 & \quad \zeta _{1}^{x} \\[5pt] \vdots & \quad \vdots & \quad \vdots & \quad \vdots & \quad \vdots & \quad \vdots \\[5pt] 0 & \quad 0 & \quad 1 & \quad \zeta _{4}^{y} & \quad -\zeta _{4}^{z} & \quad 0 \\[5pt] \end{matrix} \right ]}_{12\times 6}} \end{equation}

Positioner 1 does not install prismatic pair in x and y direction, and positioner 2 has no prismatic pair in y direction. The positional error of prismatic pairs in z direction is extremely small, and the corresponding positional error is 0. So far, the mapping relationship between the deformation of the telescopic rod and the pose error of the moving platform can be established.

4.1.2. Deformation supplementary equation based on flexibility analysis

Under the internal force of the parallel posture alignment system, the elastic deformation matrix of the telescopic rod can be expressed as:

(45) \begin{equation} \vartriangle{{\textbf{S}}_{12\times 36}}={{\textbf{F}}_{12\times 36}}{{\textbf{C}}_{36\times 36}} \end{equation}

where C represents the flexibility matrix of the telescopic rod.

(46) \begin{equation} \textbf{F}=\left [ \begin{matrix} _{x}^{1}F_{t}^{p} & \quad _{y}^{1}F_{t}^{p} & \quad _{z}^{1}F_{t}^{p} & \quad 0 & \quad \cdots & \quad 0 & \quad 0 \\[5pt] 0 & \quad 0 & \quad 0 & \quad _{x}^{1}F_{t}^{p} & \quad \cdots & \quad 0 & \quad 0 \\[5pt] \vdots & \quad \vdots & \quad \vdots & \quad \vdots & \quad \ddots & \quad \vdots & \quad \vdots \\[5pt] 0 & \quad 0 & \quad 0 & \quad 0 & \quad \cdots & \quad 0 & \quad 0 \\[5pt] 0 & \quad 0 & \quad 0 & \quad 0 & \quad \cdots & \quad _{y}^{4}F_{t}^{p} & \quad _{z}^{4}F_{t}^{p} \\[5pt] \end{matrix} \right ] \end{equation}

where $\big(^1_x F^p_t$ , $^1_y F^p_t$ , $^1_z F^p_t\big)^T$ = $^1 F^p_t$

(47) \begin{equation} \textbf{C}=\left [ \begin{matrix} \dfrac{-L_{11}^{3}}{EI} & \quad 0 & \quad 0 & \quad \cdots & \quad 0 & \quad 0 \\[5pt] 0 & \quad \dfrac{-L_{11}^{3}}{EI} & \quad 0 & \quad \cdots & \quad 0 & \quad 0 \\[5pt] \vdots & \quad \vdots & \quad \vdots & \quad \ddots & \quad \vdots & \quad \vdots \\[5pt] 0 & \quad 0 & \quad 0 & \quad \cdots & \quad \dfrac{-L_{44}^{3}}{EI} & \quad 0 \\[5pt] 0 & \quad 0 & \quad 0 & \quad \cdots & \quad 0 & \quad \dfrac{-{{L}_{4z}}}{EA} \\[5pt] \end{matrix} \right ] \end{equation}

As shown in Fig. 4(a), the follow-up prismatic pair moves under the internal force of the parallel posture alignment system and affected by the friction force of the prismatic pair. The deformation of the telescopic rod is related to the friction. The force component at the spherical hinge is similar to the friction when adjust the parallel posture alignment system with low accelerations. Therefore, the force component at the spherical hinge is used to replace the friction to calculate the deformation of telescopic rod. Then, the deformation at the top of the telescopic rod is equivalent to the position uncertainty of the follow-up prismatic pair, as shown in Fig. 4(b). In Fig. 4, the solid line represents the actual situation and the dotted line represents the ideal situation.

Figure 4. Structural deformation and positional error. (a) Deformation of telescopic rod. (b) Position uncertainty of follow-up prismatic pair.

The dynamic supplementary equation can be obtained by simultaneous Eqs. (40) and (45),

(48) \begin{equation} \vartriangle{{\textbf{d}}_{12\times 1}}+{{\textbf{F}}_{12\times 36}}{{\textbf{C}}_{36\times 36}}{{\textbf{N}}_{36\times 1}}={{\textbf{Q}}_{12\times 6}}\delta{{\textbf{M}}_{6\times 1}} \end{equation}

Through deformation coordination analysis, Eq. (48) establishes the correlation equation among $\delta$ R $^b_t$ and $\delta$ P $^b_t$ , $d^x_i$ , $d^y_i$ and $d^z_i$ . 12 linearly independent constraint equations are supplemented. The spatial position of the spherical hinge in the moving platform coordinate system remains unchanged, so Eqs. (49) and (50) are established.

(49) \begin{equation} \delta{{\textbf{S}}_{1}}+\delta{{\textbf{S}}_{3}}=\delta{{\textbf{S}}_{2}}+\delta{{\textbf{S}}_{4}} \end{equation}

(50) \begin{equation} \left ( {\lambda }_{s1}^{t}-{\lambda }_{s2}^{t}+\delta{{\textbf{S}}_{1}}-\delta{{\textbf{S}}_{2}} \right )\text{ }\centerdot \text{ }\left ( {\lambda }_{s3}^{t}-{\lambda }_{s2}^{t}+\delta{{\textbf{S}}_{3}}-\delta{{\textbf{S}}_{2}} \right )=0 \end{equation}

There are three constraint equations in Eq. (49) and one constraint equation in Eq. (50). Higher-order terms such as $\delta{S}_1 \delta{S}_2$ , $\delta{S}_2\delta{S}_3$ and $\delta{S}^2_2$ are ignored in Eq. (50). So Eq. (50) can be simplified as a first-order equation related to $\delta{S}_1$ , $\delta{S}_2$ and $\delta{S}_3$ . Eqs. (14)–(25) and Eqs. (48)–(50) contain 124 constraint equations and 124 unknowns. Therefore, the driving force can be solved.

4.2. Solution of force acting on spherical hinge

Maple [Reference Yang, Qi, Tang and Gu27] or Mathematics [Reference Martínez, Encinas, Muñoz and Dios28] can be used to solve the equations, but the equations cannot be solved in a short time. The least square method [Reference Wang, Zhou, Cheng, Ma, Chang, Miao and Chen29] and other numerical methods can be used to solve the unknown variables in the dynamic equation, which does not require complex analytical modeling. However, the accuracy of the solution is greatly affected by the initial iteration values, so it is difficult to obtain exact solutions. The unknowns in the dynamic equation can be solved by analytic analysis, that is, the mathematical expression of the unknowns can be obtained. In the analytical modeling, the driving force at the spherical hinge is first solved by taking the moving platform as the analytical object, then the force at the spherical hinge is brought into the dynamic equation of each branch and the analytical solution of the branch equation is solved.

It can be seen from Eq. (5) that the displacement of prismatic pair is only related to $z^b_t$ , $\alpha ^b_t$ and $\beta ^b_t$ . Therefore, the motion trajectory of the posture alignment mechanism is only related to $z^b_t$ , $\alpha ^b_t$ and $\beta ^b_t$ . Assuming that the trajectory of the parallel posture alignment mechanism is known, that is, $z^b_t$ , $\alpha ^b_t$ ,and $\beta ^b_t$ are known quantities. In Eqs. (14), (48), (49) and (50), the unknown is the force at the spherical hinge and the position uncertainty of the follow-up prismatic pair. The unknown number is 22 and the number of equations is 22.

In Eq. (14), only A $_f$ is the first-order function of $^1$ F $^p_t$ , $^2$ F $^p_t$ , $^3$ F $^p_t$ and $^4$ F $^p_t$ are unknowns. In Eq. (48), only the parameters and are unknown. After sorting out Eqs. (14), (48), (49) and Eqs. (50), (51) can be obtained:

(51) \begin{equation} {{\textbf{A}}_{f}}{{\textbf{X}}_{f}}={{\textbf{B}}_{f}} \end{equation}

where $A_f$ is the first-order function of $^1$ F $^p_t$ , $^2$ F $^p_t$ , $^3$ F $^p_t$ and $^4$ F $^p_t$ . $B_f$ is the function of $m_t \ddot q$ , $m_{dr}$ g, $^o$ F $^b_t$ , ${{I}}_b$ ${{\dot \omega }}^b_t$ , ${\omega }^b_t\times ({\textbf{I}}_b \omega ^b_t)$ and $^o$ M $^b_t$ . A $_f$ is a matrix of order 22, whose expression is known. Therefore, the force at each spherical joint can be solved,

(52) \begin{equation} {{\textbf{X}}_{f}}={{\left ({{\textbf{A}}_{f}} \right )}^{-1}}{{\textbf{B}}_{f}} \end{equation}

4.3. Solution of driving force based on vector mixed product

After calculating the force at the spherical joint, take the force at the spherical joint as the intermediate variable to calculate the driving force required by each active prismatic pair. Taking the force at the spherical hinge as the intermediate variable, the driving force of the branch chain can be solved. According to the principle of mixed product of vector, multiply Eq. (15) by vector n $^z_1$ and then multiply by $l^{Fd}_{3z}$ , Eq. (53) can be obtained:

(53) \begin{equation} l_{3z}^{Fd}\!\left ( \textbf{n}_{3}^{z}\times{}^{n}\textbf{F}_{3x}^{3z} \right )+l_{3z}^{Fd}\!\left ( \textbf{n}_{3}^{z}\times{}^{d}\textbf{F}_{3}^{z} \right )-l_{3z}^{Fd}\!\left ( \textbf{n}_{3}^{z}\times{}^{3}\textbf{F}_{t}^{p} \right )={{\textbf{E}}_{0}} \end{equation}

After adding Eqs. (53) to (16), Eq. (54) can be obtained:

(54) \begin{equation} {}^{s}\textbf{M}_{3x}^{3z}=\left ( l_{3z}^{Fd}+l_{3z}^{Fu} \right )\left ( \textbf{n}_{3}^{z}\times{}^{3}\textbf{F}_{t}^{p} \right ) \end{equation}

Based on Eqs. (16), (55) can be obtained by multiplying the left cross of the vector n $^z_3$ and expanding it according to the triple product of the vector:

(55) \begin{equation} {}^{n}\textbf{F}_{3x}^{3z}=\frac{1}{l_{3z}^{Fd}}\!\left \{ l_{3z}^{Fu}\!\left [ \textbf{n}_{3}^{z}\!\left ( \textbf{n}_{3}^{z}\centerdot{}^{3}\textbf{F}_{t}^{p} \right )-{}^{3}\textbf{F}_{t}^{p} \right ]-\textbf{n}_{3}^{z}\times{}^{s}\textbf{M}_{3x}^{3z} \right \} \end{equation}

Substituting Eq. (54) into Eq. (55) gives:

(56) \begin{equation} {}^{n}\textbf{F}_{3x}^{3z}={}^{3}\textbf{F}_{t}^{p}-\textbf{n}_{3}^{z}\!\left ( \textbf{n}_{3}^{z}\centerdot{}^{3}\textbf{F}_{t}^{p} \right ) \end{equation}

Additionally, substituting Eq. (56) into Eq. (23) gives:

(57) \begin{equation} {}^e{\textbf{F}}_{3x}^{3z} = -{\mu _c}\!\left |{{{}^{3}\textbf{F}_{t}^{p}-\textbf{n}_{3}^{z}\!\left ( \textbf{n}_{3}^{z}\centerdot{}^{3}\textbf{F}_{t}^{p} \right )}} \right |{\textrm{sgn}} \!\left ({v_3^z} \right ){\textbf{n}}_3^z \end{equation}

According to Eq. (15), $^d$ F $^{z}_{3}$ can be expressed as:

(58) \begin{equation} {}^{d}\textbf{F}_{3}^{z}=m_{3}^{z}\textbf{a}_{3}^{z}+{}^{3}\textbf{F}_{t}^{p}-m_{3}^{z}\textbf{g}-{}^{e}\textbf{F}_{3x}^{3z}-{}^{n}\textbf{F}_{3x}^{3z} \end{equation}

Substituting Eqs. (57) and (56) into Eq. (58) gives:

(59) \begin{equation} {}^{d}\textbf{F}_{3}^{z}=m_{3}^{z}\textbf{a}_{3}^{z}-m_{3}^{z}\textbf{g}+-{\mu _c}\!\left |{{{}^{3}\textbf{F}_{t}^{p}-\textbf{n}_{3}^{z}\!\left ( \textbf{n}_{3}^{z}\centerdot{}^{3}\textbf{F}_{t}^{p} \right )}} \right |{\textrm{sgn}} \!\left ({v_3^z} \right ){\textbf{n}}_3^z+\textbf{n}_{3}^{z}\!\left ( \textbf{n}_{3}^{z}\centerdot{}^{3}\textbf{F}_{t}^{p} \right ) \end{equation}

Based on Eqs. (17), (60) can be obtained by multiplying the left cross of the vector n $^x_3$ twice and expanding it according to the triple product of the vector:

(60) \begin{equation} {}^{n}\textbf{F}_{3y}^{3x}=-m_{3}^{x}\textbf{g}+{}^{d}\textbf{F}_{3}^{z}+{}^{e}\textbf{F}_{3x}^{3z}-\textbf{n}_{3}^{z}\!\left ( \textbf{n}_{3}^{z}\centerdot{}^{n}\textbf{F}_{3x}^{3z} \right )+{}^{n}\textbf{F}_{3x}^{3z} \end{equation}

Substituting Eqs. (57), (56) and (58) into Eq. (60) gives:

(61) \begin{equation} {}^{n}\textbf{F}_{3y}^{3x}=m_{3}^{z}\textbf{a}_{3}^{z}-\left ( m_{3}^{z}+m_{3}^{x} \right )\textbf{g}+{}^{3}\textbf{F}_{t}^{p}-\textbf{n}_{3}^{z}\!\left ( \textbf{n}_{3}^{z}\centerdot{}^{3}\textbf{F}_{t}^{p} \right ) \end{equation}

It can be seen from Eq. (17) that:

(62) \begin{equation} \textbf{a}_{3}^{x} = {\left ( m_{3}^{x}\textbf{g}-{}^{d}\textbf{F}_{3}^{z}-{}^{e}\textbf{F}_{3x}^{3z}-{}^{n}\textbf{F}_{3x}^{3z}+{}^{e}\textbf{F}_{3y}^{3x}+{}^{n}\textbf{F}_{3y}^{3x} \right )}/{m_{3}^{x}}\; \end{equation}

Substituting Eqs. (57) to (61) into Eq. (62) gives:

(63) \begin{equation} \textbf{a}_{3}^{x} = \left \langle -\textbf{n}_{3}^{x}\!\left ( \textbf{n}_{3}^{x}\centerdot{}^{3}\textbf{F}_{t}^{p} \right )-{\mu _c}\{|m_{3}^{z}\textbf{a}_{3}^{z}-\left ( m_{3}^{z}+m_{3}^{x} \right )\textbf{g}+{}^{3}\textbf{F}_{t}^{p}|+|\textbf{n}_{3}^{y}\centerdot{}^{3}\textbf{F}_{t}^{p}| \}{\textrm{sgn}} \!\left ({v_3^z} \right ){\textbf{n}}_3^z \right \rangle/m_{3}^{x} \end{equation}

Add Eqs. (15), (17) and (19), and Eq. (64) can be obtained:

(64) \begin{equation} m_{3}^{z}\textbf{g}-{}^{3}\textbf{F}_{t}^{p}+m_{3}^{x}\textbf{g}+{}^{e}\textbf{F}_{3r}^{3y}+{}^{n}\textbf{F}_{3r}^{3y}+m_{3}^{y}\textbf{g}=m_{3}^{x}\textbf{a}_{3}^{x}+m_{3}^{y}\textbf{a}_{3}^{y}+m_{3}^{z}\textbf{a}_{3}^{z} \end{equation}

Based on Eqs. (64), (65) can be obtained by multiplying the left cross of the vector n $^y_3$ twice and expanding it according to the triple product of the vector:

(65) \begin{equation} {}^{n}\textbf{F}_{3r}^{3y}=m_{3}^{z}\textbf{a}_{3}^{z}+m_{3}^{x}\textbf{a}_{3}^{x}-\left ( m_{3}^{z}+m_{3}^{x}+m_{3}^{y} \right )\textbf{g}-\textbf{n}_{3}^{y}\!\left ( \textbf{n}_{3}^{y}\centerdot{}^{3}\textbf{F}_{t}^{p} \right )+{}^{3}\textbf{F}_{t}^{p} \end{equation}

$^e$ F $^{3y}_{3r}$ and $^e$ F $^{3x}_{3y}$ can be calculated by Eqs. (24), (25), (61) and (65).

The similar method can be used to calculate the driving force required by the remaining prismatic pair of positioners under the specified trajectory.

(66) \begin{equation} {}^{d}\textbf{F}_{1}^{z}=m_{1}^{z}\textbf{a}_{1}^{z}-m_{1}^{z}\textbf{g}+\left ({}^{1}\textbf{F}_{t}^{p}\centerdot \text{ }\textbf{n}_{1}^{z} \right )\textbf{n}_{1}^{z}+{\mu _c}|{}^{1}\textbf{F}_{t}^{p}-\textbf{n}_{1}^{z}\!\left ( \textbf{n}_{1}^{z}\centerdot{}^{1}\textbf{F}_{t}^{p} \right )|{\textrm{sgn}} \!\left ({v_1^z} \right ){\textbf{n}}_1^z \end{equation}

(67) \begin{equation} {}^{d}\textbf{F}_{2}^{z}=m_{2}^{z}\textbf{a}_{2}^{z}-m_{2}^{z}\textbf{g}+\left ({}^{2}\textbf{F}_{t}^{p}\centerdot \text{ }\textbf{n}_{2}^{z} \right )\textbf{n}_{2}^{z}+{\mu _c}|{}^{2}\textbf{F}_{t}^{p}-\textbf{n}_{2}^{z}\!\left ( \textbf{n}_{2}^{z}\centerdot{}^{2}\textbf{F}_{t}^{p} \right )|{\textrm{sgn}} \!\left ({v_2^z} \right ){\textbf{n}}_2^z \end{equation}

(68) \begin{equation} {}^{d}\textbf{F}_{2}^{x}=m_{2}^{x}\textbf{a}_{2}^{x}+\left ({}^{2}\textbf{F}_{t}^{p}\centerdot \text{ }\textbf{n}_{2}^{x} \right )\textbf{n}_{2}^{x}+{\mu _c}\{|m_{3}^{z}\textbf{a}_{3}^{z}-\left ( m_{3}^{z}+m_{3}^{x} \right )\textbf{g}+{}^{3}\textbf{F}_{t}^{p}|+|\textbf{n}_{3}^{y}\centerdot{}^{3}\textbf{F}_{t}^{p}| \}{\textrm{sgn}} \!\left ({v_3^z} \right ){\textbf{n}}_3^z \end{equation}

The expressions for $^d$ F $^{z}_{4}$ and $^d$ F $^{z}_{3}$ are similar. Thus, the inverse dynamics modeling of the parallel posture alignment system with actuation redundancy can be realized. To analysis efficiency and verify the validity and applicability of the dynamics model, simulation analysis and experimental verification are carried out.

5.1. Analysis of algorithm efficiency

To compare the running time of this algorithm with the existing algorithms, the driving force calculation time is analyzed, as shown in Table II. The computer configuration used for the inverse dynamics modeling is Intel Core i7-9750H CPU 2.60 GHz with 16G RAM.

Table II. Comparison of calculation time of driving force.

It can be seen from Table II that the dynamic modeling method proposed has a shorter calculation time and high efficiency, which is 56.28% lower than Moore–Penrose [Reference Stojanović and Mosić30] and has more advantages in the design of dynamic controllers. The dynamic modeling method proposed can improve the anti-interference and anti-noise performance of the parallel posture alignment mechanism.

5.2. Analysis of validity and applicability

To demonstrate the validity and applicability of the proposed method, ADAMS software is used to calculate the difference between the algorithm proposed in this paper and the simulation value, as shown in Fig. 5(a).

Figure 5. Driving force simulation. (a) Simulation model. (b) Simulation result.

where $^{ad}$ F $^z_1$ , $^{ad}$ F $^z_2$ and $^{ad}$ F $^z_3$ represent the driving force calculated by the proposed method. $^{sd}$ F $^z_1$ , $^{sd}$ F $^z_2$ and $^{sd}$ F $^z_3$ denote the driving force calculated by ADAMS.

It can be seen from Fig. 5(b) that the maximum difference between the analytical modeling method proposed in this paper and ADAMS simulation results is 16.35N, and the maximum percentage of error is 0.97% which demonstrates the validity and applicability of the proposed method.