2.1. Mechanism configuration design

In this paper, a new 4-DOF PDM with two rotational DOFs and two translational DOFs is taken as the research object. The virtual prototype for the mechanism is represented in Fig. 1. $\boldsymbol{\$ }_{T1}$ and $\boldsymbol{\$ }_{T2}$ denote the translation along x-axis and y-axis, respectively. $\boldsymbol{\$ }_{R3}$ and $\boldsymbol{\$ }_{R4}$ represent rotation around z-axis and x-axis, respectively. The 4-DOF PDM is composed of a fixed platform, a moving platform (MP), an operating mechanism and driving mechanisms. The driving mechanisms consist of four distributed limbs, with two identical RPR limbs and two identical SPU limbs. Each limb is formed by a cylinder and a piston and the prismatic joints of limbs are actuated, which are the inputs of the system, the position and orientation of the MP is the output. The operating mechanism consists of a six-bar mechanism and a revolute joint in series. As shown in Fig. 2, $R_{1i},R_{2i},R_{3i},R_{4i}$ , and $R_{4}(i=1,2)$ represent the revolute joint; $U_{2j}(j=3,4)$ represents the universal joint; $S_{2j}$ denotes the spherical joint; $P_{1i}$ and $P_{2j}$ represent the prismatic joint. The reference frame P -xyz and the moving frame A -xyz are attached to the MP and the base, respectively. $l_{1i}$ and $l_{2j}$ denote the location of the prismatic joint in each limb. The geometric parameters are given in Appendix A.

Figure 1. Virtual prototype for the 4-DOF PDM.

Table I. Joint type and number of the 4-DOF PDM.

Figure 2. Schematic diagram of the 4-DOF PDM.

2.2. Workspace analysis

The boundary of the reachable workspace is determined with the help of boundary search method as shown in Fig. 3(a), where the z-axis is parasitic motion. Figure 3(b) shows its projection in the x–y direction. The link extensions, $l_{11},l_{12},l_{23}$ , and $l_{24}$ , are limited between −50 (mm) and 50 (mm) since the left and right grounded prismatic links are assumed to be physically constrained by the ground at those boundary angles. In addition, other link angles are also subject to certain constraints based on physical properties.

Figure 3. The reachable workspace of the 4-DOF PDM.

2.3. Over-constraint analysis

The PDM perform well in structural stiffness due to its closed-loop structure. Meanwhile, the closed-loop structure brings more over-constraints, which is necessary for the mobility and force analysis. Next, the over-constrained analysis of the 4-DOF PDM will be derived in detail. The joint type and number are shown in Table I.

The number of over-constraints on the 4-DOF PDM is calculated as follows:

(1) \begin{equation} t=3S+4U+5R+5P+A-6n=9 \end{equation}

3.1. Inverse position analysis

3.1.1. Inverse position of module-1

Inverse position analysis determines the positions of each limb when the position and orientation of the MP are given. Any vector in the frame P -xyz can map to the frame A -xyz.

(2) \begin{equation} \boldsymbol{R} =\boldsymbol{T}\boldsymbol{R}'+\boldsymbol{P} \end{equation}

in which

(3) \begin{equation} \boldsymbol{T}=\boldsymbol{R}\!\left(z_{0},\alpha \right)\boldsymbol{R}\!\left(x_{0},\gamma \right) \end{equation}

(4) \begin{equation} \boldsymbol{P}=\left[\begin{array}{l@{\quad}l@{\quad}l} p_{x} & p_{y} & p_{z} \end{array}\right]^{T} \end{equation}

where T and P denote the orientation matrix and the position vector of the MP, respectively.

For the 1st and 2nd driving limbs, the position vector $R_{3i}(i=1,2)$ with respect to the frame A -xyz can be obtained based on the homogeneous transformation. In addition, the position vector $R_{2i}$ with respect to the frame A -xyz can be expressed by $\theta _{\mathrm{1}i}$ . We can derive the closed-loop motion equation as

(5) \begin{equation} \boldsymbol{R}_{1i}\boldsymbol{R}_{2i}=\boldsymbol{R}_{1i}\boldsymbol{R}_{3i}+\boldsymbol{R}_{3i}\boldsymbol{R}_{2i} \end{equation}

Eq. (5) can be simplified as follows

(6) \begin{equation} a_{1i}\cos \theta _{1i}+b_{1i}\sin \theta _{1i}=d_{1i} \end{equation}

in which

(7) \begin{equation} \theta _{1i}=a\tan 2\!\left(b_{1i},a_{1i}\right)\pm a\tan 2\left(\sqrt{a_{1i}^{2}+b_{1i}^{2}-d_{1i}^{2}},d_{1i}\right) \end{equation}

where atan2 stands for four-quadrant inverse tangent function. $a_{1i},b_{1i}$ , and $d_{1i}$ represent the simplified values of Eq. (5), and their expressions are detailed in Appendix B.

3.1.2. Inverse position of module-2

According to the constraint relationship, the length of $l_{1i}$ can be obtained

(8) \begin{equation} l_{1i}=\left\| \boldsymbol{R}_{2i}\boldsymbol{R}_{4i}\right\| \end{equation}

Similar to Eq. (8), the position vector of $U_{2j}(j=3,4)$ with respect to the frame O -xyz can be obtained. We can derive the length of $l_{2j}$ as

(9) \begin{equation} l_{2j}=\left\| \boldsymbol{U}_{2j}\boldsymbol{S}_{2j}\right\| \end{equation}

3.2. Velocity and acceleration analysis

The displacement of the prismatic joint is given by Eqs. (8) and (9). However, to derive the inverse dynamic formulation for the 4-DOF PDM with Newton–Euler method, the positions, velocities, and accelerations of all bodies in the hybrid manipulator are required to be derived. In the following, we will derive the inverse kinematics in detail.

3.2.1. Velocity and acceleration of module-1

The angular velocity, acceleration, and linear acceleration of the MP can be obtained

(10) \begin{equation} \boldsymbol{a}_{mp}=\boldsymbol{a}_{mp}^{\tau }+\boldsymbol{a}_{mp}^{n} \end{equation}

in which

(11) \begin{equation} \begin{array}{l} \boldsymbol{a}_{mp}^{\tau }=\boldsymbol{\varepsilon }_{mp}\times \boldsymbol{r}_{{R_{4}},P}\\ \boldsymbol{a}_{mp}^{n}=\boldsymbol{\omega }_{mp}\times \left(\boldsymbol{\omega }_{mp}\times \boldsymbol{r}_{{R_{4}},P}\right) \end{array} \end{equation}

where $\boldsymbol{\omega }_{mp}$ and $\boldsymbol{\varepsilon }_{mp}$ represent the angular velocity and acceleration of the MP, respectively. $\boldsymbol{a}_{mp}^{n}$ and $\boldsymbol{a}_{mp}^{\tau }$ denote the normal and tangential acceleration of the MP, respectively.

For the operating mechanism, $\theta _{2i}$ can be written as follows:

(12) \begin{equation} \theta _{2i}=a\text{cos}\!\left(P_{{R_{3i}}}-P_{{R_{2i}}}\right)/r_{3i} \end{equation}

Furthermore, differentiating Eq. (12) with respect to time yields

(13) \begin{equation} \begin{array}{l} \dot{\theta }_{2i}=\left(v_{{R_{2i}}}-v_{{R_{3i}}}\right)/\left(r_{3i}s\theta _{2i}\right)\\[6pt] \ddot{\theta }_{2i}=\left(a_{{R_{2i}}}-a_{{R_{3i}}}-r_{3i}c\theta _{2i}\dot{\theta }_{2i}^{2}\right)/\left(r_{3i}s\theta _{2i}\right) \end{array} \end{equation}

Since the operating mechanism is a planar mechanism, the angular velocity and acceleration of $l_{3i}$ can be written as follows:

(14) \begin{equation} \begin{array}{l} \boldsymbol{\omega }_{r3i}=\left[\begin{array}{l@{\quad}l@{\quad}l} 0 & 0 & \dot{\theta }_{2i} \end{array}\right]^{T}\\[6pt] \boldsymbol{\varepsilon }_{r3i}=\left[\begin{array}{l@{\quad}l@{\quad}l} 0 & 0 & \ddot{\theta }_{2i} \end{array}\right]^{T} \end{array} \end{equation}

Similarly, the acceleration of the operating mechanism can be written as

(15) \begin{equation} \boldsymbol{a}_{r3i}=\boldsymbol{a}_{r3i}^{\tau }+\boldsymbol{a}_{r3i}^{n}+\boldsymbol{a}_{{R_{2i}}} \end{equation}

in which

(16) \begin{equation} \begin{array}{l} \boldsymbol{a}_{r3i}^{\tau }=\boldsymbol{\omega }_{r3i}\times \boldsymbol{r}_{3i}\\ \boldsymbol{a}_{r3i}^{n}=\boldsymbol{\omega }_{r3i}\times \left(\boldsymbol{\omega }_{r3i}\times \boldsymbol{r}_{3i}\right)\\ \boldsymbol{a}_{{R_{2i}}}=\boldsymbol{a}_{R_{2i}}^{\tau }+\boldsymbol{a}_{R_{2i}}^{n} \end{array} \end{equation}

3.2.2. Velocity and acceleration of module-2

The Euler angle of the 1st and 2nd driving limbs can be written as follows:

(17) \begin{equation} \alpha _{li}=a\tan 2\!\left(b_{li},a_{li}\right) \end{equation}

The angular velocity and angular acceleration of $l_{i}$ can be obtained by differentiating Eq. (17). Similar to Eq. (14), the angular velocity and acceleration vector can be expressed as

(18) \begin{equation} \begin{array}{ll} \boldsymbol{\omega }_{li} & =\left[\begin{array}{l@{\quad}l@{\quad}l} 0 & 0 & \omega _{li} \end{array}\right]^{T}\\[6pt] \ddot{\boldsymbol{\alpha }}_{li} & =\left[\begin{array}{l@{\quad}l@{\quad}l} 0 & 0 & \ddot{\alpha }_{li} \end{array}\right]^{T} \end{array} \end{equation}

Furthermore, the acceleration at the center of mass of the cylinder can be written as follows:

(19) \begin{equation} \boldsymbol{a}_{li1}\boldsymbol{=}\boldsymbol{a}_{li1}^{t}+\boldsymbol{a}_{li1}^{n} \end{equation}

in which

(20) \begin{equation} \begin{array}{l} \boldsymbol{a}_{li1}^{t}=\boldsymbol{\alpha }_{li}\times \boldsymbol{r}_{i1}\\ \boldsymbol{a}_{li1}^{n}=\boldsymbol{\omega }_{li}\times \left(\boldsymbol{\omega }_{li}\times \boldsymbol{r}_{i1}\right) \end{array} \end{equation}

Similarly, the acceleration at the center of mass of the piston can be written as

(21) \begin{equation} \boldsymbol{a}_{li2}=\boldsymbol{a}_{li2}^{\tau }+\boldsymbol{a}_{li2}^{n}+\boldsymbol{a}_{li2}^{k} \end{equation}

in which

(22) \begin{equation} \begin{array}{l} \boldsymbol{a}_{li2}^{\tau }=\boldsymbol{\alpha }_{li}\times \boldsymbol{r}_{i2}\\ \boldsymbol{a}_{li2}^{n}=\boldsymbol{\omega }_{li}\times \left(\boldsymbol{\omega }_{li}\times \boldsymbol{r}_{i2}\right)\\ \boldsymbol{a}_{li2}^{k}=2\boldsymbol{\omega }_{li}\times \boldsymbol{v}_{r} \end{array} \end{equation}

where $\boldsymbol{a}_{li2}^{k}$ denote the Coriolis acceleration of the piston in 1st and 2nd driving limbs.

For the U joint in the 3rd and 4th driving limbs, many papers assume that the angular velocity and acceleration are perpendicular to the direction along SPU limb for the convenience of derivation. In this paper, the actual angular velocity and acceleration values are obtained through geometric constraint relationship. The local coordinate frame attached to the U joint is shown in Fig. 4.

Figure 4. Coordinate frame of the U joint.

The coordinate systems of the 3rd and 4th driving limbs are shown in Fig. 4. First, we assume that the orientation matrix of the 3rd and 4th driving limbs frame Q -xyz with respect to the reference frame A -xyz can be derived in terms of three rotational angles $\alpha,\beta$ , and $\gamma$ satisfying the z-x-z convention:

(23) \begin{equation} \begin{array}{l} \boldsymbol{R}_{z}\!\left(\alpha \right)\boldsymbol{R}_{x}\!\left(\beta \right)\boldsymbol{R}_{z}\!\left(\gamma \right)\\ =\left[\begin{array}{c@{\quad}c@{\quad}c} \mathrm{c}\alpha \mathrm{c}\gamma -\mathrm{s}\alpha \mathrm{c}\beta \mathrm{s}\gamma & -\mathrm{c}\alpha \mathrm{s}\gamma -\mathrm{s}\alpha \mathrm{c}\beta \mathrm{c}\gamma & \mathrm{s}\alpha \mathrm{s}\beta \\ \mathrm{s}\alpha \mathrm{c}\gamma +\mathrm{c}\alpha \mathrm{c}\beta \mathrm{s}\gamma & -\mathrm{s}\alpha \mathrm{s}\gamma +\mathrm{c}\alpha \mathrm{c}\beta \mathrm{c}\gamma & -\mathrm{c}\alpha \mathrm{s}\beta \\ \mathrm{s}\beta \mathrm{s}\gamma & \mathrm{s}\beta \mathrm{c}\gamma & \mathrm{c}\beta \end{array}\right] \end{array} \end{equation}

where s and c represent the abbreviations of sine and cosine, respectively. In addition, $\boldsymbol{Q}_{j}$ can be obtained by the following equation:

(24) \begin{equation} \boldsymbol{Q}_{j}=\frac{\boldsymbol{U}_{2j}\boldsymbol{S}_{2j}}{\left\| \boldsymbol{U}_{2j}\boldsymbol{S}_{2j}\right\| } \end{equation}

Hence, taking Eqs. (23) and (24) into account, the Euler angle of $l_{2j}$ can be written as follows:

(25) \begin{equation} \begin{array}{l} \alpha =-a\tan 2\!\left(\boldsymbol{Q}_{j}\left(1,1\right),\boldsymbol{Q}_{j}\left(2,1\right)\right)\\ \beta =a\cos\! \left(\boldsymbol{Q}_{j}\left(3,1\right)\right) \end{array} \end{equation}

Differentiating Eq. (25) with respect to time, $\dot{\alpha }, \ddot{\alpha }, \dot{\beta }$ , and $\ddot{\beta }$ can be obtained, respectively. To derive the Euler angle $\gamma$ , it is necessary to supplement equations based on the geometric constraint relationship. Hence, we can derive $\boldsymbol{U}_{j2}$ as

(26) \begin{equation} \begin{array}{l} \boldsymbol{U}_{j2}\boldsymbol{=}\boldsymbol{Q}_{j}\times \boldsymbol{U}_{j1}\\[6pt] \boldsymbol{U}_{j2}^{e}=\frac{\boldsymbol{U}_{j2}}{\left| \boldsymbol{U}_{j2}\right| } \end{array} \end{equation}

where vector $\boldsymbol{U}_{j1}$ denotes the axis along with the MP at the U joint, vector $\boldsymbol{U}_{j2}$ denotes the axis along with the 3rd and 4th SPU limb. From Eqs. (23) to (26), $\gamma$ can be expressed as

(27) \begin{equation} \gamma =a\cos\! \left(\boldsymbol{U}_{j2}^{e}\left(3,1\right)\right)/\sin\! \left(\beta \right) \end{equation}

Differentiating Eq. (27) with respect to time, $\dot{\gamma }, \ddot{\gamma }$ can be obtained, respectively. In addition, $\boldsymbol{U}_{j3}$ can be obtained by vector cross product, which is the direction of the constrained torque at the U joint.

(28) \begin{equation} \boldsymbol{U}_{j3}=\boldsymbol{U}_{j1}\times \boldsymbol{U}_{j2} \end{equation}

Furthermore, to obtain the angular velocity of each link, we need to map the Euler angles rate to the angular velocity. The mapping principle is shown in Fig. 5.

Figure 5. Mapping of Euler angles rate to angular velocity.

According to the above transformation, Eq. (29) can be obtained

(29) \begin{equation} \left[\begin{array}{l} \omega _{x}\\ \omega _{y}\\ \omega _{z} \end{array}\right]=rot\!\left(z,\alpha \right)rot\!\left(x,\beta \right)\left[\begin{array}{l} 0\\ 0\\ \dot{\gamma } \end{array}\right]+rot\!\left(z,\alpha \right)\left[\begin{array}{l} \dot{\beta }\\ 0\\ 0 \end{array}\right]+\left[\begin{array}{l} 0\\ 0\\ \dot{\alpha } \end{array}\right] \end{equation}

By sorting out Eq. (29), the following equation can be obtained

(30) \begin{equation} \boldsymbol{\omega }_{lj}=\boldsymbol{J}_{t}\!\left[\begin{array}{l@{\quad}l@{\quad}l} \dot{\alpha }_{lj} & \dot{\beta }_{lj} & \dot{\gamma }_{lj} \end{array}\right]^{T} \end{equation}

in which

(31) \begin{equation} \boldsymbol{J}_{t}=\left[\begin{array}{c@{\quad}c@{\quad}c} 0 & c\alpha _{lj} & s\alpha _{lj}s\beta _{lj}\\ 0 & s\alpha _{lj} & -c\alpha _{lj}s\beta _{lj}\\ 1 & 0 & c\beta _{lj} \end{array}\right] \end{equation}

Furthermore, differentiating Eq. (30) with respect to time, the following equation can be obtained

(32) \begin{equation} \boldsymbol{\varepsilon }_{lj}=\boldsymbol{J}_{t}\!\left[\begin{array}{l@{\quad}l@{\quad}l} \ddot{\alpha }_{lj} & \ddot{\beta }_{lj} & \ddot{\gamma }_{lj} \end{array}\right]^{T}+\dot{\boldsymbol{J}}_{t}\!\left[\begin{array}{l@{\quad}l@{\quad}l} \dot{\alpha }_{lj} & \dot{\beta }_{lj} & \dot{\gamma }_{lj} \end{array}\right]^{T} \end{equation}

For the 1st and 2nd driving limbs, L _1i represents the vector between the two joints $R_{2i}$ and $R_{4i}$

(33) \begin{equation} \boldsymbol{L}_{1i}=\boldsymbol{r}_{2i}-\boldsymbol{r}_{4i} \end{equation}

where $\boldsymbol{r}_{2i}$ and $\boldsymbol{r}_{4i}$ represent the vectors of joint $R_{2i}$ and $R_{4i}$ with respect to frame { A }, respectively. Moreover, use Q _1i to denote the unit vector along L _1i

(34) \begin{equation} \boldsymbol{Q}_{i}=\boldsymbol{L}_{1i}/l_{1i} \end{equation}

The velocity of any point in the moving frame can be obtained by vector cross-product

(35) \begin{equation} \boldsymbol{V}=\boldsymbol{V}'+\boldsymbol{\omega }\times \boldsymbol{r} \end{equation}

where $\boldsymbol{\omega }$ and V represent the angular velocity and the linear velocity of the MP, respectively.

Differentiating Eq. (7) with respect to time yields:

(36) \begin{equation} \dot{\theta }_{1i}=\boldsymbol{J}_{0i}\dot{\boldsymbol{P}} \end{equation}

where $\dot{\boldsymbol{P}}$ denotes the velocity matrix of the MP.

Similarly, linear velocity of l _i by differentiating Eq. (8) can be obtained

(37) \begin{equation} \dot{l}_{1i}=\boldsymbol{J}_{1i}\dot{\boldsymbol{P}} \end{equation}

The linear acceleration of the MP can be obtained by the following formula

(38) \begin{equation} \boldsymbol{A}=\boldsymbol{A}'+\boldsymbol{\varepsilon }\times \boldsymbol{r}+\boldsymbol{\omega }\times \left(\boldsymbol{\omega }\times \boldsymbol{r}\right) \end{equation}

where $\boldsymbol{\varepsilon }$ denotes the angular acceleration of the MP.

Furthermore, differentiating Eq. (36) with respect to time yields:

(39) \begin{equation} \ddot{\theta }_{1i}=\boldsymbol{J}_{0i}\ddot{\boldsymbol{P}}+\dot{\boldsymbol{P}}^{T}\boldsymbol{H}_{0i}\dot{\boldsymbol{P}} \end{equation}

where $\ddot{\boldsymbol{P}}$ denotes the acceleration matrix of the MP.

Figure 6. Forces and moments analysis for the link-2.

Similarly, linear acceleration of $l_{i}$ can be obtained as time derivative of Eq. (37)

(40) \begin{equation} \ddot{l}_{1i}=\boldsymbol{J}_{1i}\ddot{\boldsymbol{P}}+\dot{\boldsymbol{P}}^{T}\boldsymbol{H}_{1i}\dot{\boldsymbol{P}} \end{equation}

For the 3rd and 4th driving limbs, the linear velocity can be calculated by Eq. (35)

(41) \begin{equation} \boldsymbol{V}_{U2j}=\boldsymbol{G}_{p}^{U_{2j}}\dot{\boldsymbol{P}} \end{equation}

(42) \begin{equation} l_{2j}^{2}=\boldsymbol{L}_{2j}\cdot \boldsymbol{L}_{2j} \end{equation}

Differentiating Eq. (42) with respect to time, the following formula can be obtained

(43) \begin{equation} l_{2j}\dot{l}_{2j}=\boldsymbol{L}_{2j}\cdot \boldsymbol{V}_{2j} \end{equation}

Hence, the velocity of the $l_{2j}$ can be expressed as follows:

(44) \begin{equation} \dot{l}_{2j}=\boldsymbol{Q}_{j}^{T}\boldsymbol{V}_{{U_{2j}}}=\boldsymbol{J}_{1j}\dot{\boldsymbol{P}} \end{equation}

in which

(45) \begin{equation} \boldsymbol{J}_{1j}=\boldsymbol{Q}_{j}^{T}\boldsymbol{G}_{p}^{U_{2j}} \end{equation}

According to the Eq. (38), the following formula can be obtained

(46) \begin{equation} \boldsymbol{A}_{{U_{2j}}}=\boldsymbol{G}_{p}^{U_{2j}}\ddot{\boldsymbol{P}}+\dot{\boldsymbol{P}}^{T}\boldsymbol{H}_{0j}\dot{\boldsymbol{P}} \end{equation}

Furthermore, linear acceleration of $l_{2j}$ can be obtained by differentiating Eq. (44)

(47) \begin{equation} \ddot{l}_{2j}=\boldsymbol{Q}_{j}^{T}\boldsymbol{A}_{{U_{2j}}}+\left(\boldsymbol{V}_{U_{2j}}^{T}\boldsymbol{V}_{{U_{2j}}}-\dot{l}_{2j}^{2}\right)/l_{2j} \end{equation}

Hence, the acceleration of the $l_{2j}$ can be written as follows:

(48) \begin{equation} \ddot{l}_{2j}=\boldsymbol{J}_{2j}\ddot{\boldsymbol{P}}+\dot{\boldsymbol{P}}^{T}\boldsymbol{H}_{1j}\dot{\boldsymbol{P}} \end{equation}

So far, all the kinematic parameters required for the inverse dynamics are determined. In the next section, attention will be paid on the deduction of dynamic equations of the 4-DOF PDM.

4.1. Forces and moments analysis for the module-1

The forces and moments analysis diagrams of link $r_{21}$ are given in Fig. 6. The reaction forces and moments at fixed points $R_{1i}$ are denoted by $\boldsymbol{f}_{e1}$ and $\boldsymbol{M}_{e1}$ . The forces and moments at moving points $R_{2i}$ are denoted by $\boldsymbol{f}_{32}$ and $\boldsymbol{M}_{32}$ , respectively.

The Newton–Euler equations for link $r_{21}$ are given as

(49) \begin{align} \sum \boldsymbol{F}_{ext}&=\boldsymbol{f}_{ci}-\boldsymbol{f}_{32}+\boldsymbol{f}_{e1}+m_{r2}\boldsymbol{g}=m_{r2}\boldsymbol{a}_{r2} \nonumber\\ \sum {}^{ci1}{\boldsymbol{n}}{_{ext}^{}}&={}^{A}{\boldsymbol{I}}{_{cr2}^{}}\dot{\boldsymbol{\omega }}_{r2}+\boldsymbol{\omega }_{r2}\times {}^{A}{\boldsymbol{I}}{_{cr2}^{}}\boldsymbol{\omega }_{r2}\nonumber\\ & \quad+ c_{r2}\!\left(-\boldsymbol{s}_{r2}\times \boldsymbol{f}_{e1}\right)+c_{r2}\!\left(\boldsymbol{s}_{r2}\times \boldsymbol{f}_{ci}\right)+c_{r2}\!\left(\boldsymbol{s}_{r2}\times -\boldsymbol{f}_{32}\right)+\boldsymbol{M}_{ci}-\boldsymbol{M}_{32}+\boldsymbol{M}_{e1} \end{align}

Similar to link $r_{21}$ , other links’ forces and moments of the operating mechanism can be obtained.

Furthermore, the forces and moments analysis for the MP are shown in Fig. 7.

Figure 7. Forces and moments analysis for the moving platform.

The Newton–Euler equations for MP are given as

(50) \begin{equation} \begin{array}{ll} \sum \boldsymbol{F}_{ext} & =\boldsymbol{f}_{4p}+\boldsymbol{f}_{U3}+\boldsymbol{f}_{U4}+m_{p}\boldsymbol{g}=m_{p}\boldsymbol{a}_{p}\\ \sum \boldsymbol{n}_{ext} & =\boldsymbol{S}_{1}\times \boldsymbol{f}_{4p}+\boldsymbol{S}_{2}\times \boldsymbol{f}_{U3}+\boldsymbol{S}_{3}\times \boldsymbol{f}_{U4}+\boldsymbol{M}_{U3}+\boldsymbol{M}_{U4}+\boldsymbol{M}_{4p}\\ & ={}^{A}{\boldsymbol{I}}{_{p}^{}}\dot{\boldsymbol{\omega }}_{p}+\boldsymbol{\omega }_{p}\times {}^{A}{\boldsymbol{I}}{_{p}^{}}\boldsymbol{\omega }_{p} \end{array} \end{equation}

4.2. Forces and moments analysis of the module-2

Assume that each driving limb consists of two parts, the cylinder and the piston, where the velocities and accelerations of their centers of masses are determined in Section 3. Assume that the centers of masses of the cylinder and the piston are located at a distance of $c_{i1}$ and $c_{i3}$ , and their masses are denoted by $m_{i1}$ and $m_{i2}$ , respectively. Moreover, consider that the cylinder and the pistons are symmetric about their axes, and their centers of masses lie at their midlengths.

The forces and moments analysis diagrams of 1st and 2nd driving limb are given in Fig. 8. The reaction forces and moments at fixed points $R_{4i}$ are denoted by $\boldsymbol{f}_{fi}$ and $\boldsymbol{M}_{fi}$ . The forces and moments at moving points $R_{2i}$ are denoted by $\boldsymbol{f}_{ci}$ and $\boldsymbol{M}_{ci}$ , and the internal forces and moments between cylinders and pistons are denoted by $\boldsymbol{f}_{mi}$ and $\boldsymbol{M}_{mi}$ , respectively.

Figure 8. Forces and moments analysis for 1st and 2nd driving limbs.

The Newton–Euler equation for 1st and 2nd driving limbs are given as

(51) \begin{align} \left({}^{A}{\boldsymbol{I}}{_{ci_{1}}^{}}+{}^{A}{\boldsymbol{I}}{_{ci_{2}}^{}}\right)\dot{\boldsymbol{\omega }}_{i}+\boldsymbol{\omega }_{i}\times \left({}^{A}{\boldsymbol{I}}{_{ci_{1}}^{}}+{}^{A}{\boldsymbol{I}}{_{ci_{2}}^{}}\right)\boldsymbol{\omega }_{i}&=\boldsymbol{M}_{fi}+m_{i2}\!\left(L_{1}-c_{i3}\right)\left(\boldsymbol{s}_{i}\times \left(\boldsymbol{g}-\boldsymbol{a}_{i2}\right)\right)\nonumber\\ & \quad +m_{i1}c_{i1}\!\left(\boldsymbol{s}_{i}\times \left(\boldsymbol{g}-\boldsymbol{a}_{i1}\right)\right)+L_{1}\!\left(\boldsymbol{s}_{i}\times -\boldsymbol{f}_{ci}\right)-\boldsymbol{M}_{ci} \end{align}

where ${}^{A}{\boldsymbol{I}}{_{ci_{1}}^{}}$ denotes the inertia matrix of the cylinder evaluated at a frame parallel to { A } located at point $c_{i1}$ . Since the cylinder is moving with respect to this frame, this matrix can be evaluated by rotation rule from the inertia matrix easily obtained about { A _i } attached to the cylinder at point $c_{i1}$ . This can be done through the following transformation:

(52) \begin{equation} {}^{A}{\boldsymbol{I}}{_{ci_{1}}^{}}={}^{A}{\boldsymbol{R}}{_{Ai}^{}}\ {}^{A}{\boldsymbol{I}}{_{ci_{1}}^{}}\ {}^{A}{\boldsymbol{R}}{_{Ai}^{T}} \end{equation}

Note the formula of vector mixed product is as follows:

(53) \begin{equation} \left(\boldsymbol{a}\times \boldsymbol{b}\right)\cdot \boldsymbol{c}=\left(\boldsymbol{b}\times \boldsymbol{c}\right)\cdot \boldsymbol{a}=\left(\boldsymbol{c}\times \boldsymbol{a}\right)\cdot \boldsymbol{b} \end{equation}

Multiply $\boldsymbol{t}_{i}$ on both sides of Eq. (51), and the following equation can be obtained

(54) \begin{align} \boldsymbol{f}_\boldsymbol{ci}^{ri}&=m_{i1}c_{i1}\boldsymbol{r}_{i}\cdot \left(\boldsymbol{g}-\boldsymbol{a}_{i1}\right)+m_{i2}\!\left(L_{1}-c_{i3}\right)\boldsymbol{r}_{i}\cdot \left(\boldsymbol{g}-\boldsymbol{a}_{i2}\right)\nonumber\\ & \quad- \boldsymbol{t}_{i}\cdot \left(\left({}^{A}{\boldsymbol{I}}{_{ci1}^{}}+{}^{A}{\boldsymbol{I}}{_{ci2}^{}}\right)\dot{\boldsymbol{\omega }}_{i}\right)-\boldsymbol{t}_{i}\cdot \left(\boldsymbol{\omega }_{i}\times \left({}^{A}{\boldsymbol{I}}{_{ci1}^{}}+{}^{A}{\boldsymbol{I}}{_{ci2}^{}}\right)\boldsymbol{\omega }_{i}\right) \end{align}

Forces and moments analysis for piston are shown in Fig. 9.

Figure 9. Forces and moments analysis for the piston.

The Newton–Euler equations for piston are given as

(55) \begin{equation} \sum \boldsymbol{F}_{ext}=m_{i2}\boldsymbol{a}_{i2}=\boldsymbol{f}_{mi}-\boldsymbol{f}_{ci}+m_{i2}\boldsymbol{g} \end{equation}

By multiplying $\boldsymbol{s}_{i}$ on both sides of Eq. (55), formulas that only include driving force and kinematic parameters can be derived as follows:

(56) \begin{equation} f_{_{ci}}^{s}=\tau _{i}+m_{i2}\boldsymbol{g}\cdot \boldsymbol{s}_{i}-m_{i2}\boldsymbol{a}_{i2}\cdot \boldsymbol{s}_{i} \end{equation}

The forces and moments analysis diagrams of 3rd and 4th driving limbs are given in Fig. 10. The reaction forces and moments at fixed points $S_{2j}$ are denoted by $\boldsymbol{f}_{sj}$ and $\boldsymbol{M}_{sj}$ . The forces and moments at moving points $U_{2j}$ are denoted by $\boldsymbol{f}_{uj}$ and $\boldsymbol{M}_{uj}$ , and the internal forces and moments between cylinders and pistons are denoted by $\boldsymbol{f}_{mj}$ and $\boldsymbol{M}_{mj}$ , respectively.

Figure 10. Forces and moments analysis for 3rd and 4th driving limbs.

The Newton–Euler equation for 3rd and 4th driving limbs are given as

(57) \begin{align} \left({}^{A}{\boldsymbol{I}}{_{cj1}^{}}+{}^{A}{\boldsymbol{I}}{_{cj2}^{}}\right)\dot{\boldsymbol{\omega }}_{j}+\boldsymbol{\omega }_{j}\times \left({}^{A}{\boldsymbol{I}}{_{cj1}^{}}+{}^{A}{\boldsymbol{I}}{_{cj2}^{}}\right)\boldsymbol{\omega }_{j}&=-L_{j}\left(\boldsymbol{s}_{j}\times \boldsymbol{f}_{uj}\right)-m_{i2}\!\left(L_{j}-c_{i3}\right)\boldsymbol{s}_{j}\times \left(\boldsymbol{a}_{i2}-\boldsymbol{g}\right)\nonumber \\ & \quad-m_{i1}L_{31}\boldsymbol{s}\times \left(\boldsymbol{a}_{i1}-\boldsymbol{g}\right)-M_{uj}\boldsymbol{t}_{uj} \end{align}

According to Eq. (56), the following equation can be obtained

(58) \begin{equation} M_{uj}=\frac{\boldsymbol{s}_{j}\cdot \left(\left({}^{A}{\boldsymbol{I}}{_{ci1}^{}}+{}^{A}{\boldsymbol{I}}{_{ci2}^{}}\right)\dot{\boldsymbol{\omega }}_{j}\right)+\boldsymbol{s}_{j}\cdot \left(\boldsymbol{\omega }_{j}\times \left({}^{A}{\boldsymbol{I}}{_{ci1}^{}}+{}^{A}{\boldsymbol{I}}{_{ci2}^{}}\right)\boldsymbol{\omega }_{j}\right)}{-\boldsymbol{s}_{j}\cdot \boldsymbol{t}_{uj}} \end{equation}

Vector cross-product formula is as follows:

(59) \begin{equation} \boldsymbol{a}\times \left(\boldsymbol{b}\times \boldsymbol{c}\right)=\left(\boldsymbol{c}\cdot \boldsymbol{a}\right)\boldsymbol{b}-\left(\boldsymbol{a}\cdot \boldsymbol{b}\right)\boldsymbol{c} \end{equation}

Furthermore, by multiplying $\boldsymbol{s}$ on both sides of Eq. (57) and combing Eq. (59), the following equation can be obtained

(60) \begin{equation} \boldsymbol{f}_{uj}=\left(\boldsymbol{s}_{j}\cdot \boldsymbol{f}_{uj}\right)+\frac{1}{L_{j}}\boldsymbol{s}_{j}\times \left(\begin{array}{l} \left(m_{j2}\!\left(L_{j}-c_{j3}\right)\boldsymbol{s}_{j}\times \left(\boldsymbol{a}_{j2}-\boldsymbol{g}\right)\right)+\left(c_{j1}m_{j1}\boldsymbol{s}_{j}\times \left(\boldsymbol{a}_{j1}-\boldsymbol{g}\right)\right)\\[5pt] +\left(\left({}^{A}{\boldsymbol{I}}{_{cj1}^{}}+{}^{A}{\boldsymbol{I}}{_{cj2}^{}}\right)\dot{\boldsymbol{\omega }}_{j}\right)+\left(\boldsymbol{\omega }_{j}\times \left({}^{A}{\boldsymbol{I}}{_{cj1}^{}}+{}^{A}{\boldsymbol{I}}{_{cj2}^{}}\right)\boldsymbol{\omega }_{j}\right)+M_{uj}\boldsymbol{t}_{uj} \end{array}\right) \end{equation}

The forces and moments analysis of piston are shown in Fig. 11.

Figure 11. Forces and moments analysis for the piston.

The Newton–Euler equations for piston segments are given as

(61) \begin{equation} \boldsymbol{f}_{mj}-\boldsymbol{f}_{uj}+m_{j2}\boldsymbol{g}=m_{j2}\boldsymbol{a}_{j2} \end{equation}

Multiplying $\boldsymbol{s}$ on both sides of Eq. (61) and substituting Eq. (60) into Eq. (61), we have

(62) \begin{align} \boldsymbol{f}_{uj}&=\left(\tau _{j}+\boldsymbol{s}_{j}\cdot m_{j2}\boldsymbol{g}-\boldsymbol{s}_{j}\cdot m_{j2}\boldsymbol{a}_{j2}\right)\boldsymbol{s}_{j}\nonumber \\ &\quad +\frac{1}{L_{j}}\boldsymbol{s}_{j}\times \left(\begin{array}{l} \left(L_{j}-c_{j2}\right)\boldsymbol{s}_{j}\times m_{j2}\!\left(\boldsymbol{g}-\boldsymbol{a}_{j2}\right)+c_{j1}\boldsymbol{s}_{j}\times m_{j1}\!\left(\boldsymbol{g}-\boldsymbol{a}_{j1}\right)\\[5pt] +\left({}^{A}{\boldsymbol{I}}{_{cj1}^{}}+{}^{A}{\boldsymbol{I}}{_{cj2}^{}}\right)\dot{\boldsymbol{\omega }}_{j}+\boldsymbol{\omega }_{j}\times \left({}^{A}{\boldsymbol{I}}{_{cj1}^{}}+{}^{A}{\boldsymbol{I}}{_{cj2}^{}}\right)\boldsymbol{\omega }_{j}+M_{uj}\boldsymbol{t}_{uj} \end{array}\right) \end{align}

To sum up, there are 84 equations based on the Newton–Euler method, but 93 unknown variables need to be solved due to the over-constraints of the 4-DOF PDM. In the next section, nine compatibility equations are introduced considering the flexibility matrix and compatible deformation.

5.1. Link’s elastic deformation

In this paper, the link is regarded as a homogeneous rod. As shown in Fig. 12, when the end link is subjected to external forces $\boldsymbol{F}\boldsymbol{=}[\begin{array}{cccccc} F_{x} & F_{y} & F_{z} & M_{x} & M_{y} & M_{z} \end{array}]^{T}$ , corresponding six-dimensional deformation will be formed, including tension, torsion, and bending.

Figure 12. Schematic diagram of deformation at the end of uniform bar with regular section.

The relation between the external force F and the corresponding small deflection X along the applied force axis can be written as follows

(63) \begin{equation} \boldsymbol{X}=\boldsymbol{CF} \end{equation}

in which

(64) \begin{equation} \boldsymbol{C}=\left[\begin{array}{cccccc} \dfrac{l}{EA} & 0 & 0 & 0 & 0 & 0\\ 0 & \dfrac{l^{3}}{3EI_{z}} & 0 & 0 & 0 & \dfrac{l^{2}}{2EI_{z}}\\ 0 & 0 & \dfrac{l^{3}}{3EI_{y}} & 0 & -\dfrac{l^{2}}{2EI_{y}} & 0\\ 0 & 0 & 0 & \dfrac{l}{GI_{p}} & 0 & 0\\ 0 & 0 & -\dfrac{l^{2}}{2EI_{y}} & 0 & \dfrac{l}{EI_{y}} & 0\\ 0 & \dfrac{l^{2}}{2EI_{z}} & 0 & 0 & 0 & \dfrac{l}{EI_{z}} \end{array}\right] \end{equation}

where C represents the flexibility matrix with respect to frame $\{\boldsymbol{O}\}$ . $I_{y},I_{z}$ , and $I_{p}$ denote cross-section inertia parameters of the rod, respectively. $l$ and $A$ are rod length and cross-sectional area of rod. E and G represent elastic modulus and shear modulus of materials, respectively.

For the two-link mechanism depicted in Fig. 13, according to the force balance condition of the rod, the Jacobian matrix J as the projection of forces $\boldsymbol{F}_{i}$ to external forces $\boldsymbol{F}$ are given

Figure 13. Deformation diagram of two link ends with regular section.

(65) \begin{equation} \boldsymbol{F}_{i}=\boldsymbol{J}_{i}\boldsymbol{F} \end{equation}

in which

(66) \begin{equation} \boldsymbol{J}_{k}=\left(\begin{array}{c@{\quad}c} {}_{ok}^{oi}{\boldsymbol{R}}{} & \mathbf{0}_{3\times 3}\\[5pt] \boldsymbol{S}\!\left({}^{oi}{\boldsymbol{P}}{_{ok}^{}}\right){}_{ok}^{oi}{\boldsymbol{R}}{} & {}_{ok}^{oi}{\boldsymbol{R}}{} \end{array}\right) \end{equation}

where ${}_{ok}^{oi}{\boldsymbol{R}}{}$ represents rotational transformation matrix between frame $\{\boldsymbol{O}_{\boldsymbol{k}}\}$ and frame $\{\boldsymbol{O}_{\boldsymbol{i}}\}$ . ${}^{oi}{\boldsymbol{P}}{_{ok}^{}}$ is the position vector with respect to the frame $\{\boldsymbol{O}_{\boldsymbol{k}}\}$ . $\boldsymbol{S}({}^{oi}{\boldsymbol{P}}{_{ok}^{}})$ denotes antisymmetric matrix of position vector ${}^{oi}{\boldsymbol{P}}{_{ok}^{}}$ .

Referring to Eq. (63), the relation between the force $\boldsymbol{F}_{i}$ and the corresponding small deflection $\boldsymbol{X}_{i}$ along the applied force axis can be written as follows

(67) \begin{equation} \boldsymbol{X}_{i}=\boldsymbol{C}_{i}\boldsymbol{F}_{i} \end{equation}

Hence, the deformation of end link $\Delta \boldsymbol{X}_{i}$ caused by the deformation of the rod i is as follows

(68) \begin{equation} \Delta \boldsymbol{X}_{i}=\boldsymbol{J}_{i}^{T}\boldsymbol{C}_{i}\boldsymbol{F}_{i} \end{equation}

Therefore, the deformation of the end link under the external force F can be derived by the sum of linear deformation of all rod i

(69) \begin{equation} \boldsymbol{X}\boldsymbol{=}\sum \Delta \boldsymbol{X}_{i}=\sum \boldsymbol{J}_{i}^{T}\boldsymbol{C}_{i}\boldsymbol{J}_{i}\boldsymbol{F} \end{equation}

5.2. Rigid-flexible coupling simulation

As indicated above, the 4-DOF PDM possesses nine passive redundant constraints. Since the SPU limb own six degrees-of-freedom, the over-constraints are caused by other links besides the SPU limbs. In this section, based on the screw theory, the over-constrained direction is analyzed. Moreover, the over-constraints were verified via the co-simulation with ABAQUS and ADAMS software. First, the flexible links are meshed by hexahedron elements in ABAQUS, and the mnf files of the meshed links are obtained via ABAQUS. Then, loading the link files, rigid links are replaced with flexible bodies in ADAMS. The simulation model is shown as Fig. 14.

Figure 14. Rigid–flexible coupling model of the 4-DOF PDM.

From the above analysis, we know that over-constraints are caused by operating mechanism and two RPR limbs. In order to supplement the deformation coordination equations, we need to further determine the direction of over-constraints. Based on the screw theory, the constraint screw of the limb is as follows:

(70) \begin{equation} \begin{array}{l} \boldsymbol{\$ }_{1}^{rr}\boldsymbol{=}\left[\begin{array}{l@{\quad}l@{\quad}l@{\quad}l@{\quad}l@{\quad}l} 0 & 0 & 1 & 0 & 0 & 0 \end{array}\right]\\[3pt] \boldsymbol{\$ }_{2}^{rr}\boldsymbol{=}\left[\begin{array}{l@{\quad}l@{\quad}l@{\quad}l@{\quad}l@{\quad}l} 0 & 0 & 0 & 1 & 0 & 0 \end{array}\right]\\[3pt] \boldsymbol{\$ }_{3}^{rr}\boldsymbol{=}\left[\begin{array}{l@{\quad}l@{\quad}l@{\quad}l@{\quad}l@{\quad}l} 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right] \end{array} \end{equation}

As shown in Fig. 15, the operating mechanism and two RPR limbs possess the constrained force along z -axis and constrained couples around x -axis, y -axis, which means each link in the operating mechanism and two RPR limb own three over-constrained motions.

Figure 15. Schematic diagram of over-constrained direction analysis.

5.3. Deformation compatibility equation

In this section, on the condition of the numbers and direction of over-constraints, we derive the deformation compatibility equation in detail. First, the operating mechanism and two RPR limbs are divided into three modules. Then derive the deformation of each module according to the link’s elastic deformation in Section 5.1. Finally, the deformation compatibility equations are derived at joints $R_{21},R_{22}$ , and $R_{32}$ .

The elastic deformation of limbs 1 and 2 at joint $R_{32}$ as shown in Fig. 16.

Figure 16. Limbs 1 and 2 deformation analysis.

Referring to Eq. (65), we have

(71) \begin{equation} {}^{B2}{\boldsymbol{F}}{_{R4}^{}}=\boldsymbol{J}_{B2,B1}{}^{B1}{\boldsymbol{F}}{_{r32,r4}^{}}+\boldsymbol{J}_{B2,A}{}^{A}{\boldsymbol{F}}{_{4p}^{}} \end{equation}

(72) \begin{equation} {}^{B3}{\boldsymbol{F}}{_{r4,r31}^{}}=\boldsymbol{J}_{B3,B2}{}^{B2}{\boldsymbol{F}}{_{R4}^{}} \end{equation}

(73) \begin{equation} {}^{B4}{\boldsymbol{F}}{_{r31,r21}^{}}=\boldsymbol{J}_{B4,B3}{}^{B3}{\boldsymbol{F}}{_{r4,r31}^{}} \end{equation}

where ${}^{B1}{\boldsymbol{F}}{_{r32,r4}^{}}$ is the force applied to link $r_{4}$ by link $r_{32}$ with respect to the frame { A }. $\boldsymbol{J}_{B2,A}$ represents the transformation matrix between frame { B ₂} and frame { A }.

Based on the Eq. (69), the elastic deformations of the links $r_{21}$ and $r_{31}$ at the joint $R_{31}$ with respect to the frame { B ₃} are

(74) \begin{equation} {}^{B3}{\boldsymbol{\delta }}{_{R31}^{}}\boldsymbol{=}\boldsymbol{C}_{1}{}^{B3}{\boldsymbol{F}}{_{r4,r31}^{}}+\boldsymbol{J}_{B4,B3}^{T}\left(\boldsymbol{C}_{1}{}^{B4}{\boldsymbol{F}}{_{r31,r21}^{}}+\boldsymbol{C}_{1}{}^{B4}{\boldsymbol{F}}{_{l11,r21}^{}}\right) \end{equation}

Hence, the elastic deformation of the links $r_{21},r_{31}$ , and $r_{4}$ at the joint $R_{32}$ with respect to the frame { A } are

(75) \begin{equation} {}^{A}{\boldsymbol{\delta }}{_{R32\_ \lim b1}^{}}\boldsymbol{=}\boldsymbol{J}_{B1,A}^{T}\left(\boldsymbol{J}_{B3,B1}^{T}{}^{B3}{\boldsymbol{\delta }}{_{R31}^{}}+\boldsymbol{J}_{B2,B1}^{T}\boldsymbol{C}_{2}{}^{B2}{\boldsymbol{F}}{_{R4}^{}}+\boldsymbol{C}_{2}{}^{B1}{\boldsymbol{F}}{_{r32,r4}^{}}\right) \end{equation}

Similarly, the elastic deformation of limb2 at $R_{32}$ is given as

(76) \begin{equation} {}^{B6}{\boldsymbol{F}}{_{r32,r22}^{}}=\boldsymbol{J}_{B6,B5}{}^{B5}{\boldsymbol{F}}{_{r32,r4}^{}} \end{equation}

(77) \begin{equation} {}^{R22}{\boldsymbol{\delta }}{_{R22}^{}}=\boldsymbol{C}_{1}{}^{B6}{\boldsymbol{F}}{_{r32,r22}^{}}+\boldsymbol{C}_{1}{}^{B6}{\boldsymbol{F}}{_{l12,r22}^{}} \end{equation}

(78) \begin{equation} {}^{A}{\boldsymbol{\delta }}{_{R32\_ \lim\! b2}^{}}=\boldsymbol{J}_{B5,A}\left(\boldsymbol{J}_{B6,B5}{}^{R22}{\boldsymbol{\delta }}{_{R22}^{}}+\boldsymbol{C}_{1}{}^{B5}{\boldsymbol{F}}{_{r4,r32}^{}}\right) \end{equation}

The deformation caused by limbs 3 and 4 at joint $R_{32}$ as shown in Fig. 17.

Figure 17. Limbs 3 and 4 deformation analysis.

Referring to Eq. (69), the elastic deformation of limb 3 at joint $R_{21}$ is given as

(79) \begin{equation} {}^{A}{\boldsymbol{\delta }}{_{R21\_ \lim\! b3}^{}}=\boldsymbol{J}_{B7,A}^{T}\left(\boldsymbol{J}_{B8,B7}^{T}\boldsymbol{C}_{4}\boldsymbol{J}_{B8,B7}\!\left(-{}^{B7}{\boldsymbol{F}}{_{l11,r21}^{}}\right)+\boldsymbol{C}_{3}\!\left(-{}^{B7}{\boldsymbol{F}}{_{l11,r21}^{}}\right)\right) \end{equation}

Similarly, the elastic deformation of limb 4 at joint $R_{21}$ is given as

(80) \begin{equation} {}^{A}{\boldsymbol{\delta }}{_{R21\_ \lim\! b4}^{}}=\boldsymbol{J}_{B4,A}^{T}\!\left(\boldsymbol{C}_{1}{}^{B4}{\boldsymbol{F}}{_{l2,l1}^{}}+\boldsymbol{C}_{1}{}^{B4}{\boldsymbol{F}}{_{l7,l1}^{}}\right) \end{equation}

The deformation caused by limbs 5 and 6 at joint $R_{32}$ as shown in Fig. 18.

Figure 18. Limbs 5 and 6 deformation analysis.

Based on the Eq. (69), the elastic deformation of limb 5 at joint $R_{22}$ are given as

(81) \begin{equation} {}^{A}{\boldsymbol{\delta }}{_{R22\_ \lim\! b5}^{}}=\boldsymbol{J}_{B9,A}^{T}\left(\boldsymbol{J}_{B10,B9}^{T}\boldsymbol{C}_{6}\boldsymbol{J}_{B10,B9}\left(-{}^{B9}{\boldsymbol{F}}{_{l12,r22}^{}}\right)+\boldsymbol{C}_{5}\!\left(-{}^{B7}{\boldsymbol{F}}{_{l12,r22}^{}}\right)\right) \end{equation}

Similarly, the elastic deformation of limb 6 at joint $R_{22}$ are given as

(82) \begin{equation} {}^{A}{\boldsymbol{\delta }}{_{R22\_ \lim\! b6}^{}}=\boldsymbol{J}_{B6,A}^{T}\left(\boldsymbol{C}_{1}{}^{B6}{\boldsymbol{F}}{_{r32,r22}^{}}+\boldsymbol{C}_{1}{}^{B6}{\boldsymbol{F}}{_{l12,r22}^{}}\right) \end{equation}

The deformations of the operating mechanism and two RPR limbs have to be compatible with each other to satisfy the geometric constraints. Hence, translational deformation along z and rotational deformation about x and y axis at joints $R_{21},R_{22},R_{32}$ should be equal, namely

(83) \begin{equation} \begin{array}{l} {}^{A}{\boldsymbol{\delta }}{_{R32\_ \lim\! b1}^{}}\left(k,1\right)={}^{A}{\boldsymbol{\delta }}{_{R32\_ \lim\! b2}^{}}\left(k,1\right)\\ {}^{A}{\boldsymbol{\delta }}{_{R21\_ \lim\! b3}^{}}\left(k,1\right)={}^{A}{\boldsymbol{\delta }}{_{R21\_ \lim\! b4}^{}}\left(k,1\right)\left(k=3,4,5\right)\\ {}^{A}{\boldsymbol{\delta }}{_{R22\_ \lim\! b5}^{}}\left(k,1\right)={}^{A}{\boldsymbol{\delta }}{_{R22\_ \lim\! b6}^{}}\left(k,1\right) \end{array} \end{equation}

So far, the complete dynamics models with constrained forces/moments of the over-constrained 4-DOF PDM with sub-closed chains are established. The actuation forces and constrained forces/moments can be obtained simultaneously, which is essential for structure design. Also, clearance and friction can be easily integrated into the system equation.

6.1. Simulation validation

According to the derivation of the above formula, the displacements, velocities, accelerations, driving forces, and all constraint forces/moments of the 4-DOF PDM are obtained. The numerical simulation results are qualitatively consistent with the theoretical results. As shown in Fig. 19, $l_{1,a}, v_{1,a}$ , and $a_{1,a}$ (i = 1, 2, 3, 4) denote the theoretical displacements, velocities, and accelerations of the slider respectively. $l_{i,n}, v_{i,n}$ , and $a_{i,n}$ (i = 1, 2, 3, 4) denote the numerical displacements, velocities, and accelerations of the slider, respectively.

Figure 19. Numerical and theoretical results of displacement/velocity/acceleration.

The results of co-simulation with ADAMS and ABAQUS are shown in Fig. 14. Figure 20 shows the active forces computed by the proposed method and ADAMS; Figure 21 shows the active forces computed by the proposed method and Lagrangian method; Figure 22(a) and (b) show the constraint forces/moments in the over-constrained direction at joints $R_{32}$ and $R_{21}$ , respectively. $f_{i\_ z}^{a},m_{i\_ x}^{a}$ and $m_{i\_ y}^{a}(i=1,2)$ represent the theoretical value and $f_{i\_ z}^{n},m_{i\_ x}^{n}$ and $m_{i\_ y}^{n}(i=1,2)$ represent the simulation value.

Figure 20. Active forces computed by the proposed method and ADAMS.

Figure 21. Active forces computed by the proposed method and Lagrangian method.

Figure 22. Partial constraint forces/moments simulation and theoretical value.

It is assumed that the cylinders and pistons are symmetrically distributed in this paper. That is, the influence of the eccentricity of the electric cylinder on the constraint torque at the U-joint is not considered. In order to verify the influence of acceleration on the constraint torque and driving force at the U-joint, the constraint torques at the U-joint are calculated when the maximum translational acceleration along y-axis is up to 50, 500, and 1000 mm/s². Moreover, the driving forces are solved with and without constraint torque at the U-joint, and the driving forces error percentage are plotted as shown in Fig. 23(b).

Figure 23. Constraint torque at U-joint and driving force error percentage under different accelerations.

6.2. Experiment validation

According to the CAD model displayed in Fig. 1, a physical prototype has been built for the motion control experiment. As shown in Fig. 24, actuators are driven by the electric piston-cylinder, which are connected to an embedded controller (CX5130) made by the company Beckhoff through an Ethercat bus. Based on the position loop mode PID controller to achieve the target trajectory, it is one of the commonly used controllers in academic research and industrial products. The motion control experiment based on PID controller is shown in Figure 25. The measured values of displacements, velocities, and accelerations of the four driving limbs are shown in Fig. 26, which are consistent with the theoretical calculation.

Figure 24. Physical prototype and control system.

Figure 25. Trajectory planning based on classical PID control.

In this paper, the control cycle of the embedded controller (CX5130) is 2 ms. As shown in Fig. 26, the measured position and velocity of the four driving limbs are consistent with the theoretical value, but the lag of phase is 20 ms. Due to the short control cycle, the position signal obtained by the encoder undergoes two differential and filtering processes, resulting in significant signal distortion. As shown in Fig. 26, the measured acceleration not only have phase lag but also have attenuation in amplitude.