3.1.1. Forward position analysis of the PM2

Given the input $\theta _{i}(i=1\sim 6)$ of the driving joints, to determine the posture of the moving platform.

Suppose the coordinates of any point S _i (i = 1 ∼ 3) on the moving platform in the moving coordinate system are denoted as S _i ’, and the origin O‘ of the moving coordinate system has coordinates S _o (x, y, z) in the base coordinate system. Then, the coordinates of any point S _i on the moving platform in the base coordinate system can be expressed as follows:

(1) \begin{equation} S_{i}=QS_{i}^{\prime}+S_{0} \end{equation}

Here, Q represents the transformation matrix from the moving coordinate system to the base coordinate system.

Due to the end POC of the moving platform being 3T3R, where α, β, and γ represent the rotating angles of the moving platform around the x, y, and z axes, respectively, Q can be expressed as:

(2) \begin{equation} Q=\left[\begin{array}{c@{\quad}c@{\quad}c} \cos \beta \cos \gamma & -\cos \beta \sin \gamma & \sin \beta \\[4pt] \mathit{\sin }\mathit{\alpha }\sin \beta \cos \gamma +\cos \alpha \sin \gamma & \cos \alpha \cos \gamma -\mathit{\sin }\mathit{\alpha }\sin \beta \sin \gamma & -\sin \alpha \cos \beta \\[4pt] \sin \alpha \sin \gamma -\mathit{\cos }\mathit{\alpha }\sin \beta \cos \gamma & \mathit{\cos }\mathit{\alpha }\sin \beta \sin \gamma +\sin \alpha \cos \gamma & \cos \alpha \cos \beta \end{array}\right] \end{equation}

The coordinates of the vertices of the moving platform in the moving coordinate system are given as:

\begin{equation*} S^{\prime}_{1}=(b,0,0), S^{\prime}_{2}=\left(-b/2,\sqrt{3}b/2,0\right), S^{\prime}_{3}=\left(-b/2,-\sqrt{3}b/2,0\right). \end{equation*}

According to Eq. (1), the coordinates of the three vertices S ₁∼S ₃ of the moving platform in the base coordinate system can be obtained as follows:

(3) \begin{equation} \left\{\begin{array}{l} x_{s1}=\mathit{b}\cos \beta \cos \gamma +x\\[4pt] y_{s1}=b\left(\mathit{\sin }\mathit{\alpha }\sin \beta \cos \gamma +\cos \alpha \sin \gamma \right)+y\\[4pt] z_{s1}=b\left(\sin \alpha \sin \gamma -\mathit{\cos }\mathit{\alpha }\sin \beta \cos \gamma \right)+z \end{array}\right. \end{equation}

(4) \begin{equation} \left\{\begin{array}{l} x_{s2}=-\dfrac{b}{2}\cos \beta \cos \gamma -\dfrac{\sqrt{3}}{2}\mathit{b}\cos \beta \sin \gamma +x\\[12pt] y_{s2}=-\dfrac{b}{2}\left(\mathit{\sin }\mathit{\alpha }\sin \beta \cos \gamma +\cos \alpha \sin \gamma \right)+\dfrac{\sqrt{3}b}{2}\left(\cos \alpha \cos \gamma -\mathit{\sin }\mathit{\alpha }\sin \beta \sin \gamma \right)+y\\[12pt] z_{s2}=-\dfrac{b}{2}\left(\sin \alpha \sin \gamma -\mathit{\cos }\mathit{\alpha }\sin \beta \cos \gamma \right)+\dfrac{\sqrt{3}b}{2}\left(\mathit{\cos }\mathit{\alpha }\sin \beta \sin \gamma +\sin \alpha \cos \gamma \right)+z \end{array}\right. \end{equation}

(5) \begin{equation} \left\{\begin{array}{l} x_{s3}=-\dfrac{b}{2}\cos \beta \cos \gamma +\dfrac{\sqrt{3}b}{2}\cos \beta \sin \gamma +x\\[12pt] y_{s3}=-\dfrac{b}{2}\left(\mathit{\sin }\mathit{\alpha }\sin \beta \cos \gamma +\cos \alpha \sin \gamma \right)-\dfrac{\sqrt{3}}{2}b\left(\cos \alpha \cos \gamma -\mathit{\sin }\mathit{\alpha }\sin \beta \sin \gamma \right)+y\\[12pt] z_{s3}=-\dfrac{b}{2}\left(\sin \alpha \sin \gamma -\mathit{\cos }\mathit{\alpha }\sin \beta \cos \gamma \right)-\dfrac{\sqrt{3}}{2}b\left(\mathit{\cos }\mathit{\alpha }\sin \beta \sin \gamma +\sin \alpha \cos \gamma \right)+z \end{array}\right. \end{equation}

1) Analysis of hybrid branch chain 1

For analysis of forward position for hybrid branch chain 1, the line A₁₃S₁ is perpendicular to the line A₂₂S₁ in hybrid branch chain 1, and the angle between the line A₁₃S₁ and A₁₃A₂₂ is $\gamma _{1}=30^{\circ }$ .

Furthermore, the angle between link A₁₃A₂₂ and the negative direction of the Z-axis is set as $\alpha _{1}$ and used as an intermediate variable.

It’s easy to know the coordinates of the following points.

$A_{11}=(a,0,a), A_{21}=(a,0,a-l_{0}), A_{12}=(a,0,a+l_{0}), A_{13}=(a+l_{1}\sin \theta _{2}\cos \theta _{1},l_{1}\sin \theta _{2}\sin \theta _{1},a+l_{0}-l_{1}\cos \theta _{2})$ ,

$A_{22}=(a+(l_{1}\sin \theta _{2}+l_{5}\sin \alpha _{1})\cos \theta _{1},(l_{1}\sin \theta _{2}+l_{5}\sin \alpha _{1})\sin \theta _{1},a+l_{0}-l_{1}\cos \theta _{2}-l_{5}\cos \alpha _{1})$ .

According to the link length equation by constraint $A_{21}A_{22}=l_{3}$ , the intermediate variable $\alpha _{1}$ is found as

(6) \begin{equation} \alpha _{1}=2\arctan \left(\frac{N_{1}+n\sqrt{N_{1}^{2}+N_{2}^{2}-N_{3}^{2}}}{N_{2}+N_{3}}\right)\left(n=\pm 1\right) \end{equation}

Where: $N_{1}=2l_{1}l_{5}\sin \theta _{2}, N_{2}=-2l_{5}(2l_{0}-l_{1}\cos \theta _{2}), N_{3}=l_{3}^{2}-l_{5}^{2}-(l_{1}\sin \theta _{2})^{2}-(2l_{0}-l_{1}\cos \theta _{2})^{2}$ .

Then the coordinates of S₁ are expressed as

(7) \begin{equation} \left\{\begin{array}{l} x_{s1}=a+\left(l_{1}\sin \theta _{2}+l_{2}\sin \left(\gamma _{1}+\alpha _{1}\right)\right)\cos \theta _{1}\\[4pt] y_{s1}=\left(l_{1}\sin \theta _{2}+l_{2}\sin \left(\gamma _{1}+\alpha _{1}\right)\right)\sin \theta _{1}\\[4pt] z_{s1}=a+l_{0}-l_{1}\cos \theta _{2}-l_{2}\cos \left(\gamma _{1}+\alpha _{1}\right) \end{array}\right. \end{equation}

2) Analysis of hybrid branch chain 2

The angle between link A₃₃A₄₂ in hybrid branch chain 2 and the negative direction of the X-axis is defined as $\alpha _{2}$ , and serves as an intermediate variable. It is easy to know the following coordinates.

$A_{31}=(a,a,0), A_{32}=(a+l_{0},a,0), A_{41}=(a-l_{0},a,0)$ ,

$A_{33}=(a+l_{0}-l_{1}\cos \theta _{4},a+l_{1}\sin \theta _{4}\cos \theta _{3},l_{1}\sin \theta _{4}\sin \theta _{3})$ ,

$A_{42}=(a+l_{0}-l_{1}\cos \theta _{4}-l_{5}\cos \alpha _{2},a+(l_{1}\sin \theta _{4}+l_{5}\sin \alpha _{2})\cos \theta _{3},(l_{1}\sin \theta _{4}+l_{5}\sin \alpha _{2})\sin \theta _{3})$ .

Based on the link length equation by constraint $A_{41}A_{42}=l_{3}$ , we get the intermediate variable $\alpha _{2}$ below

(8) \begin{equation} \alpha _{2}=2\arctan \left(\frac{M_{1}+m\sqrt{M_{1}^{2}+M_{2}^{2}-M_{3}^{2}}}{M_{2}+M_{3}}\right)\left(m=\pm 1\right) \end{equation}

Where: $M_{1}=2l_{5}l_{1}\sin \theta _{4}, M_{2}=-2l_{5}(2l_{0}-l_{1}\cos \theta _{4}), M_{3}=l_{3}^{2}-l_{5}^{2}-(l_{1}\sin \theta _{4})^{2}-(2l_{0}-l_{1}\cos \theta _{4})^{2}$ .

Then the coordinates of S₂ are expressed as

(9) \begin{equation} \left\{\begin{array}{l} x_{s2}=a+l_{0}-l_{1}\cos \theta _{4}-l_{2}\cos \left(\gamma _{1}+\alpha _{2}\right)\\[4pt] y_{s2}=a+\left(l_{1}\sin \theta _{4}+l_{2}\sin \left(\gamma _{1}+\alpha _{2}\right)\right)\cos \theta _{3}\\[4pt] z_{s2}=\left(l_{1}\sin \theta _{4}+l_{2}\sin \left(\gamma _{1}+\alpha _{2}\right)\right)\sin \theta _{3} \end{array}\right. \end{equation}

3) Analysis of hybrid branch chain 3

Similarly, let the angle between the link A₅₃A₆₂ in hybrid branch chain 3 and the negative Y-axis be $\alpha _{3}$ , which serves as an intermediate variable. It is also easy to know the following coordinates.

$A_{51}=(0,a,a), A_{52}=(0,a+l_{0},a), A_{61}=(0,\,a-l_{0},a), A_{53}=(l_{1}\sin \theta _{6}\sin \theta _{5},a+l_{0}-l_{1}\cos \theta _{6},a+l_{1}\sin \theta _{6}\cos \theta _{5})$ ,

$A_{62}=((l_{1}\sin \theta _{6}+l_{5}\sin \alpha _{3})\sin \theta _{5},\,a+l_{0}-l_{1}\cos \theta _{6}-l_{5}\cos \alpha _{3},a+(l_{1}\sin \theta _{6}+l_{5}\sin \alpha _{3})\cos \theta _{5})$ .

Based on the link length equation by constraint $A_{61}A_{62}=l_{3}$ , the intermediate variable $\alpha _{3}$ is given below

(10) \begin{equation} \alpha _{3}=2\arctan \left(\frac{Q_{1}+q\sqrt{Q_{1}^{2}+Q_{2}^{2}-Q_{3}^{2}}}{Q_{2}+Q_{3}}\right)\left(q=\pm 1\right) \end{equation}

Where: $\mathrm{Q}_{1}=2l_{5}l_{1}\sin \theta _{6}, Q_{2}=-2l_{5}(2l_{0}-l_{1}\cos \theta _{6}), Q_{3}=l_{3}^{2}-l_{5}^{2}-(l_{1}\sin \theta _{6})^{2}-(2l_{0}-l_{1}\cos \theta _{6})^{2}$ .

Then the coordinates of S₃ are expressed as

(11) \begin{equation} \left\{\begin{array}{l} x_{s3}=\left(l_{1}\sin \theta _{6}+l_{2}\sin \left(\gamma _{1}+\alpha _{3}\right)\right)\sin \theta _{5}\\[4pt] y_{s3}=a+l_{0}-l_{1}\cos \theta _{6}-l_{2}\cos \left(\gamma _{1}+\alpha _{3}\right)\\[4pt] z_{s3}=a+\left(l_{1}\sin \theta _{6}+l_{2}\sin \left(\gamma _{1}+\alpha _{3}\right)\right)\cos \theta _{5} \end{array}\right. \end{equation}

To here, the coordinate values of S₁, S₂, and S₃ are determined. Therefore, according to Eqs. (3)-(5), it is easy to know that the expression of the posture parameters of the moving platform of the 6-DOF PM is given below.

(12) \begin{equation} \left\{\begin{array}{l} x=\dfrac{x_{s1}+x_{s2}+x_{s3}}{3}\\[12pt] y=\dfrac{y_{s1}+y_{s2}+y_{s3}}{3}\\[12pt] z=\dfrac{z_{s1}+z_{s2}+z_{s3}}{3}\\[12pt] \gamma =\arctan \left[\left(x_{s3}-x_{s1}\right)/\left(\sqrt{3}\left(x_{s1}-x\right)\right)\right]\\[10pt] \beta =\arccos \left[\left(x_{s1}-x\right)/\left(b\cos \gamma \right)\right]\\[12pt] \alpha =\arctan \left[\dfrac{\sqrt{3}\left(y_{s1}-y\right)\cos \gamma -\left(y_{s2}-y_{s3}\right)\sin \gamma }{\sqrt{3}\left(y_{s1}-y\right)\sin \gamma \sin \beta +\left(y_{s2}-y_{s3}\right)\cos \gamma \sin \beta }\right] \end{array}\right. \end{equation}

Obviously, from the forward position analysis of the PM2, it can be inferred that the coordinate values of points S₁, S₂, and S₃ on the moving platform of PM2 can be obtained by solving the hybrid branch chain to which it belongs. Therefore, when any one of the points S₁, S₂, and S₃ is taken as the base point, i.e., the origin of the moving coordinate system, the PM has partial motion decoupling.

3.1.2 Inverse position analysis of the PM2

Given the position and postures of the moving platform of the PM2, to get the driving input $\theta _{i}(i=1\sim 6)$ .

1) Hybrid branch chain 1

From Eq (7), it is easy to know the following formula.

(13) \begin{equation} \theta _{1}=\arctan \left(\frac{y_{s1}}{x_{s1}-a}\right) \end{equation}

(14) \begin{equation} \theta _{2}=2\arctan \left(\frac{f_{1}+F\sqrt{f_{1}^{2}+f_{2}^{2}-f_{3}^{2}}}{f_{2}+f_{3}}\right)\left(F=\pm 1\right) \end{equation}

Where: $f_{1}=\frac{-2l_{1}y_{s1}}{\sin \theta _{1}}, f_{2}=2l_{1}\left(z_{s1}-a-l_{0}\right), f_{3}=l_{2}^{2}-l_{1}^{2}-\left(y_{s1}/\sin \theta _{1}\right)^{2}-\left(z_{s1}-a+l_{0}\right)^{2}$ .

2) Hybrid branch chain 2

From Eq. (9), we have

(15) \begin{equation} \theta _{3}=\arctan \left(\frac{z_{s2}}{y_{s2}-a}\right) \end{equation}

(16) \begin{equation} \theta _{4}=2\arctan \left(\frac{g_{1}+G\sqrt{g_{1}^{2}+g_{2}^{2}-g_{3}^{2}}}{g_{2}+g_{3}}\right)\left(G=\pm 1\right) \end{equation}

Where: $g_{1}=\frac{-2l_{11}z_{s2}}{\sin \theta _{3}}, g_{2}=2l_{1}\left(x_{s2}-a-l_{0}\right), g_{3}=l_{2}^{2}-l_{1}^{2}-\left(z_{s2}/\sin \theta _{3}\right)^{2}-\left(x_{s2}-a+l_{0}\right)^{2}$ .

3) Hybrid branch chain 3

From Eq. (11), we get

(17) \begin{equation} \theta _{5}=\arctan \left(\frac{x_{s3}}{z_{s3}-a}\right) \end{equation}

(18) \begin{equation} \theta _{6}=2\arctan \left(\frac{k_{1}+K\sqrt{k_{1}^{2}+k_{2}^{2}-k_{3}^{2}}}{k_{2}+k_{3}}\right)\left(K=\pm 1\right) \end{equation}

Where: $k_{1}=\frac{-2l_{1}x_{s3}}{\sin \theta _{5}}, k_{2}=2l_{1}\left(y_{s3}-a-l_{0}\right), f_{3}=l_{2}^{2}-l_{1}^{2}-\left(x_{s3}/\sin \theta _{5}\right)^{2}-\left(y_{s3}-a+l_{0}\right)^{2}$ .

3.2.1 Forward position analysis of the PM3

The angle between the link A₁₂A₁₃ and the positive direction of the Z-axis is denoted as $\alpha _{1}$ , and serves as an intermediate variable. It’s easy to know the coordinates of the following points.

$A_{11}=(a,0,a+l_{10}/2), A_{21}=(a,0,a-l_{10}/2), A_{12}=(a,l_{11}\sin \theta _{1},a+l_{10}/2-l_{11}\cos \theta _{1})$ ,

$A_{22}=(a,l_{13}\sin \theta _{2},a-l_{10}/2+l_{13}\cos \theta _{2}), A_{13}=(a,l_{11}\sin \theta _{1}+l_{12}\sin \alpha _{1},a+l_{10}/2-l_{11}\cos \theta _{1}-l_{12}\cos \alpha _{1})$ ,

$A_{14}=(a,l_{11}\sin \theta _{1}+(l_{12}+l_{15})\sin \alpha _{1},a+l_{10}/2-l_{11}\cos \theta _{1}-(l_{12}+l_{15})\cos \alpha _{1})$ .

According to the link length equation by constraint $A_{13}A_{22}=l_{4}$ , the intermediate variable $\alpha _{1}$ is obtained as

(19) \begin{equation} \alpha _{1}=2\arctan \left(\frac{N_{1}+n\sqrt{N_{1}^{2}+N_{2}^{2}-N_{3}^{2}}}{N_{2}+N_{3}}\right)\left(n=\pm 1\right) \end{equation}

Where: $N_{1}=2l_{12}(l_{11}\sin \theta _{1}-l_{13}\sin \theta _{2}), N_{2}=-2l_{12}(l_{10}-l_{11}\cos \theta _{1}-l_{13}\cos \theta _{2})$ ,

$N_{3}=l_{14}^{2}-l_{12}^{2}-(l_{11}\sin \theta _{1}-l_{13}\sin \theta _{2})^{2}-(l_{10}-l_{11}\cos \theta _{1}-l_{13}\cos \theta _{2})^{2}$ .

Let the rotation angle of R₁₀ from the YOZ plane be $\varphi _{1}$ , as shown in Fig. 4(b). Then the coordinates of S₁ can be expressed as

(20) \begin{equation} \left\{\begin{array}{l} x_{s1}=x_{A14}-l_{16}\sin \varphi _{1}\\[4pt] y_{s1}=y_{A14}+l_{16}\cos \varphi _{1}\cos \alpha _{1}\\[4pt] z_{s1}=z_{A14}+l_{16}\cos \varphi _{1}\sin \alpha _{1} \end{array}\right. \end{equation}

The angle $\varphi _{1}$ in the hybrid branch chain 1 can be obtained by the link length constraint $S_{1}S_{2}=\sqrt{3}b$ .

(21) \begin{equation} \varphi _{1}=2\arctan \left(\frac{W_{1}+w\sqrt{W_{1}^{2}+W_{2}^{2}-W_{3}^{2}}}{W_{2}+W_{3}}\right)\left(w=\pm 1\right) \end{equation}

Where, $W_{1}=-2l_{16}\cos \alpha _{1}(x_{A14}-x_{s2}), W_{2}=2l_{16}(y_{A14}-y_{s2})\cos \alpha _{1}+2l_{16}(z_{A14}-z_{s2})\sin \alpha _{1},$ $W_{3}=3b^{2}-$ $l_{16}^{2}-(x_{s2}-x_{A14})^{2}-(y_{s2}-y_{A14})^{2}-(z_{s2}-z_{A14})^{2}$ .

The hybrid branch chains 2 and 3 in PM3 are the same as the PM2, so the forward solution analysis of the kinematic position is the same. For details, see Eqs. (8) to (11).

Here, the coordinate values of S₁, S₂, and S₃ are obtained. Therefore, according to Eq. (12), it is easy to determine the posture parameters of the moving platform of the PM3.

Obviously, from the above analysis process, it is easy to know that the coordinate values of points S₂ and S₃ on the moving platform of the PM3 can be solved by the hybrid branch chain in which they are located. Therefore, taking any point in S₂ and S₃ as the base point, the PM3 has partial motion decoupling.

3.2.2 Inverse position analysis of the PM3

It is easy to know from the topological analysis that the link 6 in hybrid branch chain 1 is perpendicular to the link 5, as shown in Fig. 4(b). Therefore, in the triangles S₁A₁₄A₁₃ and S₁A₁₄A₁₂, we have

$S_{1}A_{13}=\sqrt{l_{16}^{2}+l_{15}^{2}}=l_{17}, S_{1}A_{12}=\sqrt{l_{16}^{2}+(l_{12}+l_{15})^{2}}=l_{18}$ .

According to the link length equation by constraint $S_{1}A_{12}=l_{18}$ , it is easy to know

(22) \begin{equation} \theta _{1}=2\arctan \left(\frac{n_{1}+N\sqrt{n_{1}^{2}+n_{2}^{2}-n_{3}^{2}}}{n_{2}+n_{3}}\right)\left(N=\pm 1\right) \end{equation}

Where: $n_{1}=-2l_{11}y_{s1}, n_{2}=-2l_{11}(a+l_{10}/2-z_{s1}), n_{3}={l_{18}}^{2}-l_{11}^{2}-(a-x_{s1})^{2}-y_{s1}^{2}-(a+l_{10}/2-z_{s1})^{2}$ .

According to the link length equation by constraint $S_{1}A_{13}=l_{17}$ , we have

(23) \begin{equation} \alpha _{1}=2\arctan \left(\frac{m_{1}+M\sqrt{m_{1}^{2}+m_{2}^{2}-m_{3}^{2}}}{m_{2}+m_{3}}\right)\left(M=\pm 1\right) \end{equation}

Where: $m_{1}=2l_{12}(l_{11}\sin \theta _{1}-y_{s1}), m_{2}=-2l_{12}(a+l_{10}/2-l_{11}\cos \theta _{1}-z_{s1}), m_{3}=l_{17}^{2}-l_{18}^{2}-{l_{12}}^{2}$ .

According to the link length equation by constraint $A_{13}A_{22}=l_{14}$ , it is easy to know

(24) \begin{equation} \theta _{2}=2\arctan \left(\frac{q_{1}+Q\sqrt{q_{1}^{2}+q_{2}^{2}-q_{3}^{2}}}{q_{2}+q_{3}}\right)\left(Q=\pm 1\right) \end{equation}

Where: $q_{1}=-2l_{13}(l_{11}\sin \theta _{1}+l_{12}\sin \alpha _{1}), q_{2}=2l_{13}(-l_{10}+l_{11}\cos \theta _{1}+l_{12}\cos \alpha _{1})$ ,

$q_{3}=l_{14}^{2}-l_{13}^{2}-(l_{11}\sin \theta _{1}+l_{12}\sin \alpha _{1})^{2}-(-l_{10}+l_{11}\cos \theta _{1}+l_{12}\cos \alpha _{1})^{2}$ .

Because the hybrid branched chains 2 and 3 in PM3 are the same as PM2, the inverse solution analysis of their positions is the same. For details, see Eqs. (15) to (18).

3.3.1. Forward position analysis of the PM4

For forward solution analysis of hybrid branch chain 1, the line A₁₃S₁ is perpendicular to the line A₂₂S₁ in hybrid branch chain 1, and the angle between the line A₁₃S₁ and A₁₃A₂₂ is $\gamma _{1}=30^{\circ }$ .

Also assign the angle between the link A₁₃A₂₂ and the negative direction of Z-axis as $\alpha _{1}$ , which is taken as the intermediate variable. It’s easy to know the coordinates of the following points.

$A_{11}=(a,0,a), A_{21}=(a,0,a-l_{10}), A_{12}=(a,0,a+l_{10}), A_{13}=(a+l_{11}\sin \theta _{2}\cos \theta _{1},l_{1}\sin \theta _{2}\sin \theta _{1},a+l_{10}-l_{11}\cos \theta _{2})$ ,

$A_{22}=(a+(l_{11}\sin \theta _{2}+l_{15}\sin \alpha _{1})\cos \theta _{1},(l_{11}\sin \theta _{2}+l_{15}\sin \alpha _{1})\sin \theta _{1},a+l_{10}-l_{11}\cos \theta _{2}-l_{15}\cos \alpha _{1})$ .

According to the link length equation by constraint $A_{21}A_{22}=l_{3}$ , the intermediate variable $\alpha _{1}$ is given by

(25) \begin{equation} \alpha _{1}=2\arctan \left(\frac{N_{1}+n\sqrt{N_{1}^{2}+N_{2}^{2}-N_{3}^{2}}}{N_{2}+N_{3}}\right)\left(n=\pm 1\right) \end{equation}

Where: .

Then we have

\begin{align*} S_{1} & =\left(a+\left(l_{11}\sin \theta _{2}+l_{12}\sin \left(\gamma _{1}+\alpha _{1}\right)\right)\cos \theta _{1},\left(l_{11}\sin \theta _{2}+l_{12}\sin \left(\gamma _{1}+\alpha _{1}\right)\right)\sin \theta _{1},a+l_{10}-l_{11}\cos \theta _{2}\right. \left.-l_{12}\cos \left(\gamma _{1}+\alpha _{1}\right)\right) \end{align*}

The hybrid branch chains 2 and 3 in the PM4 are the same as the PM1. Therefore, their forward position solution analysis is the same. For details, please refer to Eqs. (10)-(13) in Ref. [Reference Du, Li, Liu, Zhao and Shen29].

Among them, the angle $\varphi _{2}$ and $\varphi _{3}$ set in the hybrid branch chain 2 and 3 can be obtained successively by the link length constraint $S_{1}S_{2}=\sqrt{3}b$ and $S_{1}S_{3}=\sqrt{3}b$ , respectively

(26) \begin{equation} \varphi _{2}=2\arctan \left(\frac{H_{1}+h\sqrt{H_{1}^{2}+H_{2}^{2}-H_{3}^{2}}}{H_{2}+H_{3}}\right)\left(h=\pm 1\right) \end{equation}

Where, $H_{1}=2l_{6}(y_{s1}-y_{A34}), H_{2}=-2l_{6}(x_{s1}-x_{A34})\sin \alpha _{2}-2l_{6}(z_{s1}-z_{A34})\cos \alpha _{2}$ , $H_{3}=3b^{2}-l_{6}^{2}-(x_{s1}-x_{A34})^{2}-(y_{s1}-y_{A34})^{2}-(z_{s1}-z_{A34})^{2}$ .

(27) \begin{equation} \varphi _{3}=2\arctan \left(\frac{G_{1}+g\sqrt{G_{1}^{2}+G_{2}^{2}-G_{3}^{2}}}{G_{2}+G_{3}}\right)\left(g=\pm 1\right) \end{equation}

Where, $G_{1}=2l_{6}(z_{s1}-z_{A54}), G_{2}=-2l_{6}(x_{s1}-x_{A54})\cos \alpha _{3}-2l_{6}(y_{s1}-y_{A54})\sin \alpha _{3}$ , $G_{3}=3b^{2}-l_{6}^{2}-(x_{s1}-x_{A54})^{2}-(y_{s1}-y_{A54})^{2}-(z_{s1}-z_{A54})^{2}$ .

By substituting the coordinate values of S₁, S₂, and S₃ into Eq. (12), the posture parameters of the moving platform of the PM4 can be solved.

Through the position analysis of the PM4, it can be easily known that the coordinate values of point S₁ on its moving platform can be obtained by solving the hybrid chain 1 to which it belongs. Therefore, when taking S₁ as the base point, PM4 has the characteristic of partial motion decoupling.

3.3.2 Analysis of inverse position solution of the PM4

For analysis of the inverse solution of the position of the hybrid branch chain 1.

From the forward solution analysis, it is easy to know that S₁ coordinates are given below.

(28) \begin{equation} \left\{\begin{array}{l} x_{s1}=a+\left(l_{11}\sin \theta _{2}+l_{12}\sin \left(\gamma _{1}+\alpha _{1}\right)\right)\cos \theta _{1}\\[4pt] y_{s1}=\left(l_{11}\sin \theta _{2}+l_{12}\sin \left(\gamma _{1}+\alpha _{1}\right)\right)\sin \theta _{1}\\[4pt] z_{s1}=a+l_{10}-l_{11}\cos \theta _{2}-l_{12}\cos \left(\gamma _{1}+\alpha _{1}\right) \end{array}\right. \end{equation}

From Eq. (36), we have

(29) \begin{equation} \theta _{1}=\arctan \left(\frac{y_{s1}}{x_{s1}-a}\right) \end{equation}

(30) \begin{equation} \theta _{2}=2\arctan \left(\frac{f_{1}+F\sqrt{f_{1}^{2}+f_{2}^{2}-f_{3}^{2}}}{f_{2}+f_{3}}\right)\left(F=\pm 1\right) \end{equation}

where: $f_{1}=\frac{-2l_{11}y_{s1}}{\sin \theta _{1}}, f_{2}=2l_{11}\left(z_{s1}-a-l_{10}\right), f_{3}=l_{12}^{2}-l_{11}^{2}-\left(y_{s1}/\sin \theta _{1}\right)^{2}-\left(z_{s1}-a+l_{10}\right)^{2}$ .

The hybrid branch chains 2 and 3 in the PM4 are the same as the PM3. Therefore, their inverse position solution analysis is the same. See Eqs. (19)–(24) for details in Ref. [Reference Du, Li, Liu, Zhao and Shen29].

4.1.1 Forward position solution

Let the dimensional parameters of the PM2 be (unit: mm):

a = 300, l ₀ = 100, l ₁=l ₄ = 120, $l_{2}=120\sqrt{3}$ , l ₃ = 220, l ₅ = 240, $h=60\sqrt{3}$ .

According to the dimensional parameters, a three-dimensional (3-D) computer-aided design (CAD) model of the PM2 was established, and from which the initial input angles (unit: rad) of the six actuated joints $R_{i}(i=1\sim 6)$ , namely $\theta _{1}=1.2769, \theta _{2}=1.6087, \theta _{3}=1.4099, \theta _{4}=1.4488, \theta _{5}=1.6315, \theta _{6}=0.8430$ as well as the corresponding output values of the moving platform center point $O^{\prime}$ , namely $x=303.77, y=276.80, z=283.94, \alpha =0, \beta =0, \gamma =0$ , are measured. The 3-D CAD model of PM2 is shown in Fig. 6.

Table II.

Forward position solution for parallel mechanism 2.

Figure 6.

Three-dimensional computer-aided design model of parallel mechanism 2.

By substituting the initial input angles $\theta _{i}(i=1\sim 6)$ (unit: rad) of the six actuated joints obtained from the measured 3-D CAD model into the coordinates of each point and solving for $\varphi _{2}$ and $\varphi _{3}$ , the coordinate values of S₁, S₂, and S₃ can be determined. Through the forward equation Eq. (12), the output value of the moving platform center O’ point can be obtained. The PM2 has eight groups of forward solutions, four of which are real solutions, as shown in Table II.

The relative error values of the calculated theoretical values compared to the measured values of the positional parameters from the fourth group of forward position solutions, as given by Eq. (6) with n = −1, Eq. (8) with m = −1, and Eq. (10) with q = −1, are all within 0.1%. Therefore, the derivation of the symbolic forward position solution formulas is correct. Here, we calculate the relative error by taking the difference between the theoretical calculated values and the measured values and then dividing it by the measured values. The measured values can be obtained by using the ruler tool in SolidWorks software to directly measure the virtual simulation model.

4.1.2 The inverse solution example of the PM2

The output values of the center of the moving platform point $O^{\prime}$ measured in the 3-D model of the PM2 are $x=303.77, y=276.80, z=283.94, \alpha =0, \beta =0, \gamma =0$ , respectively, and which are substituted into the inverse solution Eqs. (13)–(18) to obtain the inverse solution input values corresponding to each branch chain, as shown in Table III.

According to Table III, the relative error values of inverse solution and 3-D model measurement in each hybrid branch chain are within 0.01%, indicating that the derivation of the position inverse solution formula is correct.

Table III.

Inverse position solution of the parallel mechanism 2.

Table IV.

Forward position solution for parallel mechanism 3.

4.2.1 Example of forward position solution for the PM3

Let the dimension parameter of hybrid branch chain 1 in PM3 be (unit: mm): $a=300, l_{10}=130, l_{11}=l_{13}=100, l_{12}=l_{14}=120, l_{15}=150, l_{16}=110$ . The size parameters of other members are the same as PM2.

According to these dimensional parameters, a 3-D model of the PM3 is established, and from which six initial input angles (unit: rad) of the actuated joints $R_{i}(i=1\sim 6)$ are measured, namely $\theta _{1}=2.6060, \theta _{2}=1.5273, \theta _{3}=1.4146, \theta _{4}=1.4354, \theta _{5}=1.6321, \theta _{6}=0.8126$ . The corresponding output values of the moving platform center O’ are $x=302.14, y=275.41, z=283.92, \alpha =0, \beta =0, \gamma =0$ , respectively.

By inputting the initial input angles $\theta _{i}(i=1\sim 6)$ (unit: rad) of the six actuated joints measured from the 3-D model into the coordinates of each point, the coordinate values of S₁, S₂, and S₃ can be easily obtained by solving $\alpha _{i}(i=1,2,3)$ and $\varphi _{1}$ . Using the position forward kinematics Eq. (12), the output value of the center O’ of the moving platform can be obtained. There are 16 groups of forward equations for the PM3, but only four groups of which have real solutions, as shown in Table IV.

The third group of position-real solutions corresponds to the values of the position parameter measured by the 3-D model, and the relative error value is less than 0.1%, which shows that the symbolic positional forward solution formulas are correctly derived.

Table V.

Inverse position solution of the parallel mechanism 3.

4.2.2 Example of inverse position solution of the PM3

The output values of the center of the moving platform point O’ measured in the 3-D model of the PM3 are $x=302.14, y=275.41, z=283.92, \alpha =0, \beta =0, \gamma =0$ , respectively, and are substituted into the inverse solution formula in Section 3.2.2. Inverse solution input values corresponding to each branch chain are obtained, as shown in Table V.

According to Table V, we found that the relative errors of the theoretical calculated values of inverse solution marker (^*) and the measured values of the 3D model are both within 0.01%, so the symbolic position inverse solution formula is correctly derived.

4.3.1 Example of forward position solution for the PM4

The parameters of the hybrid branch chains 2 and 3 in the PM4 are set to be the same as those in the PM1, while the parameters of the hybrid branch chain 1 are specified as follows (unit: mm):

$l_{10}=100, l_{11}=l_{14}=120, l_{12}=120\sqrt{3}, l_{13}=220, l_{15}=240$ .

According to the corresponding dimensional parameters, a 3-D CAD model of the PM4 is established, and the initial input angles (unit: rad) of the six actuated joints $R_{i}(i=1\sim 6)$ are measured from it, namely $\theta _{1}=1.3626, \theta _{2}=1.2748, \theta _{3}=2.6182, \theta _{4}=1.7171, \theta _{5}=2.3880$ , and $\theta _{6}=1.1126$ . The respective output values of the moving platform center O’ point are $x=278.61, y=277.31, z=243.53, \alpha =0, \beta =0,$ $\gamma =0$ .

According to the verification process of the PM1, the output value of the center point O’ of the moving platform is obtained through the position forward kinematics Eq. (12). There are 32 groups of position forward kinematics equations for the PM4, but only 2 groups of which have real solutions, as shown in Table VI.

The second group of positional real forward solutions corresponds to n = −1 in Eq. (25), h = −1 in Eq. (26), g = −1 in Eq. (27), and the positional parameter values measured by the 3-D model, and their relative error values are all within 0.1%. Therefore, the symbolic positional forward solution formula is derived correctly.

4.3.2 Example of inverse position solution of the PM4

The measured output value of the center O’ point of the moving platform in the 3-D model of the PM4 is substituted into the inverse solution Eqs. (28)–(30) in this paper and Eqs. (19)–(24) in Ref. [Reference Du, Li, Liu, Zhao and Shen29] to obtain the inverse input value corresponding to each branch chain, as shown in Table VII.

As can be seen from Table VII, the relative error values of the theoretical values of the PM4 and the measured values of the 3-D model are within 0.01%, which shows the symbolic position inverse solution formula is correctly derived.

Table VI.

Forward position solution for the parallel mechanism 4.

Table VII.

Inverse position solution of the parallel mechanism 4.