2.1. Geometry and structure

The proposed hybrid manipulator, as shown in Fig. 2(a), consists of two 3-DOF parallel mechanisms connected serially, making it an overall 6-DOF hybrid manipulator. All the prismatic pairs ( $\mathrm{P_2,P_3,P_4,P_5,P_6}$ ) and the $R_{2}$ revolute joint are actuated (active). The two platforms, M and H, are taken to be equilateral, with circumradii of $h_1$ and $h_2$ ( $\neq h_1$ )Footnote ¹ , respectively. The lower parallel mechanism resembles the upper parallel mechanism used by Tanev [Reference Tanev20] in his hybrid manipulator. The lower parallel mechanism connects the base to the mid-platform M and comprises one RRR and two SPS legs. Here, axis( $R_{1}$ )Footnote ² $\perp$ axis( $R_{2}$ ) $\perp$ axis( $R_{3}$ ) and also axis( $R_{1}$ ) $\perp$ $\overline{\text{$B_{1}$$B_{2}$}}$ edge of the lower triangular base. Further, axis( $R_{3}$ ) $\parallel$ $\overline{\text{$M_{1}$$M_{2}$}}$ edge of the mid-platform. The upper parallel mechanism between the mid-platform M and the upper platform H (end-effector) is the standard 3-UPU TPM (also called the Tsai [Reference Tsai, Lenarčič and Parenti-Castelli2] manipulator) having three identical UPU legs. Tsai [Reference Tsai, Lenarčič and Parenti-Castelli2] first proposed the conditions required for the 3-UPU manipulator to perform pure translation, and later Gregorio et al. [Reference Di Gregorio, Parenti-Castelli, Lenarčič and Husty1] derived the general TPM conditions using the proximal (modified) D-H convention [Reference Yoshikawa33], starting from the generic RRPRRFootnote ³ leg. This paper uses the distal (classical) D-H convention [Reference Saha34]. The distal form of the TPM conditions [Reference Di Gregorio, Parenti-Castelli, Lenarčič and Husty1, Reference Tsai, Lenarčič and Parenti-Castelli2] are imposed on the upper parallel mechanism in the following manner:

• $U_{1,1}$ $\parallel$ $U_{4,1}$ , $U_{1,2}$ $\parallel$ $U_{4,2}$ , $U_{2,1}$ $\parallel$ $U_{5,1}$ , $U_{2,2}$ $\parallel$ $U_{5,2}$ , and $U_{3,1}$ $\parallel$ $U_{6,1}$ , $U_{3,2}$ $\parallel$ $U_{6,2}$ , where $U_{i,j}$ denotes the $\textit{j}\text{th}$ revolute joint axis (while traversing a limb upward, i.e., from bottom to the top) of the $\textit{i}\text{th}$ universal joint

• For each of the three UPU legs, $\theta _3 = -\theta _2$ and $\theta _4 = -\theta _1$ , where $\theta _i$ is the rotation about the $\textit{i}\text{th}$ revolute joint of that particular limb while traversing upward

• $U_{1,2}$ , $U_{2,2}$ , and $U_{3,2}$ point toward the circumcenter of platform M

• $U_{4,1}$ , $U_{5,1}$ , and $U_{6,1}$ point toward the circumcenter of platform H

• $U_{1,2}$ and $U_{4,1}$ are $\perp$ axis( $P_{4}$ ), $U_{2,2}$ and $U_{5,1}$ are $\perp$ axis( $P_{5}$ ) leg, and $U_{3,2}$ and $U_{6,1}$ are $\perp$ axis( $P_{6}$ ) leg

The TPM conditions ensure that platform H does pure translation with respect to platform M. Thus, this structure of the hybrid manipulator allows the upper mechanism to perform pure translation motion while being in any desired orientation defined by the mid-platform M, with respect to the base. Furthermore, since the lower parallel mechanism can also translate, platform H can perform pure translation motion even in an oblique plane having some offset with respect to the base.

Figure 2.

Proposed hybrid manipulator: (a) schematic (the centers of upper joints of the lower mechanism (i.e., $R_{3}$ , $S_{4}$ , and $S_{3}$ ) coincide, respectively, with the centers of the lower joints of the upper mechanism (i.e., $U_{1}$ , $U_{2}$ , and $U_{3}$ ). However, they have been depicted as separate for clarity), and (b) $\mathrm{B_1M_1H_1}$ limb with D-H frames.

It is important to note that the choice of the lower mechanism as a 1-RRR 2-SPS mechanism provides the benefit of a straightforward and convenient interpretation for the Euler angles of the mid-platform M in terms of $R_{1}$ , $R_{2}$ , and $R_{3}$ revolute joint angles. From the point of view of controlling the manipulator, this can be pretty useful as encoder feedback from the three revolute joints would directly give the orientation. Furthermore, choosing the upper mechanism as a translational 3-UPU mechanism decouples the orientation and translation of the top platform H, as now the orientation part is entirely governed by the lower mechanism actuators. This helps it achieve a fully closed-form solution structure. In addition to the various applications mentioned in Section 1, this proposed hybrid manipulator offers much flexibility as it can be converted to a standard 3-UPU manipulator just by fixing the lower mechanism, further diversifying its use cases. Finally, as seen in Section 7, even a basic optimization leads to a pretty good design obtaining relatively high manipulability for large parts of the motion range.

2.2. D-H analysis

The $\mathrm{B_1M_1H_1}$ limb of the hybrid manipulator with all the attached D-H coordinate frames is shown in Fig. 2(b). The frame 1 ( $X_1Y_1Z_1$ ) is attached to the fixed base (link #0), while the frame 8 is attached to the top moving platform H (link #7). Two separate frames $4_l$ and $4_u$ fixed to the middle platform M (link #3) are used for convenience in the analysis, which will be discussed in subsequent sections. Since the axes of the two revolute joints in a universal joint intersect, the lengths of links #4 and #6 are taken to be zero as they are the imaginary links connecting the two intersecting revolute joints of a universal joint. Let $L_1 = \mathrm{B_1M_1},\ L_2 = \mathrm{B_2M_2},\ L_3 = \mathrm{B_3M_3},\ L_4 = \mathrm{M_1H_1},\ L_5 = \mathrm{M_2H_2}, \text{ and } L_6 = \mathrm{M_3H_3}$ .Footnote ⁴

Now, Table I represents the D-H table for the $\mathrm{B_1M_1}$ leg of the lower mechanism, and it transforms frame $4_l$ to frame 1. Table II represents the D-H table for the $\mathrm{M_1H_1}$ leg of the upper mechanism after imposing the TPM conditions as listed in Section 2.1, and it transforms frame 8 to frame $4_u$ . The intermediate (constant) rigid transformation matrix, which transforms frame $4_u$ to $4_l$ is

(1) \begin{align} \textbf{C}_u^l = \left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} \dfrac{1}{2} & 0 & -\dfrac{\sqrt{3}}{2} & 0 \\[5pt] 0 & 1 & 0 & 0 \\[5pt] \dfrac{\sqrt{3}}{2} & 0 & \dfrac{1}{2} & 0 \\[5pt] 0 & 0 & 0 & 1 \end{array}\right] \end{align}

Now, $\textbf{T}_{\textbf{1}} = f(\theta _1, \theta _2, \theta _3)$ corresponds to the D-H Table I, while $\textbf{T}_{\textbf{2}} = f(\theta _4, \theta _5, L_4)$ corresponds to the D-H Table II. Therefore, the overall transformation matrix is written using the D-H Tables I and II as (details in Section A.1):

(2) \begin{align} \textbf{T} = \textbf{T}_{\textbf{1}} \textbf{C}_u^l \textbf{T}_{\textbf{2}} = \left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} t_{11} & t_{12} & t_{13} & t_{14} \\[5pt] t_{21} & t_{22} & t_{23} & t_{24} \\[5pt] t_{31} & t_{32} & t_{33} & t_{34} \\[5pt] 0 & 0 & 0 & 1 \end{array}\right] \end{align}

Now, any position vector of a point expressedFootnote ⁵ in frame 8 can be transformed to the base frame using $\textbf{T}$ from Eq. (2) as:

(3) \begin{align}{}^{1}{\textbf{p}} = \textbf{T} \,{}^{8}{\textbf{p}} = \textbf{T}_{\textbf{1}} \textbf{C}_u^l \textbf{T}_{\textbf{2}}{}^{8}{\textbf{p}} \end{align}

It is to be noticed that the orientation partFootnote ⁶ of $\textbf{T}$ depends only on $\theta _1, \theta _2, \text{and } \theta _3$ , owing to the pure translational nature of the upper mechanism, which in turn facilitates the decoupling as shown later in Section 4.

Table I.

D-H table for lower mechanism (along $\mathrm{B_1M_1}$ limb) of the proposed hybrid manipulator.

Table II.

D-H table for upper mechanism (along $\mathrm{M_1H_1}$ limb) of the proposed hybrid manipulator.

3.1. Lower parallel (1-RRR 2-SPS) mechanism

The values of $\theta _2, L_2, L_3$ are given, and the values of the passive joint variables $\theta _1$ and $\theta _3$ are required. Now, from the geometry of the manipulator, the following vectorsFootnote ⁸ can be written:

(4) \begin{align}{}^{1}{\textbf{O}_{\textbf{1}} \textbf{B}_{\textbf{2}}} &={\left [ b_2,0,0,1 \right ]}^{\textsf{T}} \end{align}

(5) \begin{align}{}^{1}{\textbf{O}_{\textbf{1}} \textbf{B}_{\textbf{3}}} &={\left [ b_{3_x},0,b_{3_z},1 \right ]}^{\textsf{T}} \end{align}

(6) \begin{align}{}^{4_l}_{} {\textbf{O}_{\textbf{4}} \textbf{M}_{\textbf{2}}} &={\left [ 0,0,\sqrt{3}h_1,1 \right ]}^{\textsf{T}} \end{align}

(7) \begin{align}{}^{4_l}_{}{\textbf{O}_{\textbf{4}} \textbf{M}_{\textbf{3}}} &={\bigg[\sqrt{3}h_1 \sin{\dfrac{\pi }{3}},0,\sqrt{3}h_1 \cos{\dfrac{\pi }{3}},1 \bigg]}^{\textsf{T}} ={\left [ \dfrac{3h_1}{2},0,\dfrac{\sqrt{3}h_1}{2},1 \right ]}^{\textsf{T}} \end{align}

Now the vectors in Eqs. (6) and (7) are transformed to frame 1 as:

(8) \begin{align}{}^{1}_{}{\textbf{O}_{\textbf{1}}\textbf{M}_{\textbf{2}}} &= \textbf{T}_{\textbf{1}}{}^{4_l}_{}{\textbf{O}_{\textbf{4}}\textbf{M}_{\textbf{2}}} \end{align}

(9) \begin{align}{}^{1}_{}{\textbf{O}_{\textbf{1}}\textbf{M}_{\textbf{3}}} &= \textbf{T}_{\textbf{1}}{}^{4_l}{\textbf{O}_{\textbf{4}}\textbf{M}_{\textbf{3}}} \end{align}

The leg vectors ${}^{1}_{}{\textbf{L}_{\textbf{2}}}$ and ${}^{1}_{}{\textbf{L}_{\textbf{3}}}$ are written as:

(10) \begin{align}{}^{1}_{}{\textbf{L}_{\textbf{2}}} &={}^{1}_{}{\textbf{O}_{\textbf{1}}\textbf{M}_{\textbf{2}}} -{}^{1}_{}{\textbf{O}_{\textbf{1}}\textbf{B}_{\textbf{2}}} \end{align}

(11) \begin{align}{}^{1}_{}{\textbf{L}_{\textbf{3}}} &={}^{1}_{}{\textbf{O}_{\textbf{1}}\textbf{M}_{\textbf{3}}} -{}^{1}_{}{\textbf{O}_{\textbf{1}}\textbf{B}_{\textbf{3}}} \end{align}

Using the given leg lengths:

(12) \begin{align}{}^{1}_{}{\textbf{L}_{\textbf{2}}} \cdot{}^{1}_{}{\textbf{L}_{\textbf{2}}} - L_2^2 &= 0 \end{align}

(13) \begin{align}{}^{1}_{}{\textbf{L}_{\textbf{3}}} \cdot{}^{1}_{}{\textbf{L}_{\textbf{3}}} - L_3^2 &= 0 \end{align}

Using Eqs. (4)–(11), the above two equations are written in the following form:

(14) \begin{align} p_1c_1 + p_2s_1 + p_3 = 0 \end{align}

(15) \begin{align} q_1c_1 + q_2s_1 + q_3c_3 + q_4s_3 + q_5c_1c_3 + q_6s_1s_3 + q_7 = 0 \end{align}

where,

(16) \begin{align} \begin{gathered} p_1 = 2 \sqrt{3} b_2h_1s_2; \,\,\,\,\,\,\, p_2 = 2 L_1 b_2; \,\,\,\,\,\,\, p_3 = L_1^2 - L_2^2 + b_2^2 + 3 h_1^2 \\[5pt] q_1 = \sqrt{3} h_1 b_{3_x} s_2; \,\,\,\,\,\,\, q_2 = 2 L_1 b_{3_x}; \,\,\,\,\,\,\, q_3 = 3 h_1 b_{3_z}; \,\,\,\,\,\,\, q_4 = -3 h_1 L_1; \,\,\,\,\,\,\, q_5 = -3 h_1 b_{3_x} c_2 \\[5pt] q_6 = -3 h_1 b_{3_x}; \,\,\,\,\,\,\, q_7 = \sqrt{3} h_1 b_{3_z} c_2 + L_1^2 - L_3^2 + b_{3_x}^2 + b_{3_z}^2 +3 h_1^2 \end{gathered} \end{align}

Since Eq. (14) is only a function of $\theta _1$ , it can be solved by using half-angle substitution to get two solutions (corresponding to $\pm$ ) for $\theta _1$ as:

(17) \begin{align} \theta _{{1}_{1,2}} = 2 \,{\tan}^{-1}{\left ({\dfrac{p_2 \pm \sqrt{p_1^2+p_2^2-p_3^2}}{p_1-p_3}} \right )} \end{align}

Here, ${\tan}^{-1}$ gives a solution in the range $\left (-\dfrac{\pi }{2}, \dfrac{\pi }{2}\right )$ . Substituting the obtained value of $\theta _1$ in Eq. (15) gives the following equation:

(18) \begin{align} k_1c_3 + k_2s_3 + k_3 = 0 \end{align}

where,

(19) \begin{align} \begin{gathered} k_1 = q_5c_1 + q_3; \,\,\,\,\,\,\, k_2 = q_6s_1 + q_4; \,\,\,\,\,\,\, k_3 = q_1c_1 + q_2s_1 + q_7 \end{gathered} \end{align}

The solution for $\theta _3$ using Eq. (18) is given as:

(20) \begin{align} \theta _{3_{1,2}} = 2 \,{\tan}^{-1} \left ({\dfrac{k_2 \pm \sqrt{k_1^2+k_2^2-k_3^2}}{k_1-k_3}} \right ) \end{align}

In total, $2 \times 2 = 4$ solution sets of $\theta _1$ and $\theta _3$ are obtained. For each solution set, the transformation matrix $\textbf{T}_{\textbf{1}}$ can be fully determined, which completes the forward kinematics of the lower parallel mechanism.

3.2. Upper parallel 3-UPU TPM mechanism

For analyzing the 3-UPU mechanism, a similar procedure is adopted as done for the lower parallel mechanism. Here, the values of $L_4, L_5, L_6$ are given, and the values of the passive joint variables $\theta _4$ and $\theta _5$ are determined. The components of the relevant vectors obtained from the geometry of the manipulator are given below:

(21) \begin{align}{}^{4_u}_{}{\textbf{O}_{\textbf{4}}\textbf{M}_{\textbf{2}}} &={\bigg[ \sqrt{3}h_1 \cos{\dfrac{\pi }{6}},0,\sqrt{3}h_1 \sin{\dfrac{\pi }{6}},1 \bigg ]}^{\textsf{T}} ={\left [ \dfrac{3h_1}{2},0,\dfrac{\sqrt{3}h_1}{2},1 \right ]}^{\textsf{T}} \end{align}

(22) \begin{align}{}^{4_u}_{}{\textbf{O}_{\textbf{4}}\textbf{M}_{\textbf{3}}} &={\bigg [ \sqrt{3}h_1 \cos{\dfrac{\pi }{6}},0,-\sqrt{3}h_1 \sin{\dfrac{\pi }{6}},1 \bigg ]}^{\textsf{T}} ={\left [ \dfrac{3h_1}{2},0,-\dfrac{\sqrt{3}h_1}{2},1 \right ]}^{\textsf{T}} \end{align}

(23) \begin{align}{}^{8}_{}{\textbf{O}_{\textbf{8}}\textbf{H}_{\textbf{2}}} &={\bigg[ \sqrt{3}h_2 \cos{\dfrac{\pi }{6}},0,\sqrt{3}h_2 \sin{\dfrac{\pi }{6}},1 \bigg ]}^{\textsf{T}} ={\left [ \dfrac{3h_2}{2},0,\dfrac{\sqrt{3}h_2}{2},1 \right ]}^{\textsf{T}} \end{align}

(24) \begin{align}{}^{8}_{}{\textbf{O}_{\textbf{8}}\textbf{H}_{\textbf{3}}} &={\bigg[ \sqrt{3}h_2 \cos{\dfrac{\pi }{6}},0,-\sqrt{3}h_2 \sin{\dfrac{\pi }{6}},1 \bigg]}^{\textsf{T}} ={\left [ \dfrac{3h_2}{2},0,-\dfrac{\sqrt{3}h_2}{2},1 \right ]}^{\textsf{T}} \end{align}

Now, the vectors in Eqs. (23) and (24) can be transformed to frame $4_u$ as:

(25) \begin{align}{}^{4_u}_{}{\textbf{O}_{\textbf{4}}\textbf{H}_{\textbf{2}}} &= \textbf{T}_{\textbf{2}}{}^{8}_{}{\textbf{O}_{\textbf{8}}\textbf{H}_{\textbf{2}}} \end{align}

(26) \begin{align}{}^{4_u}_{}{\textbf{O}_{\textbf{4}}\textbf{H}_{\textbf{3}}} &= \textbf{T}_{\textbf{2}}{}^{8}_{}{\textbf{O}_{\textbf{8}}\textbf{H}_{\textbf{3}}} \end{align}

The leg vectors ${}^{4_u}_{}{\textbf{L}_{\textbf{5}}}$ and ${}^{4_u}_{}{\textbf{L}_{\textbf{6}}}$ can be written as:

(27) \begin{align}{}^{4_u}_{}{\textbf{L}_{\textbf{5}}} &={}^{4_u}_{}{\textbf{O}_{\textbf{4}}\textbf{H}_{\textbf{2}}} -{}^{4_u}_{}{\textbf{O}_{\textbf{4}}\textbf{M}_{\textbf{2}}} \end{align}

(28) \begin{align}{}^{4_u}_{}{\textbf{L}_{\textbf{6}}} &={}^{4_u}_{}{\textbf{O}_{\textbf{4}}\textbf{H}_{\textbf{3}}} -{}^{4_u}_{}{\textbf{O}_{\textbf{4}}\textbf{M}_{\textbf{3}}} \end{align}

The leg length constraint equations are:

(29) \begin{align}{}^{4_u}_{}{\textbf{L}_{\textbf{5}}} \cdot{}^{4_u}_{}{\textbf{L}_{\textbf{5}}} - L_5^2 &= 0 \end{align}

(30) \begin{align}{}^{4_u}_{}{\textbf{L}_{\textbf{6}}} \cdot{}^{4_u}_{}{\textbf{L}_{\textbf{6}}} - L_6^2 &= 0 \end{align}

Using Eqs. (21)–(28) in Eqs. (29) and (30), then adding Eqs. (29) and (30), and subtracting Eqs. (30) from (29), yields the following two equations:

(31) \begin{align} f_1s_5 + f_2 = 0 \end{align}

(32) \begin{align} g_1c_4c_5 + g_2 = 0 \end{align}

where

(33) \begin{align} \begin{gathered} f_1 = -2 \sqrt{3} L_4 (h_1-h_2); \,\,\,\,\,\,\, f_2 = L_6^2 - L_5^2 \\[5pt] g_1 = -6 L_4 (h_1-h_2); \,\,\,\,\,\,\, g_2 = 2 L_4^2 - L_5^2 - L_6^2 + 6 (h_1-h_2)^2 \end{gathered} \end{align}

Since Eq. (31) is only a function of $\theta _5$ , the two solutions are written as:

(34) \begin{align} \theta _{5_{1,2}} ={\sin}^{-1} \left ({-\dfrac{f_2}{f_1}}\right ) \end{align}

Now substituting the value of $\theta _5$ obtained in Eq. (32), the following equation is obtained:

(35) \begin{align} w_1c_4 + w_2 = 0 \end{align}

where,

(36) \begin{align} \begin{gathered} w_1 = g_1c_5; \,\,\,\,\,\,\, w_2 = g_2 \end{gathered} \end{align}

The two solutions for $\theta _4$ from the above equation are written as:

(37) \begin{align} \theta _{4_{1,2}} ={\cos}^{-1} \left ({-\dfrac{w_2}{w_1}}\right ) \end{align}

In total, $2 \times 2 = 4$ solution sets of $\theta _4$ and $\theta _5$ are obtained. For each solution set, the transformation matrix $\textbf{T}_{\textbf{2}}$ can be fully determined, which completes the forward kinematics of the upper parallel mechanism. However, these four distinct joint space solution sets for the upper parallel mechanism result in only two distinct poses, that is, only two distinct task space configurations. The reason for this lies in the interdependence of the four joint space solution sets, that is, say the first solution set is $\left \{ \theta _4 = \beta _4, \, \theta _5 = \beta _5 \right \}$ , then the other three solution sets are given as $\left \{ (\beta _4 - \pi ),\, (\pi -\beta _5) \right \}$ , $\left \{ -\beta _4, \, \beta _5 \right \}$ , $\left \{ (\pi - \beta _4),\, (\pi -\beta _5) \right \}$ . So rotating about the two revolute axes of the universal joint $U_{1}$ following the first solution set values for $\theta _4$ and $\theta _5$ , and then again following the second solution set values leads to the same final configuration.

Similarly, the third and the fourth solution sets lead to the same final configuration, but it is different from the one attained by the first two solution sets. Therefore, only two distinct task space configurations are obtained, as would be numerically illustrated later in Section 5.2.1 as well. This fact can also be directly deduced easily by substituting the solution sets into the symbolic form of $\textbf{T}_{\mathrm{2}}$ and observing that there are indeed only two distinct $\textbf{T}_{\mathrm{2}}$ matrices, and hence only two distinct poses. Moreover, the two obtained task space configurations are related to each other – if the upper mechanism of one of the configurations is reflected in the mid-plane, the other configuration is obtained. In other words, both task space configurations are just the reflection of the upper mechanism in the mid-plane, with the lower mechanism configuration being the same.

3.3. Determining the overall transformation matrix

Overall, $4 \times 4 = 16$ distinct solution sets of $\left ( \theta _1, \theta _3, \theta _4, \theta _5 \right )$ quadruplet of passive variables are obtained that correspond to a total of 16 possible direct kinematic solutions in joint space for the complete hybrid manipulator. However, for the above-mentioned reason in Section 3.2, there are only $4 \times 2 = 8$ direct kinematic solutions in task space.

Having determined all the passive variables and consequently evaluating the matrices $\textbf{T}_{\mathrm{1}}$ and $\textbf{T}_{\mathrm{2}}$ , the overall transformation matrix $\textbf{T}_{\mathrm{2}}$ can be determined using Eq. (2). Further, the position vector of vertex $\mathrm{H_1}$ , as well as the orientation of the upper platform, can be extracted from it, which can be used to get the position vector of any other point (including the center) of the upper platform. Therefore, this completes the forward kinematic analysis of the proposed hybrid manipulator.

4.1. Lower parallel (1-RRR 2-SPS) mechanism

Let the orientation part of $\textbf{T}_{\mathrm{num}}$ be:

(38) \begin{align} \textbf{Q}_{\mathrm{num}} = \left[\begin{array}{c@{\quad}c@{\quad}c} q_{11} & q_{12} & q_{13} \\[5pt] q_{21} & q_{22} & q_{23} \\[5pt] q_{31} & q_{32} & q_{33} \end{array}\right] \end{align}

From the components of $\textbf{T}$ described in symbolic form in Section 2.2, the components of the orientation part are extracted to get the orientation matrix $\textbf{Q}$ in symbolic form, which can then be compared with the given numerical orientation matrix $\textbf{Q}_{\mathrm{num}}$ to solve for the active and the passive variables of the lower mechanism. So using the fact that $\textbf{Q} \equiv \textbf{Q}_{\mathrm{num}}$ , and comparing their (3,1), (3,2), and (3,3) elements, the following three equations are obtained:

(39) \begin{align} q_{31} &= -\dfrac{s_2c_3}{2} -\dfrac{\sqrt{3} \, c_2}{2} \end{align}

(40) \begin{align} q_{32} &= s_2s_3 \end{align}

(41) \begin{align} q_{33} &= \dfrac{\sqrt{3} \, s_2c_3}{2} -\dfrac{c_2}{2} \end{align}

where the RHSFootnote ⁹ is known, while the LHS is unknown. Multiplying Eq. (39) by $\sqrt{3}$ and adding it to Eq. (41), followed by suitable manipulation gives

(42) \begin{align} c_2 &= -\dfrac{1}{2} \left (\sqrt{3} \, q_{31} + q_{33}\right ) \end{align}

(43) \begin{align} s_2c_3 &= -\dfrac{1}{2} \left (q_{31} - \sqrt{3} \, q_{33}\right ) \end{align}

Now, dividing the RHS of Eq. (40) by LHS of Eq. (43) (provided $s_2 \neq 0$ ) gives

(44) \begin{align} \theta _{3_{1,2}} = \arctan \left ({-\dfrac{2 \, q_{32}}{\left (q_{31} - \sqrt{3} \, q_{33}\right )}} \right ) \end{align}

Therefore, two solutionsFootnote ¹⁰ (having a difference of $\pi$ ) are obtained for $\theta _3$ . In order to get $\theta _2$ , the obtained $ \theta _3$ value is substituted in Eq. (40) to get:

(45) \begin{align} s_2 = \dfrac{q_{32}}{s_3} \end{align}

From Eqs. (42) and (45), the value of $ \theta _2$ can be obtained:

(46) \begin{align} \theta _2 = \mathrm{Atan2}{ (s_2, c_2) } \end{align}

Now, in order to get $ \theta _1$ , the (1,2) and (2,2) elements of $\textbf{Q}$ and $\textbf{Q}_{\mathrm{num}}$ are compared and the calculated values of $ \theta _2$ and $ \theta _3$ are substituted to get the following two equations in $ s_1$ and $ c_1$ :

(47) \begin{align} q_{12} &= s_1c_3 - c_1c_2s_3 \end{align}

(48) \begin{align} q_{22} &= -c_1c_3 - s_1c_2s_3 \end{align}

Eliminating $ c_1$ using Eqs. (47) and (48), an equation in $ s_1$ of the following form is obtained:

(49) \begin{align} a_1s_1 + a_2 = 0 \, \Longrightarrow s_1 = -\dfrac{a_2}{a_1} \end{align}

where the coefficients are in terms of the known quantities,

(50) \begin{align} \begin{gathered} a_1 = c_3^2 + c_2^2 s_3^2; \,\,\,\,\,\,\, a_2 = -c_3 q_{12} + c_2 s_3 q_{22} \end{gathered} \end{align}

Substituting back $ s_1$ in Eq. (47), an equation in $ c_1$ is obtained:

(51) \begin{align} b_1c_1 + b_2 = 0 \, \Longrightarrow c_1 = -\dfrac{b_2}{b_1} \end{align}

where the coefficients are in terms of the known quantities,

(52) \begin{align} \begin{gathered} b_1 = c_3^2 + c_2^2 s_3^2; \,\,\,\,\,\,\, b_2 = c_2 s_3 q_{12} + c_3q_{22} \end{gathered} \end{align}

From Eqs. (49) and (51), the value of $ \theta _1$ can be obtained:

(53) \begin{align} \theta _1 = \mathrm{Atan2}{ (s_1, c_1) } \end{align}

Finally, overall, two solutions are obtained, since for each of the two possible $ \theta _3$ values, unique solutions for $ \theta _1$ and $ \theta _2$ exist. Also, now the values of $ \theta _1, \theta _2$ , and $ \theta _3$ can be substituted into Eqs. (10) and (11) to obtain the leg length vectors, and then $L_2$ and $L_3$ can be calculated using Eqs. (12) and (13), respectively, as:

(54) \begin{align} L_2 &= \sqrt{{}^{1}_{}{\textbf{L}_{\textbf{2}}} \cdot{}^{1}_{}{\textbf{L}_{\textbf{2}}}} \end{align}

(55) \begin{align} L_3 &= \sqrt{{}^{1}_{}{\textbf{L}_{\textbf{3}}} \cdot{}^{1}_{}{\textbf{L}_{\textbf{3}}}} \end{align}

It is to be noted that, for the reasons stated earlier $^{6}$ on Page 6, the limb lengths $L_2$ and $L_3$ are only taken to be positive.Footnote ¹¹

This completes the inverse kinematics for the lower parallel mechanism since there are now the values for $ \theta _1, \theta _2, \theta _3, L_2$ and $ L_3$ . Further, the numeric $\textbf{T}_{\textbf{1}_{\mathrm{num}}}$ matrix can be obtained by substituting all these values in the symbolic form of $\textbf{T}_{\textbf{1}}$ .

4.2. Upper parallel 3-UPU TPM mechanism

Now, since the inverse kinematics is solved for the lower parallel mechanism, we can decouple the overall numeric transformation matrix $\textbf{T}_{\mathrm{num}}$ using Eq. (2) and get the transformation matrix $\textbf{T}_{\textbf{2}_{\mathrm{num}}}$ having the following form as:

(56) \begin{align} \textbf{T}_{\textbf{2}_{\mathrm{num}}} = \left ( \textbf{T}_{\textbf{1}_{\mathrm{num}}} \textbf{C}_u^l \right )^{-1} \textbf{T}_{\mathrm{num}} = \left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} 1 & 0 & 0 & r_x \\[5pt] 0 & 1 & 0 & r_y \\[5pt] 0 & 0 & 1 & r_z \\[5pt] 0 & 0 & 0 & 1 \end{array}\right] \end{align}

Moreover, the symbolic form of $\textbf{T}_{\textbf{2}}$ matrix found out using the D-H Table II is

(57) \begin{align} \textbf{T}_{\textbf{2}} = \left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} 1 & 0 & 0 & L_4c_4c_5 \\[5pt] 0 & 1 & 0 & L_4s_4c_5 \\[5pt] 0 & 0 & 1 & L_4s_5 \\[5pt] 0 & 0 & 0 & 1 \end{array}\right] \end{align}

Thus, exploiting the fact that $\textbf{T}_{\textbf{2}} \equiv \textbf{T}_{\textbf{2}_{\mathrm{num}}}$ , and comparing their (1,4), (2,4) and (3,4) elements, the following three equations are seen:

(58) \begin{align} r_x &= L_4c_4c_5 \end{align}

(59) \begin{align} r_y &= L_4s_4c_5 \end{align}

(60) \begin{align} r_z &= L_4s_5 \end{align}

Squaring and adding Eqs. (58), (59) and (60), $L_4$ is obtained as:

(61) \begin{align} L_4 = \sqrt{r_x^2 + r_y^2 + r_z^2} \end{align}

Also, dividing Eqs. (59) and (58), two solutions for $\theta _4$ are obtained as:

(62) \begin{align} \theta _{4_{1,2}} = \arctan \left ( \dfrac{r_y}{r_x} \right ) \end{align}

Now, using the values of $ L_4$ and $ \theta _4$ in Eqs. (59) and (60), $s_5$ and $c_5$ , and therefore, $\theta _5$ are found out as:

(63) \begin{align} s_5 &= \dfrac{r_z}{L_4} \end{align}

(64) \begin{align} c_5 &= \dfrac{r_y}{L_4s_4} \end{align}

(65) \begin{align} \Longrightarrow \theta _5 &= \mathrm{Atan2}{ (s_5, c_5) } \end{align}

So, again, there are overall two solutions, since for each of the two possible $ \theta _4$ values, there exist unique solutions for $ L_4$ and $ \theta _5$ . However, the two solution sets of $\theta _4$ and $\theta _5$ are of the form $\left \{ \beta _4, \, \beta _5 \right \}$ and $\left \{ (\beta _4 - \pi ),\, (\pi -\beta _5) \right \}$ , which again lead to same task space configuration, for reasons mentioned earlier in Section 3.2. Thus, only one distinct task space configuration exists corresponding to the two inverse kinematic solutions of the upper parallel mechanism.

Now, the values of $ \theta _4, \theta _5$ , and $ L_4$ can be substituted into Eqs. (27) and (28) to obtain the leg length vectors, and then $L_5$ and $L_6$ can be calculated using Eqs. (29) and (30), respectively, as:

(66) \begin{align} L_5 &= \sqrt{{}^{4_u}_{}{\textbf{L}_{\textbf{5}}} \cdot{}^{4_u}_{}{\textbf{L}_{\textbf{5}}}} \end{align}

(67) \begin{align} L_6 &= \sqrt{{}^{4_u}_{}{\textbf{L}_{\textbf{6}}} \cdot{}^{4_u}_{}{\textbf{L}_{\textbf{6}}}} \end{align}

It is to be noted that, here also, for the reasons stated earlier $^{6}$ on Page 6, the limb lengths $L_4,\, L_5$ , and $L_6$ are only taken to be positive.Footnote ¹²

Finally, $2 \times 2 = 4$ possible inverse kinematic solutions are obtained for the complete hybrid manipulator in joint space. However, only $2 \times 1 = 2$ inverse kinematic solutions are seen with distinct task space configurations. This fact will also be numerically illustrated later in Section 5.2.2.

Thus, this completes the inverse kinematic analysis of the hybrid manipulator as all the actuated joint variables $\theta _2, L_2, L_3, L_4, L_5, L_6$ have been obtained from the given pose of the end-effector.

5.1. Model

In the literature, different simulation environments have been developed and used, some of which, such as the RoboSim and RoboAnalyzer $^{\circledR }$ , are specialized in robotics. In contrast, others are generic 3D simulation software, such as Adams™, MapleSim™, and Simscape™ Multibody™ (earlier SimMechanics™). In this paper, the proposed hybrid manipulator is modeled and simulated using Simscape Multibody, which is based on Simulink $^{\circledR }$ and MATLAB $^{\circledR }$ . The advantage of using Simscape Multibody is that sensing various signals, including joint movements and displacements, torques, and forces, becomes much more accessible.

The 3D model of the hybrid manipulator is shown in Fig. 3, while Fig. 4(a) shows the complete hybrid manipulator block diagram built in Simscape Multibody. The whole model is made up of five parts: base, lower parallel mechanism, platform M, upper parallel mechanism, and platform H, where the lower mechanism and upper mechanism models, as shown in Fig. 4(b) and (c), respectively, have been built in accordance with the geometry and structure presented in Section 2.1.

Figure 3.

Simulated 3D model of the hybrid manipulator in Simscape Multibody (in Simscape Multibody models, joints are not displayed in the final simulated 3D model).

Figure 4.

Model of the proposed hybrid manipulator built in Simscape Multibody.

5.2. Numerical example

Consider the following numerical example with the chosen design parameters:

(68) \begin{align} \begin{gathered} b_2 = 40 \sqrt{3} \,\,\mathrm{cm} = 69.28 \,\,\mathrm{cm}; \,\,\,\,\,\,\,\,\,\,\, b_{3_x} = 20 \sqrt{3} \,\,\mathrm{cm} = 34.64 \,\,\mathrm{cm}; \,\,\,\,\,\,\,\,\,\,\, b_{3_z} = 60 \,\,\mathrm{cm} \\[5pt] h_1 = 40 \,\,\mathrm{cm}; \,\,\,\,\,\,\,\,\,\,\, h_2 = 30 \,\,\mathrm{cm}; \,\,\,\,\,\,\,\,\,\,\, L_1 = 60 \,\,\mathrm{cm} \end{gathered} \end{align}

5.2.1. Forward kinematics

The following active joint variable values are noted:

(69) \begin{align} \begin{gathered} \theta _2 = \dfrac{\pi }{3} \,\,\mathrm{rad}; \,\,\,\,\,\,\,\,\,\,\, L_2 = 49 \,\,\mathrm{cm}; \,\,\,\,\,\,\,\,\,\,\, L_3 = 81 \,\,\mathrm{cm} \\[5pt] L_4 = 60 \,\,\mathrm{cm}; \,\,\,\,\,\,\,\,\,\,\, L_5 = 59 \,\,\mathrm{cm}; \,\,\,\,\,\,\,\,\,\,\, L_6 = 70 \,\,\mathrm{cm} \end{gathered} \end{align}

Solving Eq. (14), two solutions are obtained as $\theta _{1_1} ={-2.7628} \,\, \mathrm{rad}$ and $\theta _{1_2} ={-1.9496} \,\, \mathrm{rad}$ . Further, solving Eq. (18) with $\theta _{1_1}$ as the solution leads to $\theta _{3_1}^1 ={-2.7336} \,\, \mathrm{rad}$ and $\theta _{3_2}^1 ={1.5209} \,\, \mathrm{rad}$ , while with $\theta _{1_2}$ as the solution, it gives $\theta _{3_1}^2 ={-2.5702} \,\, \mathrm{rad}$ and $\theta _{3_2}^2 ={1.6808} \,\, \mathrm{rad}$ .

For $\theta _5$ , Eq. (31) is solved to get $\theta _{5_1} ={0.7515} \,\, \mathrm{rad}$ and $\theta _{5_2} ={2.3901} \,\, \mathrm{rad}$ . Solving Eq. (35) with $\theta _{5_1}$ as the solution, it gives $\theta _{4_1}^1 ={1.7935} \,\, \mathrm{rad}$ and $\theta _{4_2}^1 ={-1.7935} \,\, \mathrm{rad}$ , while with $\theta _{5_2}$ as the solution, it gives $\theta _{4_1}^2 ={1.3481} \,\, \mathrm{rad}$ and $\theta _{4_2}^2 ={-1.3481} \,\, \mathrm{rad}$ . As can be seen, the solutions for $\theta _4$ are symmetric about the mid-platform M. All the 16 possible joint space solutions to the forward kinematics have been listed in Table III. Also, as was stated earlier in Section 3.2, on calculating the overall transformation matrix $\textbf{T}$ for each of the 16 solutions, only 8 distinct task space configurations are obtained, which have been depicted in Fig. 5.

Table III.

All 16 solutions for the DK of the hybrid manipulator, with each pair of it corresponding to a single configuration.

Figure 5.

Eight distinct task space solutions, with each representing a pair of joint space solutions in Table III.

Table IV.

All four solutions to the inverse kinematics of the hybrid manipulator, with each pair of it corresponding to a single configuration.

5.2.2. Inverse kinematics

Consider the $4{\mathrm{th}}$ solution $\{\theta _{1_1} ={-2.7628} \,\, \mathrm{rad}, \, \theta _{3_1}^1 ={-2.7336} \,\, \mathrm{rad}, \, \theta _{5_2} ={2.3901} \,\, \mathrm{rad}, \, \theta _{4_1}^2 ={1.3481} \,\, \mathrm{rad}\}$ of the 16 total solutions for the direct kinematics presented above. For this solution, the numerical overall transformation matrix $\textbf{T}_{\mathrm{num}}$ is

(70) \begin{align} \textbf{T}_{\mathrm{num}} = \left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} 0.9834 & 0.1551 & -0.0941 & 2.181 \\[5pt] 0.1778 & -0.9262 & 0.3324 & -4.249 \\[5pt] -0.0355 & -0.3436 & -0.9384 & -23.403 \\[5pt] 0 & 0 & 0 & 1 \end{array}\right] \end{align}

Given this matrix, the inverse kinematics involves solving for the joint variables. So, solving for $\theta _3$ , two solutions are seen as $\theta _{3_1} ={0.4080} \,\, \mathrm{rad}$ and $\theta _{3_2} ={-2.7336} \,\, \mathrm{rad}$ . Corresponding to $\theta _{3_1}$ , $\theta _2^1 ={-1.0472} \,\, \mathrm{rad}$ and $\theta _1^1 ={0.3788} \,\, \mathrm{rad}$ are seen, while for $\theta _{3_2}$ , $\theta _2^2 ={1.0472} \,\, \mathrm{rad}$ and $\theta _1^2 ={-2.7628} \,\, \mathrm{rad}$ are seen. Thus, two solutions are obtained here for lower parallel mechanism. Now, for instance, for the solution, which is $\{ \theta _{3_1} ={0.4080} \,\, \mathrm{rad}, \, \theta _2^1 ={-1.0472} \,\, \mathrm{rad}, \, \theta _1^1 ={0.3788} \,\, \mathrm{rad} \}$ , the decoupled $\textbf{T}_{\textbf{2}_{\mathrm{num}}}$ matrix is obtained as:

(71) \begin{align} \textbf{T}_{\textbf{2}_{\mathrm{num}}} = \left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} 1 & 0 & 0 & 14.124 \\[5pt] 0 & 1 & 0 & 67.390 \\[5pt] 0 & 0 & 1 & -0.274 \\[5pt] 0 & 0 & 0 & 1 \end{array}\right] \end{align}

From this, $L_4$ is obtained as ${68.855} \,\, \mathrm{cm}$ . Further, the two solutions are obtained as $\theta _{4_1} ={1.3642} \,\, \mathrm{rad}$ and $\theta _{4_2} ={-1.7774} \,\, \mathrm{rad}$ , and corresponding to them $\theta _5^1 ={-0.0039} \,\, \mathrm{rad}$ and $\theta _5^2 ={3.1455} \,\, \mathrm{rad}$ are seen, respectively. Therefore, again, there are two solutions for the upper parallel mechanism for any given solution to the lower one. Hence, four possible joint space solutions are obtained to the inverse kinematics problem of the complete hybrid manipulator and are listed in Table IV.Footnote ¹³ Also, as was stated earlier in Section 4.2, there are only two distinct configurations, as depicted in Fig. 6.

Figure 6.

Two distinct configurations, with each representing a pair of solutions in Table IV.

5.3. Simulation example: translation in a specific direction in a plane parallel to an oblique plane

Consider the following example in which a planar translation of the upper platform is desired along a specified translational direction while maintaining some constant orientation with respect to the base. As described in Section 1, four parameters can be specified, of which the offset distance and the required orientation of the mid-platform are embedded in the given $\textbf{T}_{\textbf{1}_{\mathrm{num}}}$ matrix:

(72) \begin{align} \textbf{T}_{\textbf{1}_{\mathrm{num}}} = \left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} -0.125 & 0.6495 & 0.75 & 30.0 \\[5pt] -0.6495 & -0.625 & 0.4330 & -51.961 \\[5pt] 0.75 & -0.4330 & 0.5 & 0 \\[5pt] 0 & 0 & 0 & 1 \end{array}\right] \end{align}

The desired perpendicular distance is 50 $\mathrm{cm}$ . The translational motion is expected to be along the $X_8 \equiv X_{4_u}$ axis, such that it first translates 50 $\mathrm{cm}$ along $-X_8$ axis, and then from that point onward it translates 100 $\mathrm{cm}$ along the $+X_8$ axis, effectively ending the motion at a displacement of 50 $\mathrm{cm}$ along $+X_8$ axis. Furthermore, the same design parameters were used as mentioned in Eq. (68) in Section 5.2.

For the simulation, inverse kinematics was performed on the desired motion to get the actuated joint values, which were then fed to the model. The resulting simulation is presented in Fig. 7, showing the three stages of motions. Also, the simulated translational motion measured using the Transform Sensor Block in Simscape Multibody is shown in Fig. 8, and it clearly matches the desired motion.

During the simulation, the joint limits were added appropriately to only allow for feasible motions.

Figure 7.

Simulation of the numerical example, showing the three different stages of motion.

Figure 8.

Simulated translational motion.

6.1. Type A singularity: DOF-gain singularity

For the gain singularity to occur, $\det\!(\textbf{J}_{\textbf{p}}) = 0$ should hold. From Eq. (74), it is known that:

(75) \begin{align} \det\!(\textbf{J}_{\textbf{p}}) = j_{p_{21}} \, j_{p_{33}} \, \left ( j_{p_{64}} \, j_{p_{55}} - j_{p_{54}} \, j_{p_{65}} \right ) = 0 \end{align}

So, from Eq. (75), the following different cases are seen for the gain singularity to occur:

Case 1: $j_{p_{21}} = 0$

Consider

(76) \begin{align} j_{p_{21}} = b_2 \, \left ( L_1 \, \cos\!({\theta _1}) - \sqrt{3} \, h_1 \, \sin\!({\theta _1}) \, \sin\!({\theta _2}) \right ) = 0 \end{align}

Here, $b_2 = 0$ is a design singularity, which, however, seems practically infeasible. Now, from Eqs. (76) and (14), there are two equations in $\theta _1$ , $\theta _2$ , and $L_2$ :

(77) \begin{align} w_1c_1 + w_2s_1s_2 &= 0 \end{align}

(78) \begin{align} e_1s_1 + e_2c_1s_2 + e_3 &= L_2^2 \end{align}

where the coefficients are in terms of the known quantities,

(79) \begin{align} \begin{gathered} w_1 = L_1; \,\,\,\,\,\,\, w_2 = -\sqrt{3} h_1 \\[5pt] e_1 = 2 L_1 b_2; \,\,\,\,\,\,\, e_2 = 2 \sqrt{3} b_2 h_1; \,\,\,\,\,\,\, e_3 = L_1^2 + b_2^2 + 3 h_1^2 \end{gathered} \end{align}

Solving Eq. (77) to get two solutions (with a difference of $\pi$ ) for $\theta _1$ :

(80) \begin{align} \theta _1 = -\arctan \left ( \dfrac{w_1}{w_2 \sin{\theta _2}} \right ) \end{align}

Substituting each value of $\theta _1$ in Eq. (78), two separate implicit equations are obtained for the two branches of the singularity curve in $\theta _2 \text{-} L_2$ plane. For the numerical example presented earlier in Section 5.2, the two branches of the singularity curve are depicted in Fig. 9.

Figure 9.

The two branches of the singularity curve in $\theta _2 \text{-} L_2$ plane.

Case 2: $j_{p_{33}} = 0$

(81) \begin{align} j_{p_{33}} = -\dfrac{3}{2} h_1 \left ( \left ( s_1 b_{3_x} + L_1 \right ) c_3 + \left ( b_{3_z} s_2 - b_{3_x} c_1 c_2 \right ) s_3 \right ) = 0 \end{align}

Here, again, $h_1 = 0$ is a design singularity, which is not feasible practically. Again Eq. (14) $\equiv \, f(\theta _1, \theta _2, L_2)$ is used to get two solutions for $\theta _1 \, \equiv \, f(\theta _2, L_2)$ . Substituting $\theta _1$ back in Eq. (81) $\equiv \, f(\theta _1, \theta _2, \theta _3)$ , there are now two solutions for $\theta _3 \, \equiv \, f(\theta _2, L_2)$ . Finally, putting the values of $\theta _1$ and $\theta _3$ into Eq. (15) $\equiv \, f(\theta _1, \theta _2, \theta _3, L_3)$ , an equation for a branch of the singularity surface in actuated joint space of $\theta _2$ , $L_2$ , and $L_3$ is obtained. Since $2 \times 2 = 4$ overall solutions for $\theta _1$ and $\theta _3$ exist, a total of four branches are seen of the singularity surface in $\theta _2 \text{-} L_2 \text{-} L_3$ actuated joint space. For the numerical example presented earlier in Section 5.2, the four branches of the singularity surface in $\theta _2 \text{-} L_2 \text{-} L_3$ actuated joint space are depicted in Fig. 10.

Figure 10.

The four branches (left) and the overall singularity surface (right) in $\theta _2 \text{-} L_2 \text{-} L_3$ actuated joint space.

Case 3: $\left ( j_{p_{64}} \, j_{p_{55}} - j_{p_{54}} \, j_{p_{65}} \right ) = 0$

(82) \begin{align} \left ( j_{p_{64}} \, j_{p_{55}} - j_{p_{54}} \, j_{p_{65}} \right ) = -\dfrac{3 \sqrt{3}}{2} (h_1 - h_2)^2 \, L_4^2 \, \cos ^2({\theta _5}) \, \sin\!({\theta _4}) = 0 \end{align}

Clearly, $h_1 = h_2$ gives a feasible design singularity. Additionally, the following singularities are also obtained:

1. (1) $L_4 = 0$

2. (2) $\theta _5 = \left (n + \frac{1}{2} \right ) \pi \,\, \text{ where } \,\, n = 0, \pm 1, \pm 2, \ldots$ , occurs when $\mathrm{M_1H_1}$ limb becomes parallel or antiparallel to the $\mathrm{U}_{{1}_{1}}$ axis

3. (3) $\theta _4 = n \pi \,\, \text{ where } \,\, n = 0, \pm 1, \pm 2, \ldots$ , occurs when $\mathrm{M_1H_1}$ limb comes in the plane of mid-platform M

Here, (b) yields the same singularity surface in $L_4 \text{-} L_5 \text{-} L_6$ actuated joint space as shown in Fig. 12. From (c), $ \left | \cos\!({\theta _4}) \right | = 1$ . So, using Eqs. (31) and (32):

(83) \begin{align} s_5 &= -\dfrac{f_2}{f_1} \end{align}

(84) \begin{align} c_5 &= -\dfrac{g_2}{g_1c_4} \end{align}

(85) \begin{align} \Longrightarrow s_5^2 + c_5^2 &= \left (\dfrac{f_2}{f_1}\right )^2 + \left (\dfrac{g_2}{g_1}\right )^2 = 1 \end{align}

Thus, Eq. (85) gives the singularity surface in $L_4 \text{-} L_5 \text{-} L_6$ actuated joint space. For the numerical example presented earlier in Section 5.2, the singularity surface is shown in Fig. 11.

Figure 11.

The singularity surface in $L_4 \text{-} L_5 \text{-} L_6$ actuated joint space.

Now, the “serial singularities” are analyzed occurring because the two platforms are serially connected by a serial kinematic chain. Consider the end-effector point to be at the centroid of platform H. The right-invariant angular velocity matrix ${}^{1}_{}{\boldsymbol{\Omega _e}}$ is written as:

(86) \begin{align}{}^{1}_{}{\boldsymbol{\Omega _e}} = \dot{\textbf{Q}}(t) \, \textbf{Q}(t)^{\textsf{T}} \end{align}

and extract from it the angular velocity vector of the end-effector ${}^{1}_{}{\boldsymbol{\omega _e}} ={\left [{}^{1}_{}{\boldsymbol{\Omega _{e_{3,2}}}},{}^{1}_{}{\boldsymbol{\Omega _{e_{1,3}}}},{}^{1}_{}{\boldsymbol{\Omega _{e_{2,1}}}} \right ]}^{\textsf{T}}$ since:

(87) \begin{align}{}^{1}_{}{\boldsymbol{\Omega _e}} = \left[\begin{array}{c@{\quad}c@{\quad}c} 0 & -{}^{1}_{}{\boldsymbol{\omega _{e_z}}} &{}^{1}_{}{\boldsymbol{\omega _{e_y}}} \\[5pt] {}^{1}_{}{\boldsymbol{\omega _{e_z}}} & 0 & -{}^{1}_{}{\boldsymbol{\omega _{e_x}}} \\[5pt] -{}^{1}_{}{\boldsymbol{\omega _{e_y}}} &{}^{1}_{}{\boldsymbol{\omega _{e_x}}} & 0 \end{array}\right] \end{align}

Now, the position vector of the end-effector is written as ${}^{8} \textbf{p}_{\textbf{e}} ={\left [ h_2,0,0,1 \right ]}^{\textsf{T}}$ and is transformed into base frame as ${}^{1} \textbf{p}_{\textbf{e}} = \textbf{T} \,{}^{8} \textbf{p}_{\textbf{e}}$ . Differentiating ${}^{1} \textbf{p}_{\textbf{e}}$ with respect to time, the velocity ${}^{1} \textbf{v}_{\textbf{e}}$ of end-effector ${}^{1} \textbf{v}_{\textbf{e}}$ is finally obtained as the first three entries (i.e., ignoring the fourth homogeneous coordinate) of $\dot{{}^{1} \textbf{p}_{\textbf{e}}}$ . Thus, now the twist $\textbf{\$}_e$ of the platform H about the global coordinate system at the end-effector point can be found out by concatenating ${}^{1}_{}{\boldsymbol{\omega _e}}$ and ${}^{1} \textbf{v}_{\textbf{e}}$ . Further simplification and rearranging gives the following equation:

(88) \begin{align} \textbf{\$}_e = \textbf{J}_{\textbf{s}} \, \dot{\textbf{q}}(t) \end{align}

where $\textbf{J}_{\textbf{s}} = \textbf{J}_{\textbf{s}}(q(t))$ is another $6 \times 6$ jacobian matrix with the following form (details in Section A.2):

(89) \begin{align} \textbf{J}_{\textbf{s}} = \left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 0 & j_{s_{12}} & j_{s_{13}} & 0 & 0 & 0 \\[5pt] 0 & j_{s_{22}} & j_{s_{23}} & 0 & 0 & 0 \\[5pt] 1 & 0 & j_{s_{33}} & 0 & 0 & 0 \\[5pt] j_{s_{41}} & j_{s_{42}} & j_{s_{43}} & j_{s_{44}} & j_{s_{45}} & j_{s_{46}} \\[5pt] j_{s_{51}} & j_{s_{52}} & j_{s_{53}} & j_{s_{54}} & j_{s_{55}} & j_{s_{56}} \\[5pt] 0 & j_{s_{62}} & j_{s_{63}} & j_{s_{64}} & j_{s_{65}} & j_{s_{66}} \end{array}\right] \end{align}

Combining Eqs. (73) and (88), while assuming $\textbf{J}_{\textbf{p}}$ as nonsingular:

(90) \begin{align} \textbf{\$}_e = \textbf{J}_{\textbf{s}} \, \textbf{J}_{\textbf{p}}{}^{-1} \, \textbf{D} \, \dot{\boldsymbol{\Theta}}(t) \end{align}

Again, the matrix $\textbf{J}_{\textbf{s}} \, \textbf{J}_{\textbf{p}}{}^{-1} \, \textbf{D} \equiv \textbf{J}_{\textbf{eq}}$ matrix definition given by Ghosal [Reference Ghosal35]. Thus, in accordance with the condition described in ref. [Reference Ghosal35], since $\det\!(\textbf{J}_{\textbf{eq}}) = 0$ if any one of $\textbf{J}_{\textbf{s}}$ or $\textbf{D}$ is singular, $\det\!(\textbf{J}_{\textbf{s}}) = 0$ and $\det\!(\textbf{D}) = 0$ give Type B (loss) singularity.

6.2. Type B singularity: DOF-loss singularity

From $\textbf{D}$ matrix definition, $\det\!(\textbf{D}) = L_2 \, L_3 \, L_5 \, L_6 = 0$ gives the trivial conditions $L_i = 0$ for $i = 2,3,5,6$ having obvious geometric interpretation (i.e., corresponding limb length becomes zero). Now, from Eq. (89), the determinant of $\textbf{J}_{\textbf{s}}$ matrix is

(91) \begin{align} \det\!(\textbf{J}_{\textbf{s}}) = \left ( \left ( j_{s_{44}}j_{s_{55}} - j_{s_{45}}j_{s_{54}} \right ) j_{s_{66}} + \left ( j_{s_{54}}j_{s_{65}} - j_{s_{55}}j_{s_{64}} \right ) j_{s_{46}} + \left ( j_{s_{45}}j_{s_{64}} - j_{s_{44}}j_{s_{65}} \right ) j_{s_{56}} \right ) \end{align}

(92) \begin{align} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \cdot \left ( j_{s_{12}}j_{s_{23}} - j_{s_{13}}j_{s_{22}} \right ) \end{align}

Substituting the expressions presented earlier for $j_{s_{xy}}$ and further simplifying, the condition for singularity is written as:

(93) \begin{align} \det\!(\textbf{J}_{\textbf{s}}) = L_4^2 \, \sin\!({\theta _2}) \, \cos\!({\theta _5}) = 0 \end{align}

Again, $L_4 = 0$ is the trivial case. For the case $\sin\!({\theta _2}) = 0$ , the condition on $\theta _2$ is

(94) \begin{align} \theta _2 = n \pi \,\, \text{ where } \,\, n = 0, \pm 1, \pm 2, \ldots \end{align}

This has a straightforward geometric interpretation that whenever the $R_{2}$ revolute joint (i.e., $\theta _2$ ) is actuated such that the axes of $R_{1}$ and $R_{3}$ (i.e., $Z_1$ and $Z_3$ ) become parallel or antiparallel, this type of singularity occurs.

For the case $\cos\!({\theta _5}) = 0$ , the condition on $\theta _5$ is

(95) \begin{align} \theta _5 = \left (n + \dfrac{1}{2} \right ) \pi \,\, \text{ where } \,\, n = 0, \pm 1, \pm 2, \ldots \end{align}

which implies that $\sin\!({\theta _5}) = \pm 1$ . Substituting in Eq. (31), two branches of the singularity surface in $L_4 \text{-} L_5 \text{-} L_6$ actuated joint space are seen:

(96) \begin{align} f_1 + f_2 = 0 \end{align}

(97) \begin{align} -f_1 + f_2 = 0 \end{align}

since $f_1$ and $f_2$ , as defined in Eq. (33), are $f(L_4,L_5,L_6)$ . For the numerical example presented earlier in Section 5.2, the two branches of the singularity surface are depicted in Fig. 12. Geometrically, this type of singularity is seen when $\mathrm{M_1H_1}$ limb becomes parallel or antiparallel to the $\mathrm{U}_{{1}_{1}}$ axis.

Figure 12.

The two branches of the singularity surface in $L_4 \text{-} L_5 \text{-} L_6$ actuated joint space.

Figure 13.

Manipulability index as a function of lower mechanism’s active joint variables for three different values of $\theta _2$ .

Figure 14.

Manipulability index as a function of upper mechanism’s active joint variables for three different values of $L_4$ .

Figure 15.

Manipulability index as a function of lower mechanism’s active joint variables for three different values of $\theta _2$ , for optimal parameters.

Figure 16.

Manipulability index as a function of upper mechanism’s active joint variables for three different values of $L_4$ , for optimal parameters.

6.3. Type C singularity: special case

As can be seen from Sections 6.1 and 6.2, singularities $L_4 = 0$ and $\theta _5 = \left (n + \dfrac{1}{2} \right ) \pi$ satisfy the DOF-gain and DOF-loss conditions simultaneously. Hence, they are termed as the “special case” of singularities. This concludes the determination of the singularity conditions for the proposed hybrid manipulator.