3.1. Forward kinematics modeling

In the study of the kinematics of complex spatial mechanisms, screw theory has the advantages of a clear modeling structure and it avoids computational singularities. It has extensive and expansive applications in the field of advanced spatial mechanisms. Screw theory was used to establish a kinematics model of the parking reconfiguration of the metamorphic robot.

The support mode of the robot, as shown in Fig. 2(b), is taken as the reference posture. The base coordinate system S is established with the center of the closed convex polygon formed by the contacts between a support foot and the ground as the origin. The X-axis points in the movement direction of the robot. The Y-axis points to the left of the movement direction of the robot, i.e., perpendicular to the paper and pointing outward. The Z-axis points vertically upward. The front frame is the end mechanism of the system.

The tool coordinate system T is established at the hinge point of the front frame and the base of the electric push rod. The directions of the axes of T and S coincide. A schematic diagram of the joint motions of the metamorphic robot is shown in Fig. 3. The parameters are shown in Table I.

Figure 3.

Schematic diagram of the joint motions of the metamorphic robot.

Table I.

Structural parameters of the metamorphic robot.

In the support mode of the robot, the center of mass of the end mechanism of the system (that is, the front frame) is at the coordinates $(x_{5}^{0},y_{5}^{0},z_{5}^{0})$ in S, and its initial configuration matrix is as follows:

(1) \begin{align}\begin{array}{c} g_{ST}\left(0\right)=\left[\begin{array}{c@{\quad}c} \begin{array}{c@{\quad}c} 1 & 0\\[5pt] 0 & 1 \end{array} & \begin{array}{c@{\quad}c} 0 & x_{5}^{0}\\[5pt] 0 & y_{5}^{0} \end{array}\\[5pt] \begin{array}{c@{\quad}c} 0 & 0\\[5pt] 0 & 0 \end{array} & \begin{array}{c@{\quad}c} 1 & z_{5}^{0}\\[5pt] 0 & 1 \end{array} \end{array}\right]\in SE\left(3\right), \end{array}\end{align}

where SE (3) represents the special Euclidean group in the Lie group, which describes the three-dimensional configuration space of a rigid body.

According to the structural analysis, joints 1–5 are all rotating joints, and the unit vector axis, $\omega _{i}$ (i = 1–5), is defined as follows:

(2) \begin{align} \begin{array}{c} \omega _{i}=\left[\omega _{x}\; \omega _{y}\; \omega _{z}\right]^{T}=\left[\begin{array}{c@{\quad}c@{\quad}c} 0 & -1 & 0 \end{array}\right]^{T}. \end{array}\end{align}

The skew-symmetric matrix, $\hat{\omega }_{i}$ (i = 1–5), is expressed as follows:

(3) \begin{align} \begin{array}{c} \hat{\omega }_{i}=\left[\begin{array}{c@{\quad}c@{\quad}c} 0 & -\omega _{z} & \omega _{y}\\[5pt] \omega _{z} & 0 & -\omega _{x}\\[5pt] -\omega _{y} & \omega _{x} & 0 \end{array}\right]=\left[\begin{array}{c@{\quad}c@{\quad}c} 0 & 0 & -1\\[5pt] 0 & 0 & 0\\[5pt] 1 & 0 & 0 \end{array}\right]\in so(3), \end{array}\end{align}

where so (3) represents the vector space of all third-order skew-symmetric matrices.

When the ith rod rotates by an angle $\theta _{i}$ at a constant speed about the unit vector axis of $\hat{\omega }_{i}$ , the following is obtained:

(4) \begin{align} \begin{array}{c} R(\omega _{i},\theta _{i})=e^{{\hat{\omega }_{i}}{\theta _{i}}}\,=\left[\begin{array}{c@{\quad}c@{\quad}c} \cos \theta _{i} & 0 & -\sin \theta _{i}\\[5pt] 0 & 1 & 0\\[5pt] \sin \theta _{i} & 0 & \cos \theta _{i} \end{array}\right]\in SO(3), \end{array}\end{align}

where $R(\omega _{i},\theta _{i})$ represents the posture matrix of the ith rod, SO (3) represents the special orthogonal group in the Lie group, also known as the three-dimensional rotation group, and $\theta _{i}$ is a variable angle of the joint and is positive for the clockwise rotation around the Y-axis.

For an arbitrary point $p_{i}$ on the axis of a joint, the following is obtained:

(5) \begin{align} \begin{array}{c} \begin{array}{c} p_{1}=\left[\begin{array}{c} l_{1x}\\[5pt] 0\\[5pt] l_{1z} \end{array}\right], p_{2}=\left[\begin{array}{c} l_{2x}\\[5pt] 0\\[5pt] l_{2z} \end{array}\right], p_{3}=\left[\begin{array}{c} l_{3x}\\[5pt] 0\\[5pt] l_{3z} \end{array}\right],\\[5pt] p_{4}=\left[\begin{array}{c} l_{4x}\\[5pt] 0\\[5pt] l_{4z} \end{array}\right], p_{5}=\left[\begin{array}{c} l_{5x}\\[5pt] 0\\[5pt] l_{5z} \end{array}\right].\, \end{array} \end{array}\end{align}

The coordinates of each motion screw, $\xi _{i}=\left[\begin{array}{c} v_{i}\\ \omega _{i} \end{array}\right]=\left[\begin{array}{c} -\omega _{i}\times p_{i}\\\omega _{i} \end{array}\right]$ (i = 1–5) can be expressed as follows:

(6) \begin{align} \begin{array}{c} \begin{array}{c} \xi _{1}=\left[\begin{array}{c} \begin{array}{c} l_{1z}\\[5pt] 0\\[5pt] -l_{1x} \end{array}\\[5pt] \begin{array}{c} 0\\[5pt] -1\\[5pt] 0 \end{array} \end{array}\right];\; \xi _{2}=\left[\begin{array}{c} \begin{array}{c} l_{2z}\\[5pt] 0\\[5pt] -l_{2x} \end{array}\\[5pt] -\begin{array}{c} 0\\[5pt] 1\\[5pt] 0 \end{array} \end{array}\right];\; \xi _{3}=\left[\begin{array}{c} \begin{array}{c} l_{3z}\\[5pt] 0\\[5pt] -l_{3x} \end{array}\\[5pt] -\begin{array}{c} 0\\[5pt] 1\\[5pt] 0 \end{array} \end{array}\right];\\[5pt] \xi _{4}=\left[\begin{array}{c} \begin{array}{c} l_{4z}\\[5pt] 0\\[5pt] -l_{4x} \end{array}\\[5pt] -\begin{array}{c} 0\\[5pt] 1\\[5pt] 0 \end{array} \end{array}\right];\; \xi _{5}=\left[\begin{array}{c} \begin{array}{c} l_{5z}\\[5pt] 0\\[5pt] -l_{5x} \end{array}\\[5pt] -\begin{array}{c} 0\\[5pt] 1\\[5pt] 0 \end{array} \end{array}\right]. \end{array} \end{array}\end{align}

Similarly, introducing the motion screw $\hat{\xi }_{i}=\left[\begin{array}{c@{\quad}c} \hat{\omega }_{i} & v_{i}\\ 0 & 0 \end{array}\right]_{4\times 4}\in se(3)$ , the $\hat{\xi }_{i}\theta _{i}$ matrix in the exponential form is as follows:

(7) \begin{align} \begin{array}{c} e^{{\hat{\xi }_{i}}{\theta _{i}}}=\begin{cases} \left[\begin{array}{c@{\quad}c} e^{{\hat{\omega }_{i}}{\theta _{i}}} & \left(I-e^{{\hat{\omega }_{i}}{\theta _{i}}}\right)\left(\omega _{i}\times v_{i}\right)+\omega _{i}{\omega _{i}}^{T}v_{i}\theta _{i}\\[5pt] 0 & 1 \end{array}\right]\\[5pt] \left(\omega _{i}\neq 0\right)\\[5pt] \,\left[\begin{array}{c@{\quad}c} I & v_{i}\theta _{i}\\[5pt] 0 & 1 \end{array}\right]\left(\omega _{i}=0\right)\,\, \end{cases}. \end{array}\end{align}

According to (7), $e^{{\hat{\xi }_{i}}{\theta _{i}}}$ (i = 1–5) can be calculated as follows:

(8) \begin{align} \begin{array}{c} e^{\widehat{\xi _{i}}{\theta _{i}}}=\left[\begin{array}{c@{\quad}c} \begin{array}{c@{\quad}c} \cos \theta _{i} & 0\\[5pt] 0 & 1 \end{array} & \begin{array}{c@{\quad}c} -\sin \theta _{i} & p_{ix}\\[5pt] 0 & 0 \end{array}\\[5pt] \begin{array}{c@{\quad}c} \sin \theta _{i} & 0\\[5pt] 0 & 0 \end{array} & \begin{array}{c@{\quad}c} \cos \theta _{i} & p_{iz}\\[5pt] 0 & 1 \end{array} \end{array}\right], \end{array}\end{align}

where

\begin{equation*} \begin{array}{c} p_{ix}=\,l_{iz}\sin \theta _{i}-l_{ix}\left(\cos \theta _{i}-1\right),\\[5pt] p_{iz}=-l_{ix}\sin \theta _{i}-l_{iz}\left(\cos \theta _{i}-1\right). \end{array} \end{equation*}

The combination of the movements of all the joints yields the kinematics model of the end mechanism (i.e., the front frame of the robot) of the metamorphic robot system:

(9) \begin{align} g_{ST}(\theta )=e^{{\hat{\xi }_{1}}{\theta _{1}}}e^{{\hat{\xi }_{2}}{\theta _{2}}}e^{{\hat{\xi }_{3}}{\theta _{3}}}e^{{\hat{\xi }_{4}}{\theta _{4}}}e^{{\hat{\xi }_{5}}{\theta _{5}}}g_{ST}(0). \end{align}

Equations (1)–(8) are combined, and we define $c_{i}=\cos \theta _{i}$ , $s_{i}=\sin \theta _{i}$ , $c_{ij\ldots n}=\cos (\theta _{i}+\theta _{j}+\ldots +\theta _{n}),$ and $s_{ij\ldots n}=\sin (\theta _{i}+\theta _{j}+\ldots +\theta _{n})$ .

Equation (9) can be written as follows:

(10) \begin{align} \begin{array}{c} \begin{array}{c} g_{ST}\left(\theta \right)=e^{{\hat{\xi }_{1}}{\theta _{1}}}e^{{\hat{\xi }_{2}}{\theta _{2}}}e^{{\hat{\xi }_{3}}{\theta _{3}}}e^{{\hat{\xi }_{4}}{\theta _{4}}}e^{{\hat{\xi }_{5}}{\theta _{5}}}g_{ST}\left(0\right)\\[5pt] =\left[\begin{array}{c@{\quad}c} \begin{array}{c@{\quad}c} c_{12345} & 0\\[5pt] 0 & 1 \end{array} & \begin{array}{c@{\quad}c} -s_{12345} & p_{x}\\[5pt] 0 & y_{5}^{0} \end{array}\\[5pt] \begin{array}{c@{\quad}c} s_{12345} & 0\\[5pt] 0 & 0 \end{array} & \begin{array}{c@{\quad}c} c_{12345} & p_{z}\\[5pt] 0 & 1 \end{array} \end{array}\right]\, \end{array}, \end{array}\end{align}

where

\begin{align*} p_{x}&=\left(x_{5}^{0}-l_{5x}\right)c_{12345}+\left(l_{5z}-z_{5}^{0}\right)s_{12345}+\left(l_{5x}-l_{4x}\right)c_{1234}+\left(-l_{5z}\right)s_{1234}+\left(l_{4x}-l_{3x}\right)c_{123}\\[5pt] & \quad +\left(l_{3z}-l_{4z}\right)s_{123}+\left(l_{3x}-l_{2x}\right)c_{12}+\left(l_{2z}-l_{3z}\right)s_{12}+\left(l_{1z}-l_{2z}\right)s_{1}+l_{2x}c_{1}, \\[5pt] p_{z} & = (z_{5}^{0}-l_{5z})c_{12345}+(x_{5}^{0}-l_{5x})s_{12345}+(l_{5z}-l_{4z})c_{1234}+(l_{5x}-l_{4x})s_{1234}+(l_{4z}-l_{3z})c_{123} \\[5pt] & \quad +(l_{4x}-l_{3x})s_{123}+(l_{3z}-l_{2z})c_{12}+(l_{3x}-l_{2x})s_{12}+(l_{2z}-l_{1z})c_{1}+ l_{2x}s_{1}+l_{1z} \end{align*}

3.2. Level-lifting motion design

The geometric relationships show that if the rear frame remains level during the reconfiguration process, then

(11) \begin{align} \begin{array}{c} \theta _{1}+\theta _{2}+\theta _{3}=0. \end{array}\end{align}

If the front frame being lifted remains level, then

(12) \begin{align} \begin{array}{c} \theta _{4}+\theta _{5}=0. \end{array}\end{align}

The following is obtained by substituting (11) and (12) into (10):

(13) \begin{align} \begin{array}{c} \begin{array}{c} g_{ST}\left(\theta \right)=e^{{\hat{\xi }_{1}}{\theta _{1}}}e^{{\hat{\xi }_{2}}{\theta _{2}}}e^{{\hat{\xi }_{3}}{\theta _{3}}}e^{{\hat{\xi }_{4}}{\theta _{4}}}e^{{\hat{\xi }_{5}}{\theta _{5}}}g_{ST}\left(0\right)\\[5pt] =\left[\begin{array}{c@{\quad}c} \begin{array}{c@{\quad}c} 1 & 0\\[5pt] 0 & 1 \end{array} & \begin{array}{c@{\quad}c} 0 & p_{x}\\[5pt] 0 & y_{5}^{0} \end{array}\\[5pt] \begin{array}{c@{\quad}c} 0 & 0\\[5pt] 0 & 0 \end{array} & \begin{array}{c@{\quad}c} 1 & p_{z}\\[5pt] 0 & 1 \end{array} \end{array}\right], \end{array} \end{array}\end{align}

where

\begin{align*} p_{x}& =\left(l_{5x}-l_{4x}\right)c_{4}+\left(l_{4z}-l_{5z}\right)s_{4}+\left(l_{3x}-l_{2x}\right)c_{12}+\left(l_{2z}-l_{3z}\right)s_{12}+\left(-l_{2z}\right)s_{1}+l_{2x}c_{1}+\left(x_{5}^{0}-l_{5x}\right)\\[5pt] & \quad +\left(l_{4x}-l_{3x}\right),+\left(x_{5}^{0}-l_{5x}\right)+\left(l_{4x}-l_{3x}\right),\\[5pt] p_{z}&=(l_{5z}-l_{4z})c_{4}+(l_{5x}-l_{4x})s_{4}+(l_{3z}-l_{2z})c_{12}+(l_{3x}-l_{2x})s_{12}+(l_{2z}-l_{1z})c_{1}+l_{2x}s_{1} \\[5pt] & \quad +(z_{5}^{0}-l_{5z})+(l_{4z}-l_{3z})+l_{1z} \end{align*}

In addition, the front frame level-lifting mechanism contains two driving links, the lift motor, and the electric push rod. For the front frame to remain level during lifting, the angle of rotation of the lifting rod about the axis of the lifting joint, $\theta _{4}$ , and the stroke of electric push rod, $\theta _{7}$ , must satisfy a certain relationship. The analysis indicates that the lifting mechanism can be regarded as a planar closed-chain mechanism composed of the lifting rods, front frame, and electric push rod. The relationship between $\theta _{4}$ and $\theta _{7}$ can be then solved and analyzed by equating the end postures determined by every branch of the kinematic chain of the closed-chain mechanism [Reference Zhang, Fu and Wang27–Reference Maradkar and Manivannan29]. As shown in Fig. 4, a base coordinate system S ′ is established at the hinge point of the lower part of the V-shaped lifting rod and the rear frame. The tool coordinate system T ′ is established at the hinge point of the V-shaped lifting rod and the electric push rod. The initial state is still the state as shown in Fig. 3. The coordinates of the origin of T ′ in S ′ is $(x',0,z')$ , and it follows that

(14) \begin{align} \begin{array}{c} \begin{array}{c} g_{S'T'}\left(0\right) \end{array}=\left[\begin{array}{c@{\quad}c} \begin{array}{c@{\quad}c} 1 & 0\\[5pt] 0 & 1 \end{array} & \begin{array}{c@{\quad}c} 0 & x'\\[5pt] 0 & 0 \end{array}\\[5pt] \begin{array}{c@{\quad}c} 0 & 0\\[5pt] 0 & 0 \end{array} & \begin{array}{c@{\quad}c} 1 & z'\\[5pt] 0 & 1 \end{array} \end{array}\right], \end{array}\end{align}

Figure 4.

Relationship between the rotation angle of the lifting joint and the stroke of the electric push rod.

Similar to joints 4 and 5, joint 6 is a rotating joint. The unit vector of the rotation axis of joint 6 is as follows:

(15) \begin{align} \begin{array}{c} w_{6}=[0-10]^{T}. \end{array}\end{align}

Similarly, for an arbitrary point on the axis of the screw, $p_{6}$ ,

(16) \begin{align} \begin{array}{c} p_{6}=\left[\begin{array}{c@{\quad}c@{\quad}c} l_{6x} & 0 & l_{6z} \end{array}\right]^{T}, \end{array}\end{align}

(17) \begin{align} \begin{array}{c} \xi _{6}=\left[\begin{array}{c@{\quad}c@{\quad}c} \begin{array}{c@{\quad}c} l_{6z} & 0 \end{array} & \begin{array}{c@{\quad}c} -l_{6x} & 0 \end{array} & \begin{array}{c@{\quad}c} -1 & 0 \end{array} \end{array}\right]^{T}. \end{array}\end{align}

From (7), it follows that

(18) \begin{align} \begin{array}{c} e^{{\hat{\xi }_{6}}{\theta _{6}}}=\left[\begin{array}{c@{\quad}c} \begin{array}{c@{\quad}c} \cos \theta _{6} & 0\\[5pt] 0 & 1 \end{array} & \begin{array}{c@{\quad}c} -\sin \theta _{6} & p_{6x}\\[5pt] 0 & 0 \end{array}\\[5pt] \begin{array}{c@{\quad}c} \sin \theta _{6} & 0\\[5pt] 0 & 0 \end{array} & \begin{array}{c@{\quad}c} \cos \theta _{6} & p_{6z}\\[5pt] 0 & 1 \end{array} \end{array}\right], \end{array}\end{align}

where

\begin{align*}p_{6x}&=l_{6\mathrm{z}}\sin \theta _{6}-l_{6x}(\cos \theta _{6}-1),\\[5pt] p_{6z}&=-l_{6x}\sin \theta _{6}-l_{6z}(\cos \theta _{6}-1). \end{align*}

Joint 7 is a slip joint. The unit vector of the translation axis of joint 7 is as follows:

(19) \begin{align} \begin{array}{c} v7=[{\nu _{7x}}\quad 0\quad {\nu _{7z}}]^{T}, \end{array}\end{align}

where $v_{7x}$ and $v_{7z}$ are the components of the unit linear velocity vector of joint 7 on the X- and Z-axes, respectively, and ${v_{7x}}^{2}{+v_{7z}}^{2}=1$ .

The following is also obtained:

(20) \begin{align} \begin{array}{c} \xi _{7}=\left[\begin{array}{c@{\quad}c} \begin{array}{c@{\quad}c} v_{7x} & 0 \end{array} & \begin{array}{c@{\quad}c@{\quad}c@{\quad}c} v_{7\mathrm{z}} & 0 & 0 & 0 \end{array} \end{array}\right]^{T}. \end{array}\end{align}

Similarly, from (7), it follows that

(21) \begin{align} \begin{array}{c} e^{{\hat{\xi }_{7}}{\theta _{7}}}=\left[\begin{array}{c@{\quad}c} \begin{array}{c@{\quad}c} 1 & 0\\[5pt] 0 & 1 \end{array} & \begin{array}{c@{\quad}c} 0 & \theta _{7}\nu _{7x}'\\[5pt] 0 & 0 \end{array}\\[5pt] \begin{array}{c@{\quad}c} 0 & 0\\[5pt] 0 & 0 \end{array} & \begin{array}{c@{\quad}c} 1 & \theta _{7}\nu _{7z}\\[5pt] 0 & 1 \end{array} \end{array}\right]. \end{array}\end{align}

Substituting the exponential matrix corresponding to the two branch chains into the exponential product equation yields the structural equation of the level-lifting mechanism:

(22) \begin{align} \begin{array}{c} g_{S'T'}(\theta )=e^{{\hat{\xi }_{4}}{\theta _{4}}}e^{{\hat{\xi }_{5}}{\theta _{5}}}e^{{\hat{\xi }_{6}}{\theta _{6}}}e^{{\hat{\xi }_{7}}{\theta _{7}}}g_{S'T'}(0)=e^{{\hat{\xi }_{4}}{\theta _{4}}}g_{S'T'}(0). \end{array}\end{align}

Analysis shows that

(23) \begin{align} \begin{array}{c} (e^{{\hat{\xi }_{4}}{\theta _{4}}})^{-1}g_{S'T'}(\theta )[g_{S'T'}(0)]^{-1}=e^{{\hat{\xi }_{5}}{\theta _{5}}}e^{{\hat{\xi }_{6}}{\theta _{6}}}e^{{\hat{\xi }_{7}}{\theta _{7}}}I. \end{array}\end{align}

where I is the fourth-order unit matrix.

Combining (8), (12), (18), (21), and (23) yields the following:

(24) \begin{align} \begin{array}{c} \begin{cases} l_{6x}\cos \theta _{4}-l_{6x}+l_{6z}\sin \theta _{4}+\theta _{7}v_{7x}=0\\[5pt] -\,l_{6z}+l_{6z}\cos \theta _{4}-l_{6x}\sin \theta _{4}+\theta _{7}v_{7z}=0 \end{cases}. \end{array}\end{align}

The following is further obtained:

(25) \begin{align} \begin{array}{c} \begin{cases} \theta _{7}=\sqrt{a^{2}+b^{2}}\\[5pt] a=l_{6x}\cos \theta _{4}-l_{6x}+l_{6z}\sin \theta _{4}\\[5pt] b=-l_{6z}+l_{6z}\cos \theta _{4}-l_{6x}\sin \theta _{4} \end{cases}. \end{array}\end{align}

Using the equations presented above, a simulation model was built using MATLAB/Simulink. The simulation results are shown in Fig. 4. In support mode of the robot, the homogeneous coordinates of the centers of mass of the ankle, calf, thigh, rear frame, lower lifting rod (including support rod), front frame, electric push rod, and upper lifting rod are $c_{i}=[x_{i}^{0}y_{i}^{0}z_{i}^{0}1]^{T}$ (i = 1–7). Based on the kinematics model of reconfiguration while parked and the level lifting model, the homogeneous coordinates of the center of the mass of the ith part of the robot, $C_{i}$ , in the reconfiguration process, can be obtained as follows:

Ankle:

(26) \begin{align} \begin{array}{c} C_{0}=[{x_{0}}\quad {y_{0}}\quad {z_{0}}\quad 1]^{T}=c_{0}, \end{array}\end{align}

Calf:

(27) \begin{align} \begin{array}{c} C_{1}=[{x_{1}}\quad{y_{1}}\quad{z_{1}}\quad 1]^{T}=e^{{\hat{\xi }_{1}}{\theta _{1}}}c_{1}, \end{array}\end{align}

Thigh:

(28) \begin{align} \begin{array}{c} C_{2}=[{x_{2}}\quad{y_{2}}\quad{z_{2}}\quad 1]^{T}=e^{{\hat{\xi }_{1}}{\theta _{1}}}e^{{\hat{\xi }_{2}}{\theta _{2}}}c_{2}, \end{array}\end{align}

Rear frame:

(29) \begin{align} \begin{array}{c} C_{3}=\left[x_{3}\quad y_{3}\quad z_{3}\quad 1\right]^{T}=e^{{\hat{\xi }_{1}}{\theta _{1}}}e^{{\hat{\xi }_{2}}{\theta _{2}}}{e^{{\hat{\xi }_{3}}{\theta _{3}}}}c_{3}, \end{array}\end{align}

Lower lifting rod:

(30) \begin{align} \begin{array}{c} C_{4}=\left[x_{4}\quad y_{4}\quad z_{4}\quad 1\right]^{T}=e^{{\hat{\xi }_{1}}{\theta _{1}}}e^{{\hat{\xi }_{2}}{\theta _{2}}}{e^{{\hat{\xi }_{3}}{\theta _{3}}}}{e^{{\hat{\xi }_{4}}{\theta _{4}}}}c_{4}, \end{array}\end{align}

Front frame:

(31) \begin{align} \begin{array}{c} C_{5}=\left[x_{6}\quad y_{6}\quad z_{6}\quad 1\right]^{T}=e^{{\hat{\xi }_{1}}{\theta _{1}}}e^{{\hat{\xi }_{2}}{\theta _{2}}}{e^{{\hat{\xi }_{3}}{\theta _{3}}}}{e^{{\hat{\xi }_{4}}{\theta _{4}}}}{e^{{\hat{\xi }_{5}}{\theta _{5}}}}c_{5}, \end{array}\end{align}

Electric push rod:

(32) \begin{align} \begin{array}{c} C_{6}=\left[x_{6}\quad y_{6}\quad z_{6}\quad 1\right]^{T}=e^{{\hat{\xi }_{1}}{\theta _{1}}}e^{{\hat{\xi }_{2}}{\theta _{2}}}{e^{{\hat{\xi }_{3}}{\theta _{3}}}}{e^{{\hat{\xi }_{4}}{\theta _{4}}}}{e^{{\hat{\xi }_{5}}{\theta _{5}}}}{e^{{\hat{\xi }_{6}}{\theta _{6}}}}c_{6}, \end{array}\end{align}

Upper lifting rod:

(33) \begin{align} \begin{array}{c} C_{7}=\left[x_{7}\quad y_{7}\quad z_{7}\quad 1\right]^{T}=e^{{\hat{\xi }_{1}}{\theta _{1}}}e^{{\hat{\xi }_{2}}{\theta _{2}}}{e^{{\hat{\xi }_{3}}{\theta _{3}}}}{e^{{\hat{\xi }_{4}}{\theta _{4}}}}{e^{{\hat{\xi }_{5}}{\theta _{5}}}}{e^{{\hat{\xi }_{6}}{\theta _{6}}}}{e^{{\hat{\xi }_{7}}{\theta _{7}}}}c_{7}. \end{array}\end{align}

By combining (26)–(33), during the reconfiguration process, the homogeneous coordinates of the center of mass of the entire robot are obtained as follows:

(34) \begin{align} \begin{array}{c} \begin{array}{c} C=\left[X_{P}\quad Y_{P}\quad Z_{P}\quad 1\right]^{T}\\[5pt] =(C_{0}m_{0}+C_{1}m_{1}+C_{2}m_{2}+C_{3}m_{3}+C_{4}m_{4}\\[5pt] +C_{5}m_{5}+C_{6}m_{6}+C_{7}m_{7})/(m_{0}+m_{1}+\\[5pt] m_{2}+m_{3}+m_{4}+m_{5}+m_{6}+m_{7}). \end{array} \end{array}\end{align}

4.1. Design of motion rule of rotating joint

To avoid impact and damage to the robot and on-board equipment by the motions during reconfiguration of the metamorphic robot, it is necessary for the motions to be smooth. To meet the design requirement that the changes in speed, acceleration, and jerk must be smooth, the angular displacement model of the rotating joint was selected as a fifth-order polynomial [Reference Tan and Wang30]. The relationship between the angular displacement of a rotating joint, $\theta _{i}$ (i = 1–5), and time t can be expressed as follows:

(35) \begin{align} \begin{array}{c} \theta _{i}=a_{0}+a_{1}t+a_{2}t^{2}+a_{3}t^{3}+a_{4}t^{4}+a_{5}t^{5}. \end{array}\end{align}

The boundary conditions of the angular displacement, angular speed, and angular acceleration from the beginning to the end of the motion of a joint are as follows:

(36) \begin{align} \left\{\begin{array}{l} {\theta _{i}\left(0\right)=0}\\[5pt] {\theta _{i}\left(\tau \right)=\Delta \theta _{i}}\\[5pt] {\dot{\theta }_{i}\left(0\right)=0}\\[5pt] {\dot{\theta }_{i}\left(\tau \right)=0}\\[5pt] {\ddot{\theta }_{i}\left(0\right)=0}\\[5pt] {\ddot{\theta }_{i}\left(\tau \right)=0} , \end{array} \right. \end{align}

where $\dot{\theta }_{i}$ , $\ddot{\theta }_{i}$ , and $\Delta \theta _{i}$ represent the angular speed, angular acceleration, and rotation angle of a rotating joint, respectively, during the reconfiguration time $\tau$ .

The following is obtained by combining (35) and (36):

(37) \begin{align} \left\{ \begin{array}{l} {\theta _{i}\left(0\right)=a_{0}=0} \\[5pt] {\theta _{i}\left(\tau \right)=a_{0}+a_{1}\tau +a_{2}\tau ^{2}+a_{3}\tau ^{3}+a_{4}\tau ^{4}+a_{5}\tau ^{5}=\Delta \theta _{i}}\\[5pt] {\dot{\theta }_{i}\left(0\right)=a_{1}=0}\\[5pt] {\dot{\theta }_{i}\left(\tau \right)=a_{1}+2a_{2}\tau +3a_{3}\tau ^{2}+4a_{4}\tau ^{3}+5a_{5}\tau ^{4}=0}\\[5pt] {\ddot{\theta }_{i}\left(0\right)=2a_{2}=0}\\[5pt] {\ddot{\theta }_{i}\left(\tau \right)=2a_{2}+6a_{3}\tau +12a_{4}\tau ^{2}+20a_{5}\tau ^{3}=0}. \end{array} \right. \end{align}

The following is obtained by converting (37) into the form of a matrix equation:

(38) \begin{align} \begin{array}{c} \left[\begin{array}{c@{\quad}c@{\quad}c} \begin{array}{c@{\quad}c} 1 & 0\\[5pt] 1 & \tau \end{array} & \begin{array}{c@{\quad}c} 0 & 0\\[5pt] \tau ^{2} & \tau ^{3} \end{array} & \begin{array}{c@{\quad}c} 0 & 0\\[5pt] \tau ^{4} & \tau ^{5} \end{array}\\[5pt] \begin{array}{c@{\quad}c} 0 & 1\\[5pt] 0 & 1 \end{array} & \begin{array}{c@{\quad}c} 0 & 0\\[5pt] 2\tau & 3\tau ^{2} \end{array} & \begin{array}{c@{\quad}c} 0 & 0\\[5pt] 4\tau ^{3} & 5\tau ^{4} \end{array}\\[5pt] \begin{array}{c@{\quad}c} 0 & 0\\[5pt] 0 & 0 \end{array} & \begin{array}{c@{\quad}c} 2 & 0\\[5pt] 2 & 6\tau \end{array} & \begin{array}{c@{\quad}c} 0 & 0\\[5pt] 12\tau ^{2} & 20\tau ^{3} \end{array} \end{array}\right]\left[\begin{array}{c} a_{0}\\[5pt] a_{1}\\[5pt] a_{2}\\[5pt] a_{3}\\[5pt] a_{4}\\[5pt] a_{5} \end{array}\right]=\left[\begin{array}{c} 0\\[5pt] \Delta \theta _{i}\\[5pt] 0\\[5pt] 0\\[5pt] 0\\[5pt] 0 \end{array}\right], \end{array}\end{align}

from which the following is obtained:

\begin{equation*} \begin{array}{c} \left[\begin{array}{c} a_{0}\\[5pt] a_{1}\\[5pt] a_{2}\\[5pt] a_{3}\\[5pt] a_{4}\\[5pt] a_{5} \end{array}\right]=\left[\begin{array}{c@{\quad}c@{\quad}c} \begin{array}{c@{\quad}c} 0 & 0 \end{array} & \begin{array}{c@{\quad}c} 0 & \dfrac{10\Delta \theta _{i}}{\tau ^{3}} \end{array} & \begin{array}{c@{\quad}c} -\dfrac{15\Delta \theta _{i}}{\tau ^{4}} & \dfrac{6\Delta \theta _{i}}{\tau ^{5}} \end{array} \end{array}\right]^{T}. \end{array} \end{equation*}

The motion model of the angular displacement of a rotating joint is expressed by

(39) \begin{align} \begin{array}{c} \theta _{i}=\dfrac{10\Delta \theta _{i}}{\tau ^{3}}t^{3}-\dfrac{15\Delta \theta _{i}}{\tau ^{4}}t^{4}+\dfrac{6\Delta \theta _{i}}{\tau ^{5}}t^{5}. \end{array}\end{align}

From this, motion models are derived for the angular speed,

(40) \begin{align} \begin{array}{c} \dot{\theta }_{i}=\dfrac{30\Delta \theta _{i}}{\tau ^{3}}t^{2}-\dfrac{60\Delta \theta _{i}}{\tau ^{4}}t^{3}+\dfrac{30\Delta \theta _{i}}{\tau ^{5}}t^{4}, \end{array}\end{align}

the angular acceleration,

(41) \begin{align} \begin{array}{c} \ddot{\theta }_{i}=\dfrac{60\Delta \theta _{i}}{\tau ^{3}}t-\dfrac{180\Delta \theta _{i}}{\tau ^{4}}t^{2}+\dfrac{120\Delta \theta _{i}}{\tau ^{5}}t^{3}, \end{array}\end{align}

and the angular jerk,

(42) \begin{align} \begin{array}{c} \dddot {\theta }_{i}=\dfrac{60\Delta \theta _{i}}{\tau ^{3}}-\dfrac{360\Delta \theta _{i}}{\tau ^{4}}t+\dfrac{360\Delta \theta _{i}}{\tau ^{5}}t^{2}. \end{array}\end{align}

For the case shown in Fig. 4 in which joints 2 and 4 turn clockwise around the Y-axis and joints 1, 3, and 5 turn counterclockwise around the Y-axis, the motion rule for the ith rotating joint is revised as follows:

(43) \begin{align} \begin{array}{c} \theta _{i}=\pm \left(\dfrac{10\Delta \theta _{i}}{\tau ^{3}}t^{3}-\dfrac{15\Delta \theta _{i}}{\tau ^{4}}t^{4}+\dfrac{6\Delta \theta _{i}}{\tau ^{5}}t^{5}\right), \end{array}\end{align}

where the positive sign is for joints 2 and 4 and the negative sign is for joints 1, 3, and 5.

According to the reconfiguration planning of the metamorphic robot, the reconfiguration transitions from support mode to humanoid mode within $\tau =10\,\text{s}$ . The ankle, knee, and lifting joints rotate about their respective joint axes by $\Delta \theta _{1}$ = 81°, $\Delta \theta _{2}$ = 154°, and $\Delta \theta _{4}$ = 72°, respectively. According to (43), the motion rules of the rotation angle of the ankle joint $\theta _{1}$ , the rotation angle of the knee joint $\theta _{2}$ , and the rotation angle of the lifting joint $\theta _{4}$ , can be obtained.

The positions of the centers of mass of the front frame (the end mechanism) and the entire robot can be obtained by substituting $\theta _{1}$ , $\theta _{2}$ , and $\theta _{4}$ into (31) and (34). The second and third derivatives of the variables in (31) and (34) yield the changes in the acceleration and jerk, respectively, of the centers of mass of the front frame and the entire robot. The results are shown in Fig. 5–8. Following the rules of rotational motion for the rotating joints given by (43), the front frame and the entire robot moved smoothly without encountering any shocks in the reconfiguration process. The accelerations and jerks of the centers of mass of the front robot frame and the entire robot in the X-direction were smaller than those in the Z-direction.

Figure 5.

Acceleration of the center of mass of the front frame in the X- and Z-directions.

Figure 6.

Jerk of the center of mass of the front frame in the X- and Z-directions.

Figure 7.

Acceleration of the center of mass of the entire robot in the X- and Z-directions.

Figure 8.

Jerk of the center of mass of the entire robot in the X- and Z-directions.

4.2. Stability criterion of reconfiguration of metamorphic robot

During the process of reconfiguring the metamorphic robot, the components of the robot are subjected to gravity and the inertial forces of the reconfiguration motions. The stability criterion of the reconfiguration process is that the ZMP of the robot is in the support region (the foot region). The ZMP is the intersection of the lines of action of the gravitational, inertial, and frictional forces on the supporting ground during the reconfiguration process, and it must fall within the support region.

The ZMP in a static state is calculated by

(44) \begin{align} \begin{array}{c} \begin{array}{c} X_{zmp}=\dfrac{\sum _{i=1}^{n}m_{i}\left(\ddot{z}_{i}+g\right)x_{i}-\sum _{i=1}^{n}m_{i}\left(\ddot{x}_{i}+g\right)z_{i}}{\sum _{i=1}^{n}m_{i}\left(\ddot{z}_{i}+g\right)},\\[15pt] Y_{zmp}=\dfrac{\sum _{i=1}^{n}m_{i}\left(\ddot{z}_{i}+g\right)y_{i}-\sum _{i=1}^{n}m_{i}\left(\ddot{y}_{i}+g\right)z_{i}}{\sum _{i=1}^{n}m_{i}\left(\ddot{z}_{i}+g\right)}, \end{array} \end{array}\end{align}

where $g$ is the gravitational acceleration; $m_{i}$ is the mass of the ith component; $x_{i}, y_{i}$ , and $z_{i}$ are the coordinates of the center of mass of the ith component in the base coordinate system; $\ddot{x}_{i}, \ddot{y}_{i}$ , and $\ddot{z}_{i}$ are the acceleration values of the center of mass of the ith member in the base coordinate system; $X_{ZMP}$ is the ZMP value corresponding to the X-coordinate; and $Y_{ZMP}$ is the ZMP value corresponding to the Y-coordinate.

The pattern of variation of the ZMP in S during the reconfiguration process is obtained by taking the second derivative of the variables in (26)–(33) and then substituting the results into (44).

4.3. Optimal design of supporting foot landing position

In the reconfiguration process of the metamorphic robot from car mode to humanoid mode, the leg mechanism will be deployed first until the feet are on the ground to form a support mode. The forward kinematics of the robot are analyzed based on the determined foot landing position. Therefore, the foot landing position in support mode has a significant effect on the stability of the reconfiguration of the robot. To improve the stability during reconfiguration, the design of the foot landing positions must be optimized.

During parking, the base coordinate system O is established with the rotation center of the hip joint as the origin, and the coordinate axes of the system point in the same directions as the coordinate axes in S, as shown in Fig. 3. The schematic diagram of the leg mechanism is shown in Fig. 9(a). During the reconfiguration process of the robot, the support region for the robot remains unchanged and is always a closed convex polygon surrounded by the foot support region. The support mode in Fig. 2(b) was taken as the initial mode of the robot and considered with Fig. 3, and the support region of the robot during the reconfiguration process is shown in Fig. 9(b).

Figure 9.

Schematic diagram of the optimal design of the foot landing position: (a) schematic of the leg mechanism and (b) support region for the reconfiguration process.

In Fig. 9(a), x is the horizontal distance between the foot landing position and the origin of O, h is the distance from the origin of O to the ground. In Fig. 9(b), the closed convex polygon ABCDEF is the support region in the robot’s reconfiguration process. S corresponds to the base coordinate system in Fig. 3, and its origin is also the center point, O ₁, of the closed polygon ABCDEF. $x_{f}$ and $x_{r}$ are the front and rear boundaries of the support region along X, respectively, where $x_{f }$ = − $x_{r}$ = 102 mm.

Analysis of the landing motion of the supporting feet shows that, due to the structural design of the leg mechanism and the motion interference, there are two position limits in the X-direction, as shown in Fig. 9(a): the position limit, $x_{1}$ , determined by the interference between the thigh and the hip joint component, and the position limit, $x_{2}$ , determined by interference generated by the thigh and the ground. Thus,

(45) \begin{align} \begin{array}{c} x_{1}\leq x\leq x_{2}, \end{array}\end{align}

where $x_{1}=-90\,\mathrm{mm}$ and $x_{2}=210\,\mathrm{mm}$ .

Figure 9(a) shows that the relationship between the foot landing position and the initial angles $\theta _{1}^{0}$ , $\theta _{2}^{0}$ , and $\theta _{3}^{0}$ of the joints of the leg mechanism in the robot’s support mode are as follows:

(46) \begin{align} \begin{array}{c} \begin{cases} l_{0}+l_{1}\cos \theta _{1}^{0} -l_{2}\cos \theta _{3}^{0} =h\\[5pt] l_{1}\sin \theta _{1}^{0} -l_{2}\sin \theta _{3}^{0} =x\\[5pt] \theta _{2}^{0}=\theta _{3}^{0}-\theta _{1}^{0} \end{cases}, \end{array}\end{align}

where $\theta _{1}^{0}$ , $\theta _{2}^{0}$ , and $\theta _{3}^{0}$ are the angle between the calf rod and the Z-axis, the angle between the calf rod and the thigh rod, and the angle between the thigh rod and the Z-axis, respectively, when the robot is in support mode.

When $x$ is given, substituting it into (46) yields two sets of solutions for $\theta _{1}^{0}$ , $\theta _{2}^{0}$ , and $\theta _{3}^{0}$ . The solution with $\theta _{3}^{0}$ < 0 (the thigh is on the left side of the Z-axis) is discarded (due to restrictions of the thigh movement, $\theta _{3}^{0}$ is always positive).

The rotated angles of each joint of the leg mechanism during the reconfiguration process are as follows:

(47) \begin{align} \begin{array}{c} \begin{cases} \Delta \theta _{1}=\theta _{1}^{0}(x)\\[5pt] \Delta \theta _{2}=180^{\circ}-\theta _{2}^{0}(x)\\[5pt] \Delta \theta _{3}=\theta _{3}^{0}(x)-90^{\circ} \end{cases}. \end{array}\end{align}

The metamorphic robot’s reconfiguration involves movements on the sagittal plane, so the position of each component changes greatly along the X-axis. Therefore, during reconfiguration, with the deviation of the coordinates on the X-axis of the ZMP, $X_{zmp}(t,x)$ , from the center of the support region, as the evaluation metric to optimize the positions, the following optimization model for the foot landing position is obtained:

(48) \begin{align} \min f\left(x\right)=\int _{0}^{T}\left[X_{zmp}\left(t,x\right)-0\right]^{2}dt, \\[5pt] \mathrm{s}.\mathrm{t}.\begin{cases} x_{1}\leq x\leq x_{2}\\[5pt] x_{r}\leq X_{zmp}\left(t,x\right) \leq x_{f} \end{cases}, \end{align}

where the optimization variable x is the foot position in the horizontal direction in O in support mode, $X_{zmp}(t, x)$ is the value of $X_{zmp}$ of the robot with the foot landing position at x in the horizontal direction of O at time t, and T is the reconfiguration time.

The corresponding joint rotation angle $\theta _{i}$ can be calculated using (46), (47), and (43) based on the given foot landing position $x$ . By substituting $\theta _{i}$ into (26)–(33), the center of mass of each moving component of the robot can be obtained. By taking the second derivative, the acceleration of the center of mass of each moving component can be obtained. By substituting the position and acceleration of the center of mass of each moving component into (44) to calculate the ZMP, $X_{zmp}(t,x)$ can be obtained.

Equation (48) is a single-objective optimization problem, which can be optimized using particle swarm optimization (PSO) [Reference Kennedy and Eberhart31–Reference Yan, Lu, Liu and Deng34]. In this study, the number of particle swarms was selected as 20, the number of population update iterations was 800, the acceleration coefficients were both 2, and the optimal value was selected by linearly decreasing the inertia weight from 0.9 to 0.4. The flow chart of the PSO algorithm is shown in Fig. 10.

Figure 10.

Flowchart of the particle swarm optimization (PSO) algorithm.

The optimal foot landing position was $x$ = 87.5 mm. Substituting this into (45) yields the rotation angles of the joints, when the robot was in support mode: $\theta _{1}^{0}$ = 17.5°, $\theta _{2}^{0}$ = 21.5°, and $\theta _{3}^{0}$ = 4°. Furthermore, the ZMP variation curve of the metamorphic robot during the reconfiguration process can be obtained, and the results are shown in Fig. 11.

Figure 11.

Zero-moment point (ZMP) of the entire robot during the reconfiguration process after the optimization.

With the optimal foot landing, the changes of the centers of mass of the front frame and the entire robot are shown in Fig. 12. The comparison of the stability results before and after the optimization of the foot landing position are shown in Fig. 13. After the foot landing position was optimized, the ZMP was always inside the support region during the robot’s reconfiguration process. During the first 8 s of the reconfiguration process, compared with the unoptimized value, the optimized $X_{zmp}$ values were further away from the lower boundary of the support region and closer to the center of the support region. The results of $X_{zmp}$ before and after the optimization agreed only during the last 8–10 s of the reconfiguration process. Therefore, the optimized metamorphic robot was more stable during the reconfiguration process.

Figure 12.

Trajectories of the centers of mass of the (a) front frame and (b) entire robot in the ZX plane.

Figure 13.

$X_{zmp}$ for the entire robot during the reconfiguration before and after the optimization.