2.1. Model assumptions

Figure 1 top shows the overview of the RW-type walker. This model is the same as the CRW, in which two identical eight-legged RWs with a symmetrical shape are combined via a rigid body frame. The relative angles between the two adjacent leg frames are $\alpha = \frac{\pi }{4}$ [rad], and the end positions of the rear and foreleg frames are connected by massless rods to always perfectly synchronize the absolute leg angles. In the following, the parallelogram link in contact with the ground, filled with pink in the figure, is called the stance leg, the lower massless rod of it touching the ground is called the sole, and the ground-contact points of the rear/fore RW are called the rear/fore foot. A reaction wheel is attached to the center of the body frame via an active joint, and this is used to appropriately control the ratio of the vertical ground reaction forces acting at the rear and forefeet.

Figure 1 bottom left shows the coordinate system of the robot. The coordinate system has $X$ -axis horizontal forward and $Z$ -axis vertical up with the origin fixed to the ground. $(x,z)$ is the rear foot position and $(x',z')$ is the fore foot position. $\theta _1$ and $\theta _3$ are the absolute angles of the rear and fore RWs relative to the vertical up direction, and $\theta _2$ and $\theta _4$ are those of the body frame and reaction wheel with respect to the horizontal forward direction. The robot has four masses: the mass of the rear RW is $m_1$ , that of the body frame is $m_2$ , that of the fore RW is $m_3$ , and that of the reaction wheel is $m_4$ . The inertia moment around $m_j$ is $I_j \ (j \in \{1,2,3,4\})$ . $L_1$ is the radius of the rear RW, $L_2$ is half the length of the body frame, and $L_3$ is the radius of the fore RW. We assume that the two RWs are the same, that is, $m_1 = m_3$ , $I_1 = I_3$ and $L_1 = L_3$ , and that they move in perfect synchronization with each other or $\theta _1 \equiv \theta _3$ and $\dot{\theta }_1 \equiv \dot{\theta }_3$ always hold during motion. In the following, these assumptions are used as necessary in the equations and explanations.

As illustrated in Figure 1 bottom right, the robot has two control inputs; $u_1$ is the rotation torque between the rear RW and body frame used to control the stance-leg motion, and $u_2$ is the rotation torque between the body frame and reaction wheel to control the ratio of rear and fore vertical ground reaction forces, $F_z$ and $F^{\prime}_z$ . Actually, the rear RW is heavier by the amount of the actuator for applying $u_1$ , but the same thing is attached to the fore RW as a counterweight to balance it. Since the control torque $u_1$ can act as a rotation torque to the rear RW around point P or that to the fore RW around P’ as shown in the figure, it would be possible to control the stance leg as 1-DOF robot arm when the ZMP is in the sole or $F_z \gt 0$ and $F^{\prime}_z \gt 0$ .

2.2. Equation of motion, constraint conditions, and reduced dynamics

Let $\boldsymbol{q} = \left [ \begin{array}{cccccc} x & z & \theta _1 & \theta _2 & \theta _3 & \theta _4 \end{array} \right ]^{\textrm{T}}$ be the generalized coordinate vector. The robot equation of motion then becomes

(1) \begin{equation}\boldsymbol{M} \ddot{\boldsymbol{q}} +\boldsymbol{h} =\boldsymbol{S}\boldsymbol{u} +\boldsymbol{J}_c^{\textrm{T}}{\boldsymbol{\lambda }}_c, \end{equation}

where $\boldsymbol{M} \in{{\mathbb{R}}}^{6 \times 6}$ is the inertia matrix, $\boldsymbol{h} \in{{\mathbb{R}}}^6$ is the nonlinear velocity and gravity vector, and $\boldsymbol{S}\boldsymbol{u} \in{{\mathbb{R}}}^6$ is the control input vector. They are detailed as follows.

(2) \begin{eqnarray} &\boldsymbol{M} = \left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} m & 0 & m L_1 \cos \theta _1 & -m L_2 \sin \theta _2 & 0 & 0 \\ 0 & m & -m L_1 \sin \theta _1 & -m L_2 \cos \theta _2 & 0 & 0 \\ m L_1 \cos \theta _1 & -m L_1 \sin \theta _1 & I_1 + m L_1^2 & m L_1 L_2 \sin (\theta _1 - \theta _2) & 0 & 0 \\ -m L_2 \sin \theta _2 & -m L_2 \cos \theta _2 & m L_1 L_2 \sin (\theta _1 - \theta _2) & I_2 + m L_2^2 & 0 & 0 \\ 0 & 0 & 0 & 0 & I_3 & 0 \\ 0 & 0 & 0 & 0 & 0 & I_4 \end{array} \right ] & \end{eqnarray}

(3) \begin{eqnarray} &\boldsymbol{h} = \left [ \begin{array}{c} - m \left ( L_1 \dot{\theta }_1^2 \sin \theta _1 + L_2 \dot{\theta }_2^2 \cos \theta _2 \right ) \\ -m \left ( L_1 \dot{\theta }_1^2 \cos \theta _1 - L_2 \dot{\theta }_2^2 \sin \theta _2 - g \right ) \\ -m L_1 \left ( L_2 \dot{\theta }_2^2 \cos ( \theta _1 - \theta _2) + g \sin \theta _1 \right ) \\ m L_2 \left ( L_1 \dot{\theta }_1^2 \cos (\theta _1 - \theta _2) - g \cos \theta _2 \right ) \\ 0 \\ 0 \end{array} \right ], \ \ \boldsymbol{S} = \left [ \begin{array}{c@{\quad}c} 0 & 0 \\ 0 & 0 \\ 1 & 0 \\ -1 & 1 \\ 0 & 0 \\ 0 & -1 \end{array} \right ], \ \ \boldsymbol{u} = \left [ \begin{array}{c} u_1 \\ u_2 \end{array} \right ] & \end{eqnarray}

The velocity constraint conditions that the rear and forefeet are in contact with the ground without sliding are described as

(4) \begin{align} \dot{x} &= 0, \end{align}

(5) \begin{align} \dot{z} &= 0, \end{align}

(6) \begin{align} \dot{x}' &= \frac{\textrm{d}}{\textrm{d}t} \left ( x + L_1 \sin \theta _1 + 2 L_2 \cos \theta _2 - L_3 \sin \theta _3 \right ) = 0, \end{align}

(7) \begin{align} \dot{z}' &= \frac{\textrm{d}}{\textrm{d}t} \left ( z + L_1 \cos \theta _1 - 2 L_2 \sin \theta _2 - L_3 \cos \theta _3 \right ) = 0. \end{align}

Summarizing these four equations becomes

(8) \begin{equation}\boldsymbol{J}_c \dot{\boldsymbol{q}} = \textbf{0}_{4 \times 1}, \end{equation}

where

(9) \begin{equation}\boldsymbol{J}_c = \left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & L_1 \cos \theta _1 & -2L_2 \sin \theta _2 & -L_3 \cos \theta _3 & 0 \\ 0 & 1 & -L_1 \sin \theta _1 & -2L_2 \cos \theta _2 & L_3 \sin \theta _3 & 0 \end{array} \right ]. \end{equation}

On a slippery road surface, to perfectly synchronize the rear and fore RWs, we must introduce the condition $\dot{\theta }_1 = \dot{\theta }_3$ derived from the massless rod instead of the conditions $\dot{x} = 0$ and $\dot{x}' = 0$ [Reference Asano, Chen and Liu22, Reference Asano23]. On a non-slip road surface, however, this is not necessary because the above four conditions include the physical meaning of it.

By solving Eqs. (1) and (8) for ${\boldsymbol{\lambda }}_c$ , we obtain

(10) \begin{equation}{\boldsymbol{\lambda }}_c = \left [ \begin{array}{c} F_x \\ F_z \\ F^{\prime}_x \\ F^{\prime}_z \end{array} \right ] = -\boldsymbol{X}_c^{-1} \left (\boldsymbol{J}_c\boldsymbol{M}^{-1} \left (\boldsymbol{S}\boldsymbol{u} -\boldsymbol{h} \right ) + \dot{\boldsymbol{J}}_c \dot{\boldsymbol{q}} \right ), \end{equation}

where $\boldsymbol{X}_c \;:\!=\;\boldsymbol{J}_c\boldsymbol{M}^{-1}\boldsymbol{J}_c^{\textrm{T}}$ . By substituting Eq. (10) into Eq. (1) and arranging it, we obtain the simplified robot equation of motion as

(11) \begin{equation}\boldsymbol{M} \ddot{\boldsymbol{q}} =\boldsymbol{Y}_c \left (\boldsymbol{S}\boldsymbol{u} -\boldsymbol{h} \right ) -\boldsymbol{J}_c^{\textrm{T}}\boldsymbol{X}_c^{-1} \dot{\boldsymbol{J}}_c \dot{\boldsymbol{q}}, \end{equation}

where $\boldsymbol{Y}_c \;:\!=\;\boldsymbol{I}_6 -\boldsymbol{J}_c^{\textrm{T}}\boldsymbol{X}_c^{-1}\boldsymbol{J}_c\boldsymbol{M}^{-1}$ .

The robot model has six generalized coordinates, but it has four constraints, so it is essentially 2-DOF, $\theta _1$ and $\theta _4$ . The essential (reduced) dynamics then becomes

(12) \begin{equation} \left [ \begin{array}{c@{\quad}c} I_5 & 0 \\ 0 & I_4 \end{array} \right ] \left [ \begin{array}{c}{\ddot{\theta }}_1 \\{\ddot{\theta }}_4 \end{array} \right ] + \left [ \begin{array}{c} -m g L_1 \sin \theta _1 \\ 0 \end{array} \right ] = \left [ \begin{array}{c} u_1 \\ -u_2 \end{array} \right ], \end{equation}

where $I_5 \;:\!=\; 2 I_1 + m L_1^2$ and $m \;:\!=\; 2 m_1 + m_2 + m_4$ are the robot’s total mass. Eq. (12) seems like a decoupled system, but it is shown to be equivalent to an URW capable of ZMP control through transformation of the control input as described later.

2.3. Collision equation

On the assumption that the rear feet take off immediately after the forefeet collide with the ground, the perfect inelastic collision equation is derived as

(13) \begin{equation}\boldsymbol{M} \dot{\boldsymbol{q}}^+ =\boldsymbol{M} \dot{\boldsymbol{q}}^- +\boldsymbol{J}_I^{\textrm{T}}{\boldsymbol{\lambda }}_I. \end{equation}

Here, the superscripts “ $-$ ” and “ $+$ ” denote immediately before and immediately after impact. The velocity constraint conditions of the forefeet immediately after impact are specified as

(14) \begin{align} \frac{\textrm{d}}{\textrm{d}t} \left ( x + L_1 \sin \theta _1 - L_1 \sin \left ( \theta _1 - \alpha \right ) \right )^+ &= 0, \end{align}

(15) \begin{align} \frac{\textrm{d}}{\textrm{d}t} \left ( z + L_1 \cos \theta _1 - L_1 \cos \left ( \theta _1 - \alpha \right ) \right )^+ &= 0, \end{align}

(16) \begin{align} \frac{\textrm{d}}{\textrm{d}t} \left ( x + 2 L_2 + L_3 \sin \theta _3 - L_3 \sin \left ( \theta _3 - \alpha \right ) \right )^+ &= 0, \end{align}

(17) \begin{align} \frac{\textrm{d}}{\textrm{d}t} \left ( z + L_3 \cos \theta _3 - L_3 \cos \left ( \theta _3 - \alpha \right ) \right )^+ &= 0. \end{align}

Note, however, that since leg exchange is not taken into account here, $\theta _1$ and $\theta _3$ themselves remain at their values immediately before the collision but their time derivatives are the values immediately after the collision. Summarizing the above four equations becomes

(18) \begin{equation}\boldsymbol{J}_I \dot{\boldsymbol{q}}^+ = \textbf{0}_{4 \times 1}, \end{equation}

where

(19) \begin{equation}\boldsymbol{J}_I = \left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 1 & 0 & L_1 \cos \theta _1^- - L_1 \cos \left ( \theta _1^- - \alpha \right ) & 0 & 0 & 0 \\ 0 & 1 & -L_1 \sin \theta _1^- + L_1 \sin \left ( \theta _1^- - \alpha \right ) & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & L_3 \cos \theta _3^- - L_3 \cos \left ( \theta _3^- - \alpha \right ) & 0 \\ 0 & 1 & 0 & 0 & -L_3 \sin \theta _3^- + L_3 \sin \left ( \theta _3^- - \alpha \right ) & 0 \end{array} \right ]. \end{equation}

By solving Eqs. (13) and (18) for the velocity vector immediately after impact, we obtain

(20) \begin{equation} \dot{\boldsymbol{q}}^+ = \left (\boldsymbol{I}_6 -\boldsymbol{M}^{-1}\boldsymbol{J}_I^{\textrm{T}} \left (\boldsymbol{J}_I\boldsymbol{M}^{-1}\boldsymbol{J}_I^{\textrm{T}} \right )^{-1}\boldsymbol{J}_I \right ) \dot{\boldsymbol{q}}^-. \end{equation}

After this, with the update of the ground-contact point, we must reset the first and second components of $\dot{\boldsymbol{q}}^+$ as $\dot{x}^+ = 0$ and $\dot{z}^+ = 0$ . The update formula for the angular velocity of the stance leg is obtained by extracting the relevant part from Eq. (20) and is described as

(21) \begin{equation} \dot{\theta }_1^+ = \xi \dot{\theta }_1^-, \ \ \xi = \frac{m L_1^2 \cos \alpha + 2 I_1}{m L_1^2 + 2 I_1}. \end{equation}

Here, $\xi$ is a positive constant less than $1$ . Finally, the position coordinates are reset to

(22) \begin{equation} x^+ = x^- + 2 L_1 \sin \frac{\alpha }{2}, \ \ z^+ = 0, \ \ \theta _1^+ = \theta _3^+ = \theta _1^- - \alpha = \theta _3^- - \alpha = - \frac{\alpha }{2}. \end{equation}

3.1. Target conditions and determination of control input

For achieving the desired walking motion, the stance-leg motion must be generated with the highest priority. First, we set the stance-leg angle to the control output which is described as $\theta _1 = \left [ \begin{array}{cccccc} 0\; & 0\; & 1\; & 0\; & 0\; & 0 \end{array} \right ]\boldsymbol{q} \;=\!:\; \boldsymbol{C}_1\boldsymbol{q}$ . We then consider the following target condition so that the second-order derivative with respect to time is equal to the angular acceleration command signal, $v[i]$ .

(23) \begin{eqnarray}{\ddot{\theta }}_1 &=&\boldsymbol{C}_1 \ddot{\boldsymbol{q}} \equiv v[i] \nonumber \\ &=&\boldsymbol{C}_1\boldsymbol{M}^{-1}\boldsymbol{Y}_c \left (\boldsymbol{S}\boldsymbol{u} -\boldsymbol{h} \right ) -\boldsymbol{C}_1\boldsymbol{M}^{-1}\boldsymbol{J}_c^{\textrm{T}}\boldsymbol{X}_c^{-1} \dot{\boldsymbol{J}}_c \dot{\boldsymbol{q}} \end{eqnarray}

Second, we consider the following condition so that the vertical ground reaction forces of the rear and forelegs are always equal, that is, the ZMP is always positioned at the center point of the sole.

(24) \begin{equation} 0 \equiv F_z - F^{\prime}_z = \left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c} 0 & 1 & 0 & -1 \end{array} \right ]{\boldsymbol{\lambda }}_c \;=\!:\; \boldsymbol{C}_2{\boldsymbol{\lambda }}_c \end{equation}

By substituting Eq. (10) into Eq. (24), we obtain

(25) \begin{equation}\boldsymbol{C}_2\boldsymbol{X}_c^{-1}\boldsymbol{J}_c\boldsymbol{M}^{-1}\boldsymbol{S}\boldsymbol{u} =\boldsymbol{C}_2\boldsymbol{X}_c^{-1} \left (\boldsymbol{J}_c\boldsymbol{M}^{-1}\boldsymbol{h} + \dot{\boldsymbol{J}}_c \dot{\boldsymbol{q}} \right ). \end{equation}

The target condition is then determined as ${\boldsymbol{\Phi }}\boldsymbol{u} ={\boldsymbol{\Gamma }}$ where

(26) \begin{align}{\boldsymbol{\Phi }} &= \left [ \begin{array}{c}\boldsymbol{C}_1\boldsymbol{M}^{-1}\boldsymbol{Y}_c\boldsymbol{S} \\\boldsymbol{C}_2\boldsymbol{X}_c^{-1}\boldsymbol{J}_c\boldsymbol{M}^{-1}\boldsymbol{S} \end{array} \right ] = \left [ \begin{array}{c@{\quad}c} \frac{1}{m L_1^2 + 2 I_1} & 0 \\ -\frac{1}{L_2} & \frac{1}{L_2} \end{array} \right ], \end{align}

(27) \begin{align} {\boldsymbol{\Gamma }} &= \left [ \begin{array}{c} v[i] +\boldsymbol{C}_1\boldsymbol{M}^{-1} \left (\boldsymbol{Y}_c\boldsymbol{h} +\boldsymbol{J}_c^{\textrm{T}}\boldsymbol{X}_c^{-1} \dot{\boldsymbol{J}}_c \dot{\boldsymbol{q}} \right ) \\\boldsymbol{C}_2\boldsymbol{X}_c^{-1} \left (\boldsymbol{J}_c\boldsymbol{M}^{-1}\boldsymbol{h} + \dot{\boldsymbol{J}}_c \dot{\boldsymbol{q}} \right ) \end{array} \right ] = \left [ \begin{array}{c} v[i] - \frac{m g L_1 \sin \theta _1}{mL_1^2 + 2I_1} \\ 0 \end{array} \right ]. \end{align}

The control input solved is then solved as

(28) \begin{equation}\boldsymbol{u} ={\boldsymbol{\Phi }}^{-1}{\boldsymbol{\Gamma }} = \left [ \begin{array}{c} 1 \\ 1 \end{array} \right ] \left ( (mL_1^2 + 2I_1) v[i] - m g L_1 \sin \theta _1 \right ) . \end{equation}

From this, it is immediately clear that $u_1 = u_2$ , but this result is derived from solving ${\boldsymbol{\Phi }}\boldsymbol{u} ={\boldsymbol{\Gamma }}$ so as to satisfy the condition $F_z = F^{\prime}_z$ . We will discuss this issue again in the next section.

3.2. Stance-leg motion generation

3.2.1. Phase I

The first phase starting from a static standing posture until immediately before impact is defined as Phase I. In this phase, the angular acceleration command signal $v$ is set to a positive constant value $v[0]$ , that is, ${\ddot{\theta }}_1 = v [0]$ . The angular velocity and angular position at $t$ then become

(29) \begin{equation} \dot{\theta }_1 (t) = v[0] t, \ \ \theta _1 (t) = \frac{v[0] t^2}{2}. \end{equation}

Let $T_1$ be the target period of this phase. The target terminal angular position in this phase then satisfies

(30) \begin{equation} \theta _1 \left ( T_1^- \right ) = \frac{v[0] T_1^2}{2} = \frac{\alpha }{2}. \end{equation}

By solving Eq. (24) for $v[0]$ , we obtain

(31) \begin{equation} v[0] = \frac{\alpha }{T_1^2}. \end{equation}

The terminal angular velocity in this phase also becomes

(32) \begin{equation} \dot{\theta }_1 \left ( T_1^- \right ) = v[0] T_1 = \frac{\alpha }{T_1}. \end{equation}

3.2.2. Phase II

The second phase starting immediately after impact is defined as Phase II. The angular position and angular velocity of the stance leg immediately after impact are reset to

(33) \begin{align} \theta _1 \left ( T_1^+ \right ) &= \theta _1 \left ( T_1^- \right ) - \alpha = -\frac{\alpha }{2}, \end{align}

(34) \begin{align} \dot{\theta }_1 \left ( T_1^+ \right ) &= \xi \dot{\theta }_1 \left ( T_1^- \right ), \end{align}

and these are the initial condition in this phase. The state-space realization of ${\ddot{\theta }}_1 = v[i]$ becomes

(35) \begin{equation} \frac{\textrm{d}}{\textrm{d}t} \left [ \begin{array}{c} \theta _1 \\ \dot{\theta }_1 \end{array} \right ] = \left [ \begin{array}{c@{\quad}c} 0 & 1 \\ 0 & 0 \end{array} \right ] \left [ \begin{array}{c} \theta _1 \\ \dot{\theta }_1 \end{array} \right ] + \left [ \begin{array}{c} 0 \\ 1 \end{array} \right ] v[i]. \end{equation}

Let $T_2$ be the target period of this phase, and $\theta _1$ be the control output. We then consider DODC [Reference Asano and Xiao4] for settling $\theta _1$ and $\dot{\theta }_1$ to zero in two steps. The system of Eq. (30) is discretized with the control period to $T_2/2$ as

(36) \begin{equation} \left [ \begin{array}{c} \theta _1[i+1] \\ \dot{\theta }_1[i+1] \end{array} \right ] = \left [ \begin{array}{c@{\quad}c} 1 & \frac{T_2}{2} \\ 0 & 1 \end{array} \right ] \left [ \begin{array}{c} \theta _1[i] \\ \dot{\theta }_1[i] \end{array} \right ] + \left [ \begin{array}{c} \frac{T_2^2}{8} \\ \frac{T_2}{2} \end{array} \right ] v[i]. \end{equation}

We denote this system as $\boldsymbol{x}_s [i+1] =\boldsymbol{A}_s\boldsymbol{x}_s [i] +\boldsymbol{B}_s v[i]$ . Note that the state variables of the discretized system were defined as

(37) \begin{align}\boldsymbol{x}_s [1] &= \left [ \begin{array}{c} \theta _1 \left ( T_1^+ \right ) \\ \dot{\theta }_1 \left ( T_1^+ \right ) \end{array} \right ] = \left [ \begin{array}{c} -\frac{\alpha }{2} \\ \frac{\alpha \xi }{T_1} \end{array} \right ], \end{align}

(38) \begin{align}\boldsymbol{x}_s [i+1] &= \left [ \begin{array}{c} \theta _1 \left ( T_1^+ + i \frac{T_2}{2} \right ) \\ \dot{\theta }_1 \left ( T_1^+ + i \frac{T_2}{2} \right ) \end{array} \right ] \ \ \left ( i \geq 1 \right ). \end{align}

The feedback gain $\boldsymbol{F}$ that assigns all poles of the closed-loop system $\boldsymbol{A}_s +\boldsymbol{B}_s\boldsymbol{F}$ at zero can be obtained as $\boldsymbol{F} = \left [ \begin{array}{cc} -\frac{4}{T_2^2}\; & -\frac{3}{T_2} \end{array} \right ]$ . The angular acceleration command signals in this phase are then determined as

(39) \begin{align} v[1] &=\boldsymbol{F}\boldsymbol{x}_s [1] \ \ \left ( T_1^+ \leq t \lt T_1^+ + \frac{T_2}{2} \right ), \end{align}

(40) \begin{align} v[2] &=\boldsymbol{F}\boldsymbol{x}_s [2] \ \ \left ( T_1^+ + \frac{T_2}{2} \leq t \lt T_1^+ + T_2 \right ). \end{align}

According to the method of DODC, $\boldsymbol{x}_s [3] = \textbf{0}_{2 \times 1}$ is achieved.

4.1. Relationship between control inputs and zero-moment point

The difference between the vertical ground reaction force of the rear leg and that of the fore leg satisfies the following relationship.

(41) \begin{equation} F_z - F^{\prime}_z = \frac{u_1 - u_2}{L_2} \end{equation}

Therefore, we can understand that $F_z \equiv F^{\prime}_z$ is equivalent to $u_1 \equiv u_2$ . Considering Eq. (41), the control input vector for the reduced system can be arranged to

(42) \begin{equation} \left [ \begin{array}{c} u_1 \\ -u_2 \end{array} \right ] = \left [ \begin{array}{c} u_1 \\ -u_1 \end{array} \right ] + \left [ \begin{array}{c} 0 \\ u_1 - u_2 \end{array} \right ] = \left [ \begin{array}{c} 1 \\ -1 \end{array} \right ] u_1 + \left [ \begin{array}{c} 0 \\ \left ( F_z - F^{\prime}_z \right ) L_2 \end{array} \right ], \end{equation}

and this shows that $F_z \equiv F^{\prime}_z$ does not hold only when $u_1 \equiv u_2$ does not hold. The second term of Eq. (42) means the degree of freedom of control input caused by increasing the number of the stance legs to two.

4.2. Approximate analytical solution of $T_1$

In this subsection, we derive an approximate analytical solution of $T_1$ for stabilizing the zero dynamics of the reaction wheel when $u_1 \equiv u_2$ . The equation of motion of the reaction wheel is

(43) \begin{equation} I_4{\ddot{\theta }}_4 = -u_2, \end{equation}

and the time integral for one step where $u_1 = u_2$ becomes

(44) \begin{equation} I_4 \left ( \dot{\theta }_{4} (T) - \dot{\theta }_{4} (0) \right ) = - \int _{0}^{T} u_1 \ \textrm{d}t, \end{equation}

where $T \;:\!=\; T_1 + T_2$ is the step period. Eq. (44) clearly shows that the necessary condition for the initial and terminal angular velocities to be equal becomes

(45) \begin{equation} J\;:\!=\; \int _{0}^{T} u_1 \ \textrm{d}t = 0. \end{equation}

An approximate analytical solution of $J$ can be derived by using the linearized reduced model [Reference Asano and Kawai8] and is specified as

(46) \begin{equation} J= \frac{\alpha I_5 (1-\xi )}{T_1} - \frac{\alpha m g L_1}{12 T_1} \left ( 2 T_1^2 - 3 T_1 T_2 + T_2^2 \xi \right ). \end{equation}

By solving $J=0$ for $T_1$ , we obtain

(47) \begin{equation} T_1= \frac{3 T_2}{4} + \sqrt{ \frac{ mgL_1 T_2^2 \left ( 9 - 8 \xi \right ) + 96 I_5 \left ( 1 - \xi \right ) }{16mgL_1}}. \end{equation}

Table I.

Physical and control parameters.

Figure 2.

Simulation results of limit cycle walking where $A_\Delta =0$ and $b_\Delta = 0$ .

Figure 3.

Snapshots of generated gait in Figure 2.

Figure 4.

Simulation results of limit cycle walking where $A_\Delta = -0.0195577$ and $b_{\Delta } = 0.00060233$ .

Figure 5.

Magnified views of Figure 4(d).

Figure 6.

Phase-plane plot of generated motions.

4.3. Correction of resultant terminal state of reaction wheel

The state-space realization of Eq. (43) becomes

(48) \begin{equation} \frac{\textrm{d}}{\textrm{d}t} \left [ \begin{array}{c} \theta _4 \\ \dot{\theta }_4 \end{array} \right ] = \left [ \begin{array}{c@{\quad}c} 0 & 1 \\ 0 & 0 \end{array} \right ] \left [ \begin{array}{c} \theta _4 \\ \dot{\theta }_4 \end{array} \right ] + \left [ \begin{array}{c} 0 \\ -I_4^{-1} \end{array} \right ] u_2, \end{equation}

and we denote Eq. (48) as

(49) \begin{equation} \dot{\boldsymbol{x}}_r =\boldsymbol{A}_r\boldsymbol{x}_r +\boldsymbol{B}_r u_2. \end{equation}

In this paper, we consider adding a sinusoidal control input with an offset to $u_2$ as follows.

(50) \begin{equation} u_2 = u_1 + A_{\Delta } \sin \left ( \frac{2 \pi t}{T} \right ) + b_{\Delta } \end{equation}

When $A_{\Delta } = 0$ and $b_{\Delta } = 0$ , due to the gap between the linearized and nonlinear models, the resultant terminal angular velocity of the reaction wheel becomes off zero only a little, whereas the resultant terminal angular velocity is not immediately determinable. Let $\Delta \theta _4$ and $\Delta \dot{\theta }_4$ be the resultant terminal values of the angular position and angular velocity of the reaction wheel when starting from the origin on the phase plane. Then, the analytical solution of Eq. (49) with $\boldsymbol{x}_r (0) = \textbf{0}_{2 \times 1}$ where $A_{\Delta } = 0$ and $b_{\Delta } = 0$ or $u_2 = u_1$ should satisfy

(51) \begin{equation} \left [ \begin{array}{c} \Delta \theta _4 \\ \Delta \dot{\theta }_4 \end{array} \right ] = \int _{0}^{T}{{\textrm{e}}}^{\boldsymbol{A}_r \left ( T - t \right )}\boldsymbol{B}_r u_1 \ \textrm{d}t. \end{equation}

To control these values to zero, the sinusoidal control input should satisfy the following relationship.

(52) \begin{equation} \left [ \begin{array}{c} -\Delta \theta _4 \\ -\Delta \dot{\theta }_4 \end{array} \right ] = \int _{0}^{T}{{\textrm{e}}}^{\boldsymbol{A}_r \left ( T - t \right )}\boldsymbol{B}_r \left ( A_\Delta \sin \left ( \frac{2 \pi t}{T} \right ) + b_\Delta \right ) \textrm{d}t \end{equation}

By solving this for $A_\Delta$ and $b_\Delta$ , we obtain

(53) \begin{equation} A_\Delta = \frac{I_4 \pi \left ( 2 \Delta \theta _4 - T \Delta \dot{\theta }_4 \right )}{T^2}, \ \ b_\Delta = \frac{I_4 \Delta \dot{\theta }_4}{T}. \end{equation}