Symmetry in legged locomotion: a new method for designing stable periodic gaits

Abstract

This paper introduces a method for achieving stable periodic walking for legged robots. This method is based on producing a type of odd-even symmetry in the system. A hybrid system with such symmetries is called a symmetric hybrid system (SHS). We discuss the properties of an SHS and, in particular, will show that an SHS can have an infinite number of synchronized periodic orbits. We describe how controllers can be obtained to make a legged robot an SHS. Then the stability of the synchronized periodic orbits of this SHS is studied, where the notion of self-synchronization is introduced. We show that such self-synchronized periodic orbits are neutrally stable in kinetic energy. As the final step in the process of achieving asymptotically stable periodic walking, we show how by introducing asymmetries (such as energy loss at impact) in the system, the synchronized periodic orbits of this SHS can be turned into asymptotically stable periodic orbits. Many numerical examples, including an 8-DOF 3D biped with 2 degrees of underactuation, are studied to demonstrate the effectiveness of the method.

Appendices

Appendix 1

Proposition 32

Suppose that the SHS has a synchronized solution $(\varPhi (t), \varPsi (t))$ with initial conditions $(\varPhi (0), \varPsi (0)) = (-\phi _0, \psi _0)$ and $({\dot{\varPhi }}(0), {\dot{\varPsi }}(0)) = ({\dot{\varPhi }}_0, {\dot{\varPsi }}_0)$. Suppose that $t = t_m$ is the time for which $\varPhi (t_m) = 0$ and ${\dot{\varPsi }}(t_m) = 0$ such that ${\dot{\varPhi }}(t_m) \ne 0$ . Let $\chi ({\dot{\phi }}_0,{\dot{\psi }}_0,t)$ be the flow map of the $(\phi _0,\psi _0)$-invariant SHS. Therefore, $\chi ({\dot{\phi }}_0,{\dot{\psi }}_0,0) = (-\phi _0,\psi _0,{\dot{\phi }}_0,{\dot{\psi }}_0)$ for every $({\dot{\phi }}_0,{\dot{\psi }}_0) \in {\mathcal {T}}_{(-\phi _0,\psi _0)}{\mathcal {Q}}$. Let $\chi = (\phi ,\psi ,{\dot{\phi }},{\dot{\psi }})$. Since, $(\varPhi (t), \varPsi (t))$ is synchronized,

$$\begin{aligned} \phi ({\dot{\varPhi }}_0, {\dot{\varPsi }}_0, t_m) = 0,~ {\dot{\psi }}({\dot{\varPhi }}_0, {\dot{\varPsi }}_0, t_m) = 0. \end{aligned}$$

If the Jacobian of $(\phi ,{\dot{\psi }})$ with respect to $({\dot{\phi }}_0,{\dot{\psi }}_0)$ is invertible at $({\dot{\varPhi }}_0, {\dot{\varPsi }}_0, t_m)$, then there exists a smooth function $L:{\mathcal {T}}_{(-\phi _0,\psi _0)}{\mathcal {Q}}\rightarrow {\mathbb {R}}^{m+n-1}$ such that if $L({\dot{\phi }}_0, {\dot{\psi }}_0) = 0$, the solution starting from $(-\phi _0,\psi _0)$ with initial velocity $({\dot{\phi }}_0, {\dot{\psi }}_0)$ is synchronized.

Proof

Based on the definition of a synchronized solution, we are interested in the values of $({\dot{\phi }}_0, {\dot{\psi }}_0, t)$ for which

$$\begin{aligned} \phi ({\dot{\phi }}_0,{\dot{\psi }}_0, t) = 0,~ {\dot{\psi }}({\dot{\phi }}_0,{\dot{\psi }}_0, t) = 0. \end{aligned}$$

Since $({\dot{\varPhi }}_0, {\dot{\varPsi }}_0, t_m)$ is a solution to this system, we have

$$\begin{aligned} \phi ({\dot{\varPhi }}_0, {\dot{\varPsi }}_0, t_m) = 0, ~{\dot{\psi }}({\dot{\varPhi }}_0, {\dot{\varPsi }}_0, t_m) = 0. \end{aligned}$$

Because the Jacobian of $(\phi ,{\dot{\psi }})$ with respect to $({\dot{\phi }}_0,{\dot{\psi }}_0)$ is invertible at $({\dot{\varPhi }}_0, {\dot{\varPsi }}_0, t_m)$, by implicit function theorem there exists a smooth function F defined in a neighborhood of $t_m$ such that in this neighborhood

$$\begin{aligned} \phi (F(t),t) = 0, ~{\dot{\psi }}(F(t),t) = 0. \end{aligned}$$

(60)

The function $t\mapsto F(t)$ defines a smooth curve in the manifold ${\mathcal {T}}_{(-\phi _0,\psi _0)}{\mathcal {Q}}$, where $F(t_m) = ({\dot{\varPhi }}_0, {\dot{\varPsi }}_0)$. Differentiating the two equations in (60) with respect to t at $t = t_m$, since ${\dot{\varPhi }}(t_m) \ne 0$, we can show that ${\dot{F}}(t_m) \ne 0$. Therefore, the parametrization $t\mapsto F(t)$ is a regular parametrization in a neighborhood of $t_m$, and, hence, the image of $t\mapsto F(t)$ defines an embedded 1-dimensional submanifold ${\mathcal {K}}$ of ${\mathcal {T}}_{(-\phi _0,\psi _0)}{\mathcal {Q}}$ (Lee 2003). Thus, there exists a smooth function $L:{\mathcal {T}}_{(-\phi _0,\psi _0)}{\mathcal {Q}}\rightarrow {\mathbb {R}}^{m+n-1}$ with rank $m+n-1$ such that

$$\begin{aligned} {\mathcal {K}} = \{({\dot{\phi }}_0,{\dot{\psi }}_0) \in {\mathcal {T}}_{(-\phi _0,\psi _0)}{\mathcal {Q}} | L({\dot{\phi }}_0,{\dot{\psi }}_0) = 0\}. \end{aligned}$$

Consequently, if $L({\dot{\phi }}_0, {\dot{\psi }}_0) = 0$, the solution starting from $(-\phi _0,\psi _0)$ with initial velocity $({\dot{\phi }}_0, {\dot{\psi }}_0)$ is synchronized. Function L is called the synchronization measure at $(-\phi _0, \psi _0)$ for the SHS.$\square$

Proposition 33

Let ${\mathcal {Q}}$ denote the configuration space of the $(x_0, y_0)$-invariant 3D LIP biped. Assume that the switching surface is

$$\begin{aligned} {\mathcal {S}} = \{(x,y,{\dot{x}},{\dot{y}}) | h(x,y) = h(x_0, y_0) \}, \end{aligned}$$

where $h:Q\rightarrow {\mathbb {R}}$ is a smooth function and $\partial h/\partial y (x_0, y_0)\ne 0$. Let $({\dot{x}}_*, {\dot{y}}_*)$ be a fixed point of the restricted Poincaré map $P: {\mathcal {T}}_{(-x_0, y_0)}{\mathcal {Q}} \rightarrow {\mathcal {T}}_{(-x_0, y_0)}{\mathcal {Q}}$. The eigenvalues of P at $({\dot{x}}_*, {\dot{y}}_*)$ are $\{\lambda , 1\}$ with

$$\begin{aligned} \lambda = -1+\frac{2\omega ^2(y_0^2-Cx_0^2)}{CE_x^*-E_y^*}, \end{aligned}$$

(61)

where

$$\begin{aligned} C =\frac{y_0}{x_0} \left( \frac{\partial h}{\partial y}(x_0, y_0)\right) ^{-1} \frac{\partial h}{\partial x}(x_0, y_0), \end{aligned}$$

and

$$\begin{aligned} E_x^* = {\dot{x}}_*^2-\omega ^2 x_0^2, ~ E_y^* = {\dot{y}}_*^2 - \omega ^2 y_0^2 \end{aligned}$$

are the orbital energies in the x and y directions.

Proof

Let $L:{\mathcal {T}}_{(-x_0, y_0)}{\mathcal {Q}} \rightarrow {\mathbb {R}}$ be the synchronization measure of the 3D LIP. By Proposition 15, $\lambda = \partial L_1/\partial L_0(0,K^*)$, where $K^* = (1/2) (({\dot{x}}_*)^2+({\dot{y}}_*)^2)$. Assume that $(-x_0, y_0, {\dot{x}}_*+\delta {\dot{x}}_0, {\dot{y}}_*+\delta {\dot{y}}_0)$ is the initial state of a solution of the system at the beginning of the step, and let $(-x_0, y_0, {\dot{x}}_*+\delta {\dot{x}}_1, {\dot{y}}_*+\delta {\dot{y}}_1)$ be its state at the beginning of the next step. Define $L_0 = L({\dot{x}}_*+\delta {\dot{x}}_0, {\dot{y}}_*+\delta {\dot{y}}_0)$ and $L_1 = L({\dot{x}}_*+\delta {\dot{x}}_1, {\dot{y}}_*+\delta {\dot{y}}_1)$. We have

$$\begin{aligned} \lambda = \lim _{L_0 \rightarrow 0} \frac{L_1}{L_0}. \end{aligned}$$

Denote the state of the system right before the transition by $(x_0 + \delta x_1 , y_0 + \delta y_1, {\dot{x}}_*+ \delta {\dot{x}}_1, -({\dot{y}}_*+\delta {\dot{y}}_1))$. Since $L = {\dot{x}}{\dot{y}}-\omega ^2 xy$ is a conserved quantity for the 3D LIP in the continuous phase of motion, we have

$$\begin{aligned} L_0 = -({\dot{x}}_*+ \delta {\dot{x}}_1)({\dot{y}}_*+\delta {\dot{y}}_1) -\omega ^2 (x_0+\delta x_1)(y_0+\delta y_1). \end{aligned}$$

Moreover, since the system is $(x_0,y_0)$-invariant,

$$\begin{aligned} L_1 = ({\dot{x}}_*+ \delta {\dot{x}}_1)({\dot{y}}_*+\delta {\dot{y}}_1)+\omega ^2 x_0y_0. \end{aligned}$$

Adding this equation to the previous one,

$$\begin{aligned} L_1+L_0 = -\omega ^2 (x_0\delta y_1 + y_0 \delta x_1). \end{aligned}$$

(62)

By definition of the switching surface, $h(x_0+\delta x_1, y_0+\delta y_1) = h(x_0, y_0)$, from which,

$$\begin{aligned} \frac{\partial h}{\partial x}(x_0, y_0) \delta x_1 = - \frac{\partial h}{\partial y}(x_0, y_0) \delta y_1. \end{aligned}$$

(63)

By definition of C, $\delta y_1 = -C\frac{x_0}{y_0} \delta x_1$. Substituting this into Eq. (62) results in

$$\begin{aligned} L_1+L_0 = -\omega ^2 \left( -C\frac{x_0^2}{y_0}+ y_0\right) \delta x_1. \end{aligned}$$

Therefore, from the equation above,

$$\begin{aligned} \lim _{L_0\rightarrow 0} \frac{L_1}{L_0} = -1-\omega ^2 \left( -C\frac{x_0^2}{y_0}+ y_0\right) \lim _{L_0 \rightarrow 0} \frac{\delta x_1}{L_0}. \end{aligned}$$

(64)

Thus, to find the limit on the left-hand side we need only find the limit on the right-hand side. Since in the continuous phase of motion the orbital energies, ${\dot{x}}^2-\omega ^2 x^2$ and ${y}^2-\omega ^2 y^2$, are conserved quantities, we have

$$\begin{aligned} \left( {\dot{x}}_*+\delta {\dot{x}}_0\right) ^2-\omega ^2 x_0= & {} \left( {\dot{x}}_*+\delta {\dot{x}}_1\right) ^2-\omega ^2 \left( x_0+\delta x_1\right) ,\\ \left( {\dot{y}}_*+\delta {\dot{y}}_0\right) ^2-\omega ^2 y_0= & {} \left( {\dot{y}}_*+\delta {\dot{y}}_1\right) ^2-\omega ^2 \left( y_0+\delta y_1\right) . \end{aligned}$$

From these two equations and definition of C,

$$\begin{aligned} {\dot{x}}_*\left( \delta {\dot{x}}_1-\delta {\dot{x}}_0\right) = \omega ^2 \delta x_1, ~{\dot{y}}_*\left( \delta {\dot{y}}_1-\delta {\dot{y}}_0\right) = -C\omega ^2 \frac{x_0}{y_0} \delta x_1.\nonumber \\ \end{aligned}$$

(65)

Since $L = {\dot{x}}{\dot{y}}-\omega ^2 xy$ is a conserved quantity in the continuous phase of the motion, we can write $L_0$ in terms of the states at the beginning of step, that is, $(-x_0, y_0, {\dot{x}}_*+\delta {\dot{x}}_0, {\dot{y}}_*+{\dot{y}}_0)$ or at end of the step, that is, $(x_0+\delta x_1, y_0+\delta y_1, {\dot{x}}_*+\delta {\dot{x}}_1, {\dot{y}}_*+{\dot{y}}_1)$. Hence,

$$\begin{aligned} L_0= & {} \left( {\dot{x}}_*+\delta {\dot{x}}_0\right) \left( {\dot{y}}_*+\delta {\dot{y}}_0\right) +\omega ^2 x_0 y_0,\\ L_0= & {} -\left( {\dot{x}}_*+ \delta {\dot{x}}_1\right) \left( {\dot{y}}_*+\delta {\dot{y}}_1\right) -\omega ^2 \left( x_0+\delta x_1\right) \left( y_0+\delta y_1\right) . \end{aligned}$$

From (65) and the two equations above,

$$\begin{aligned} \frac{2L_0}{\delta x_1} = -\omega ^2 \frac{x_0y_0\left( -C{\dot{x}}_*^2+{\dot{y}}_*^2\right) +\left( {\dot{x}}_*{\dot{y}}_*\right) \left( -Cx_0^2+y_0^2\right) }{y_0{\dot{x}}_*{\dot{y}}_*}. \end{aligned}$$

Substituting this into Eq. (64), we have

$$\begin{aligned} \lim _{L_0\rightarrow 0}\frac{L_1}{L_0}= & {} -1+ \frac{1}{\omega ^2}\\&\cdot&\frac{2\left( -Cx_0^2+ y_0^2\right) \left( {\dot{x}}_*{\dot{y}}_*\right) }{x_0y_0\left( -C{\dot{x}}_*^2+{\dot{y}}_*^2\right) +\left( {\dot{x}}_*{\dot{y}}_*\right) \left( -Cx_0^2+y_0^2\right) }. \end{aligned}$$

The limit on the left-hand side is $\lambda$. Since $L({\dot{x}}_*,{\dot{y}}_*) = 0$, we have ${\dot{x}}_*{\dot{y}}_* = -\omega ^2 x_0 y_0$. Therefore, if we replace ${\dot{x}}_*{\dot{y}}_*$ with $-\omega ^2 x_0 y_0$ in the equation above, we obtain

$$\begin{aligned} \lambda= & {} -1+\frac{1}{\omega ^2}\\&\cdot \frac{2\left( -Cx_0^2+ y_0^2\right) \left( -\omega ^2 x_0y_0\right) }{x_0y_0\left( -C{\dot{x}}_*^2+{\dot{y}}_*^2\right) +\left( -\omega ^2 x_0 y_0\right) \left( -Cx_0^2+y_0^2\right) }. \end{aligned}$$

After simplification

$$\begin{aligned} \lambda =-1+\frac{2\omega ^2\left( y_0^2-Cx_0^2\right) }{C{\dot{x}}_*^2-{\dot{y}}_*^2+\omega ^2\left( y_0^2-Cx_0^2\right) }. \end{aligned}$$

By definition of $E_x^*$ and $E_y^*$ this equation is equivalent to Eq. (61).$\square$

Corollary 34

In Proposition 33, if $h(x,y) = x^2+a^2y^2$ then

$$\begin{aligned} \lambda = -1+\frac{2\omega ^2\left( a^2y_0^2-x_0^2\right) }{E_x^*-a^2E_y^*}. \end{aligned}$$

In particular, if $a = 1$,

$$\begin{aligned} \lambda = -1+\frac{2\omega ^2\left( y_0^2-x_0^2\right) }{E_x^*-E_y^*}. \end{aligned}$$

Appendix 2 Proof of Proposition 20

Lemma 35

Let $\varSigma = (X, {\mathcal {Q}})$ be a second order hybrid system and $(\phi , \psi )$ be a local coordinate system defined on a symmetric neighborhood ${\mathcal {N}} \subset {\mathcal {Q}}$. Suppose that $(\phi , \psi , {v}_\phi , {v}_\psi )$ is a coordinate system defined on ${\mathcal {TN}}$, the tangent bundle of ${\mathcal {N}}$, and assume that $G: {\mathcal {TN}} \rightarrow {\mathcal {TN}}$, which maps $(\phi , \psi , v_{\phi }, v_{\psi })$ to $(-\phi , \psi , v_{\phi }, -v_{\psi })$, is a well-defined function. If

$$\begin{aligned} {\dot{\phi }}(-\phi , \psi , v_{\phi }, -v_{\psi })= & {} {\dot{\phi }}(\phi , \psi , v_{\phi }, v_{\psi }),\nonumber \\ {\dot{\psi }}(-\phi , \psi , v_{\phi }, -v_{\psi })= & {} -{\dot{\psi }}(\phi , \psi , v_{\phi }, v_{\psi }), \end{aligned}$$

(66)

and the vector field X satisfies the equality

$$\begin{aligned} X \circ G = - \mathrm {d} G \cdot X, \end{aligned}$$

(67)

where $\mathrm {d} G$ is the Jacobian of G, then $\varSigma$ is a symmetric system in the coordinates $(\phi ,\psi )$.

Proof

From (66), in the coordinates $(\phi , \psi , {\dot{\phi }}, {\dot{\psi }})$, function G maps $(\phi , \psi , {\dot{\phi }}, {\dot{\psi }})$ to $(-\phi , \psi , {\dot{\phi }}, -{\dot{\psi }})$. Let $X = (X_q, X_{{\dot{q}}})$, and suppose that in the coordinates $(\phi , \psi ,{\dot{\phi }}, {\dot{\psi }})$, $X_{{\dot{q}}} = (f, g)$ for smooth functions f and g. We need to show that f and g satisfy Eqs. (8) and (9). However, these equations immediately follow by writing (67) in the coordinates $(\phi , \psi , {\dot{\phi }}, {\dot{\psi }})$.$\square$

The above lemma is a particular example of what is called time reversal symmetry in (Altendorfer et al. 2004).

Proof

(Proof of Proposition 20) Let ${\varvec{r}}_G$ denote the position vector of the COM of the 8-DOF 3D Biped, and let ${\varvec{H}}$ denote the total angular momentum of the system about the point of contact. We have

$$\begin{aligned} \frac{d {\varvec{H}}}{dt} = M{\varvec{r}}_G\times {\varvec{g}}, \end{aligned}$$

(68)

where M is the total mass of the biped, and ${\varvec{g}}$ is the vector of gravity. Let $(x_G, y_G, z_G)$ denote the coordinates of ${\varvec{r}}_G$ in a cartesian coordinate system attached to the inertial frame I, centered at the point of contact. Assume that $H_x$ is the second component of ${\varvec{H}}$ and $H_y$ is its first component (Normally, one might assign $H_x$ to the first component and $H_y$ to the second component; however, this choice may cause confusion with the notation of Lemma 35. To avoid this confusion, we assign $H_x$ to the second component of ${\varvec{H}}$ and $H_y$ to its first component).

One can check that $(x_G, y_G, H_x, H_y)$ is a coordinate system for the zero dynamics manifold and the coordinate systems $(x_G, y_G, H_x, H_y)$ and $(x_G, y_G, {\dot{x}}_G, {\dot{y}}_G)$ satisfy (66). As a result, there exist smooth one-to-one functions $f_x$ and $f_y$ such that on the zero dynamics manifold,

$$\begin{aligned} {\dot{x}}_G= & {} f_x(x_G, y_G, H_x, H_y),\nonumber \\ {\dot{y}}_G= & {} f_y(x_G, y_G, H_x, H_y), \end{aligned}$$

(69)

and

$$\begin{aligned} f_x(-x_G, y_G, H_x, -H_y)= & {} f_x(x_G, y_G, H_x, H_y),\nonumber \\ f_y(-x_G, y_G, H_x, -H_y)= & {} -f_y(x_G, y_G, H_x, H_y). \end{aligned}$$

(70)

Moreover, from Eq. (68),

$$\begin{aligned} {\dot{H}}_x= & {} Mg x_G,\nonumber \\ {\dot{H}}_y= & {} -Mg y_G. \end{aligned}$$

(71)

Equation (69) together with Eq. (71) define the equations of motion on the zero dynamics. Therefore, if $X = (f_x, f_y, Mgx_G, -Mgy_G)$ and $z = (x_G, y_G, H_x, H_y)$, then the system ${\dot{z}} = X(z)$ defines the equations of motion on the zero dynamics manifold. Hence, to show that this system is symmetric, according to Lemma 35, if G is the function that maps $(x_G, y_G, H_x, H_y)$ to $(-x_G, y_G, H_x, -H_y)$, it suffices to show that $X \circ G = - \mathrm {d} G \cdot X$. It immediately follows that

$$\begin{aligned} X \circ G (x) = \left[ \begin{array}{c} f_x(-x_G, y_G, H_x, -H_y)\\ f_y(-x_G, y_G, H_x, -H_y)\\ -Mgx_G\\ -Mgy_G \end{array} \right] , \end{aligned}$$

and

$$\begin{aligned} \mathrm {d}G \cdot X(x)= \left[ \begin{array}{c} -f_x(x_G, y_G, H_x, H_y)\\ f_y(-x_G, y_G, H_x, H_y)\\ Mgx_G\\ Mgy_G \end{array} \right] . \end{aligned}$$

However, the above two equations together with Eq. (70) prove that $X \circ G = - \mathrm {d} G \cdot X$. Hence, the zero dynamics is a symmetric system in the coordinates $(x_G, y_G)$.

Finally, from kinematics equations on the zero dynamics manifold, the coordinate systems $(q_1, q_2)$ and $(x_G, y_G)$ are equivalent in the sense of Proposition 6. As a result, the zero dynamics is a symmetric system in the coordinates $(q_1, q_2)$ as well.$\square$