Walking of biped with passive exoskeleton: evaluation of energy consumption

Abstract

The paper aims to theoretically show the feasibility and efficiency of a passive exoskeleton for a human walking and carrying a load. The human is modeled using a planar bipedal anthropomorphic mechanism. This mechanism consists of a trunk and two identical legs; each leg consists of a thigh, shin, and foot (massless). The exoskeleton is considered also as an anthropomorphic mechanism. The shape and the degrees of freedom of the exoskeleton are identical to the biped (to human)—the topology of the exoskeleton is the same as of the biped (human). Each model of the human and exoskeleton has seven links and six joints. The hip-joint connects the trunk and two thighs of the two legs. If the biped is equipped with an exoskeleton, then the links of this exoskeleton are attached to the corresponding links of the biped and the corresponding hip, knee, and ankle joints coincide. We compare the walking gaits of a biped alone (without exoskeleton) and of a biped equipped with exoskeleton; for both cases the same load is transported. The problem is studied in the framework of a ballistic walking model. During ballistic walking of the biped with exoskeleton, the knee of the support leg is locked, but the knee of the swing leg is unlocked. The locking and unlocking can be realized in the knees of the exoskeleton by any mechanical brake devices without energy consumption. There are no actuators in the exoskeleton. Therefore, we call it a passive exoskeleton. The walking of the biped consists of alternating single- and double-support phases. In our study, the double-support phase is assumed instantaneous. At the instant of this phase, the knee of the previous swing leg is locked and the knee of the previous support leg is unlocked. Numerical results show that during the load transport the human with the exoskeleton spends less energy than human alone. For transportation of a load with mass 40 kg, the economy of the energy is approximately 28%, if the length of the step and its duration are equal to 0.5 m and 0.5 s, respectively.

Appendix: Energy consumption at the instantaneous double-support phase

If the impulsive torques described by Dirac delta-functions are applied at the interlink joints of the biped, then each interlink angular velocity undergoes a discontinuous change.

For a sake of clarity and without loss of generality, we consider only one actuated joint of the biped. The joint variable is defined with $\theta$. An impulsive torque $\varGamma$ applied at this joint is defined to be as follows:

$$ \varGamma (t)=I\delta (t-T). $$

(45)

Expression (45) describes the Dirac delta-function; the magnitude $I$ describes the intensity of the torque $\varGamma$ such that

$$ \displaystyle \int _{T^{-}}^{T^{+}}I\delta (t-T) \,\mathrm{d} t=I. $$

We want to evaluate the energy expended during the operation of the impulsive torque (45). The chosen energy criterion is as follows:

$$ W = \displaystyle \int _{T^{-}}^{T{+}} \bigl\vert I\delta (t-T) (t) \dot{\theta }(t) \bigr\vert \,\mathrm{d} t. $$

(46)

Now instead of the delta-function (45) let us consider the following distributed in time piecewise constant function:

$$ \varGamma_{\Delta }(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \dfrac{I}{2\Delta } & \text{if}\ t \in [T-\Delta ,T+\Delta ], \\ \\ 0 & \text{if}\ t\notin [T-\Delta ,T+\Delta ]. \\ \end{array}\displaystyle \right . $$

(47)

Here $\Delta =\mbox{const}>0$; function (47) is shown in Fig. 13. If $\Delta \to 0$, then function $\varGamma_{\Delta }(t)$ “tends” to Dirac delta-function (45).

Fig. 13

Function $\varGamma_{\Delta }(t)$

Let us assume that in each interlink joint the distributed in time torque (similar to (47)) is applied. And let value $\Delta$ be the same for each joint.

The joint velocity $\dot{\theta }$ undergoes discontinuity at the instant $T$ (as in each joint), when the impulsive torques (similar to torque (45)) are applied in the joints of our biped. Let equality $\dot{\theta }=\dot{\theta }^{-}$ be valid just before the applying of the impulsive torques and equality $\dot{\theta }= \dot{\theta }^{+}$ be valid just after the applying of these impulsive torques.

If interval $[T-\Delta ,~T+\Delta ]$ is “small”, then velocity $\dot{\theta }(t)$ can be distributed in this interval by the following way (function $O(\Delta )$ is a magnitude of the first order with respect to magnitude $\Delta$)

$$ \dot{\theta }_{\Delta }(t)=\dot{\theta }^{-}+ \dfrac{\dot{\theta }^{+}- \dot{\theta }^{-}}{2\Delta }(t-T+\Delta )+O(\Delta ). $$

(48)

If $\Delta \to 0$, then, according to expression (48), $O(\Delta )\to 0$ and $\dot{\theta }_{\Delta }(T-\Delta )\to \dot{\theta }^{-}$, $\dot{\theta }(T+\Delta )\to \dot{\theta }^{+}$.

Now let us consider instead of (46) the following integral using expressions (47) and (48):

$$ W_{\Delta }=\displaystyle \int _{T-\Delta }^{T+\Delta } \bigl\vert \varGamma_{ \Delta }(t)\dot{\theta }_{\Delta }(t) \bigr\vert \,\mathrm{d} t= \biggl\vert \dfrac{I}{2 \Delta } \biggr\vert \displaystyle \int _{T-\Delta }^{T+\Delta } \biggl\vert \dot{\theta }^{-}+\dfrac{\dot{\theta }^{+}-\dot{\theta }^{-}}{2\Delta }(t-T+\Delta )+O(\Delta ) \biggr\vert \,\mathrm{d} t. $$

(49)

To calculate integral (49), we consider two cases. The first case is as follows:

$$ \dot{\theta }^{-}\dot{\theta }\geqslant 0 \quad \text{and} \quad \dot{\theta }^{-}\neq 0 \quad \text{or} \quad \dot{ \theta }^{+}\neq 0. $$

(50)

It is possible to show that under condition (50),

$$ \textstyle\begin{array}{ccl} \lim_{\Delta \to 0}W_{\Delta }&=&\lim_{\Delta \to 0} \biggl\vert \dfrac{I}{2\Delta } \biggr\vert \displaystyle \int _{T-\Delta }^{T+ \Delta } \biggl\vert \dot{\theta }^{-}+\dfrac{\dot{\theta }^{+}- \dot{\theta }^{-}}{2\Delta }(t-T+\Delta )+O(\Delta )\biggr\vert \,\mathrm{d} t \\ \\ &=& \biggl\vert \dfrac{I}{2\Delta } \biggr\vert \displaystyle \int _{T-\Delta }^{T+ \Delta } \biggl\vert \dot{\theta }^{-}+\dfrac{\dot{\theta }^{+}- \dot{\theta }^{-}}{2\Delta }(t-T+\Delta ) \biggr\vert \,\mathrm{d} t = \biggl\vert I\dfrac{ \dot{\theta }^{+}+\dot{\theta }^{-}}{2} \biggr\vert . \end{array} $$

(51)

The expression of the last integral in (51) contains value $\Delta$. However, the result of its calculation does not depend on this value and this result looks like (34) or (37) for the joint $i$.

Let us consider now the second case, when

$$ \dot{\theta }^{-}\dot{\theta }^{+}< 0. $$

(52)

Function (48) becomes zero at the instant

$$ t_{0}=T-\Delta +\dfrac{2\dot{\theta }^{-}}{\dot{\theta }^{-}- \dot{\theta }^{+}}\Delta +O\bigl( \Delta^{2}\bigr). $$

(53)

Function (48) has different signs in intervals $[T-\Delta ,~t_{0})$ and $(t_{0},T+\Delta ]$. Therefore, integral (49) can be written as follows:

$$ \textstyle\begin{array}{ccl} W_{\Delta } &=& \biggl\vert \dfrac{I}{2\Delta } \biggr\vert \biggl\vert \displaystyle \int _{T-\Delta }^{t_{0}} [ \dot{\theta }^{-}+\dfrac{ \dot{\theta }^{+}-\dot{\theta }^{-}}{2\Delta }(t-T+\Delta )+O(\Delta ) ] \,\mathrm{d} t \\ \\ &&{} -\displaystyle \int _{t_{0}}^{T+\Delta } \biggl[ \dot{\theta }^{-}+\dfrac{ \dot{\theta }^{+}-\dot{\theta }^{-}}{2\Delta }(t-T+\Delta )+O(\Delta ) \biggr] \,\mathrm{d} t \biggr\vert . \end{array} $$

(54)

Straightforward calculations show that

$$ \textstyle\begin{array}{ccl} \lim_{\Delta \to 0}W_{\Delta }&=& \biggl\vert \dfrac{I}{2\Delta } \biggr\vert \biggl\vert \displaystyle \int _{~T-\Delta }^{t_{0} ^{*}}\biggl[ \dot{\theta }^{-}+\dfrac{\dot{\theta }^{+}-\dot{\theta } ^{-}}{2\Delta }(t-T+\Delta ) \biggr] \,\mathrm{d} t \\ \\ &&{} -\displaystyle \int _{t_{0}^{*}}^{T+\Delta } \biggl[ \dot{\theta } ^{-}+\dfrac{\dot{\theta }^{+}-\dot{\theta }^{-}}{2\Delta }(t-T+\Delta ) \biggr] \,\mathrm{d} t \biggr\vert = \biggl\vert \dfrac{I}{2} \dfrac{( \dot{\theta }^{+})^{2}+(\dot{\theta }^{-})^{2}}{\dot{\theta }^{+}- \dot{\theta }^{-}} \biggr\vert . \end{array} $$

(55)

Here $t_{0}^{*}=T-\Delta +\dfrac{2\dot{\theta }^{-}}{\dot{\theta } ^{-}-\dot{\theta }^{+}}\Delta$. The expression of each integral in (55) contains value $\Delta$, but the calculations show that the difference between these two integrals does not depend on this value. The result of these calculations looks like (35) and (38) for the joint $i$.