Dynamic modeling and control of hopping robot in planar space

Abstract

The paper presents two-mass inverted pendulum (TMIP) model and its control scheme for hopping robot. Unlike the conventional spring-loaded inverted pendulum (SLIP) model, the proposed TMIP model is able to provide the functions of energy storing and releasing by using a linear actuator. Also it becomes more accurate comparing to the conventional SLIP model by taking the foot mass into consideration. Furthermore how to determine both takeoff angle and velocity for hopping is analytically suggested to accomplish the desired stride and height of hopping robot. The control method for the TMIP model is also presented in the paper. Finally, the effectiveness of the proposed model and control scheme is verified through the simulation.

Appendix

1.1 Derivation of Eq. (1)

Let us obtain both the kinetic and potential energy in stance phase as follows:

$$\begin{aligned} {\text {T}}= & {} \frac{1}{2}{m_\mathrm{b}} \left( {l_{a} + l_{f}} \right) ^{2} \dot{\theta }^{2} + \frac{1}{2}{m_\mathrm{f}} l_{f}^{2} \dot{\theta }^{2} + \frac{1}{2}{m_\mathrm{b}} \dot{l}_{a}^{2} \end{aligned}$$

(22)

$$\begin{aligned} \hbox {V}= & {} m_\mathrm{b} ({l_\mathrm{a} +l_\mathrm{f}})g\hbox {cos}\theta +m_\mathrm{f} l_\mathrm{f} g\hbox {cos}\theta \end{aligned}$$

(23)

where T and V are the kinetic and potential energy during the stance phase, respectively. Now let us define the Lagrangian as

$$\begin{aligned} {\mathcal {L}}=T-V. \end{aligned}$$

(24)

Then the Euler-Lagrange equation of motion can be calculated as

$$\begin{aligned}&\frac{\hbox {d}}{{\hbox {d}}t}\left( \frac{\partial {\mathcal {L}}}{\partial \dot{\theta }}\right) -\frac{\partial {\mathcal {L}}}{\partial \theta }=\tau _r , \end{aligned}$$

(25)

$$\begin{aligned}&\frac{\hbox {d}}{{{\hbox {d}}t}}\left( {\frac{{\partial {\mathcal {L}}}}{{\partial \dot{l}_{a}}}} \right) - \frac{{\partial {\mathcal {L}}}}{{\partial l_{a}}} = f_{l} , \end{aligned}$$

(26)

by collecting above results and rearranging them, we can get the following matrix-vector equation:

$$\begin{aligned}&\left[ {\begin{array}{cc} {m_\mathrm{b}} \left( {l_{a} + l_{f}} \right) ^{2} + m_\mathrm{f} l_{f} ^{2} &{} 0 \\ 0 &{} {m_\mathrm{b}} \\ \end{array}}\right] \left[ {\begin{array}{c} {\ddot{\theta }} \\ {\ddot{l}_{a}} \\ \end{array}}\right] \nonumber \\&\quad + \left[ {\begin{array}{cc} {m_\mathrm{b} \left( {l_{a} + l_{f}} \right) \dot{l}_{a}} &{} {m_\mathrm{b} \left( {l_{a} + l_{f}} \right) \dot{\theta }} \\ { - m_\mathrm{b} \left( {l_{a} + l_{f}}\right) \dot{\theta }} &{} 0 \\ \end{array}} \right] \left[ {\begin{array}{c} {\dot{\theta }} \\ {\dot{l}_{a}} \\ \end{array}} \right] \nonumber \\&\quad + \left[ {\begin{array}{c} {-m_\mathrm{b} \left( {l_{a} + l_{f}} \right) g\hbox {sin}\theta - m_\mathrm{f} l_{f} g\hbox {sin}\theta } \\ {m_\mathrm{b} g\hbox {cos}\theta } \\ \end{array}} \right] = \left[ {\begin{array}{c} {\tau _{r}} \\ {f_{l}} \\ \end{array}} \right] . \end{aligned}$$

(27)

Above equation can be rewritten as a compact form:

$$\begin{aligned} M(q)\ddot{q} + C({q,\dot{q}})\dot{q} + g(q) = \tau \end{aligned}$$

(28)

where $q=\left[ {{\begin{array}{cc} \theta &{} {l_\mathrm{a}} \\ \end{array}}} \right] ^{T}$ is the state vector described in the polar coordinate. Also, if we consider the external force by the ground, the equation (28) is modified as

$$\begin{aligned} M(q)\ddot{q} + C({q,\dot{q}})\dot{q} + g(q) = \tau + J^{T} F_{{ext}} \end{aligned}$$

(29)

where

$$\begin{aligned} J=\left[ {{\begin{array}{cc} -l_c \hbox {cos}\theta &{} {-\hbox {sin}\theta } \\ -l_c \hbox {sin}\theta &{} {\hbox {cos}\theta } \\ \end{array}}} \right] . \end{aligned}$$