How a serpentine tail assists agile motions of kangaroo rats: a dynamics and control approach

Abstract

Kangaroo rat is a good representative for general bipedalism with a serpentine tail. Modeling and analyzing the kangaroo rat motion helps to understand the serpentine tail functionalities in agile motions of bipedal mobile platforms, and this understanding is expected to lay the foundation for the future development of such robotic systems. This paper analyzes the kangaroo rat motions through dynamic modeling and control. The system dynamic model is established using the inertia matrix method, and two typical serpentine tail models are considered: a continuum tail model where the tail is modeled as several constant curvature arcs, and an articulated tail model where the tail is discretized into rigid links. Regularized contact model is used to compute the ground reaction force (GRF). To automatically plan the tail motion, numerical optimal control techniques (i.e., direct collocation method) are utilized. Partial feedback linearization is then used to track the designed tail trajectory. Based on the formulated dynamic model and motion controller, two representative tail functions (airborne righting and supporting) were simulated and analyzed. The results validated the proposed modeling and control framework and showed the nontrivial functionalities of the serpentine tail in helping the kangaroo rat to achieve agile motions. Moreover, comparative studies on the two tail models and the tail segmentations were performed to analyze the model differences. The results demonstrated that the articulated tail model is a good approximation of the continuum tail model, and more tail segments and links enhance the kangaroo rat’s ability to deliberately adjust its motion.

Appendices

Appendix A: Articulated tail kinematics

The velocities, Jacobians, acceleration, and MOI for each link of the articulated tail model are computed recursively using Eqs. (A1–A13). Eq. (A14) computes the torso MOI. ${\mathbf{u}}_{x,y}$ is an x-dimension unit column vector with 1 on the y-th entry.

$$ {{\varvec{\upomega}}}_{j} = \left\{ {\begin{array}{*{20}l} {{{\varvec{\upomega}}}_{b} , j = 0} \\ {{{\varvec{\upomega}}}_{j - 1} + \dot{\alpha }_{i} {\mathbf{x}}_{j - 1} + \dot{\beta }_{i} {\mathbf{z}}_{j} , j > 0} \\ \end{array} } \right. $$

(A1)

$$ {\mathbf{v}}_{j,com} = {\mathbf{v}}_{j,jnt} + {\mathbf{v}}_{j,j2c} $$

(A2)

$$ {\mathbf{v}}_{j,jnt} = \left\{ {\begin{array}{*{20}l} {{\mathbf{v}}_{b} + {{\varvec{\upomega}}}_{b} \times {\mathbf{p}}_{b2t} , j = 1} \\ {{\mathbf{v}}_{j - 1,jnt} + {\mathbf{v}}_{j - 1,j2j} , j > 1} \\ \end{array} } \right. $$

(A3)

$$ \left\{ {\begin{array}{*{20}c} {{\mathbf{v}}_{j,j2c} = {{\varvec{\upomega}}}_{j} \times {\mathbf{p}}_{j,j2c} } \\ {{\mathbf{v}}_{j,j2j} = {{\varvec{\upomega}}}_{j} \times {\mathbf{p}}_{j,j2j} } \\ \end{array} } \right. $$

(A4)

$$ {\mathbf{J}}_{j,\omega } = \left\{ {\begin{array}{*{20}l} {\left[ {\begin{array}{*{20}c} {0_{3 \times 3} } & {{\mathbf{I}}_{3 \times 3} } & {0_{3 \times 2m} } \\ \end{array} } \right], j = 0} \\ {{\mathbf{J}}_{j - 1,\omega } + {\mathbf{x}}_{j - 1} {\mathbf{u}}_{d,2i + 5}^{T} + {\mathbf{z}}_{j} {\mathbf{u}}_{d,2i + 6}^{T} , j > 0} \\ \end{array} } \right. $$

(A5)

$$ {\mathbf{J}}_{j,com} = {\mathbf{J}}_{j,jnt} + {\mathbf{J}}_{j,j2c} $$

(A6)

$$ {\mathbf{J}}_{j,jnt} = \left\{ {\begin{array}{*{20}l} {\left[ {\begin{array}{*{20}c} {{\mathbf{I}}_{3 \times 3} } & { - {\tilde{\mathbf{p}}}_{b2t} } & {0_{3 \times 2m} } \\ \end{array} } \right], j = 1} \\ {{\mathbf{J}}_{j - 1,jnt} + {\mathbf{J}}_{j - 1,j2j} , j > 1} \\ \end{array} } \right. $$

(A7)

$$ \left\{ {\begin{array}{*{20}c} {{\mathbf{J}}_{j,j2c} = - {\tilde{\mathbf{p}}}_{j,j2c} {\mathbf{J}}_{j,\omega } } \\ {{\mathbf{J}}_{j,j2j} = - {\tilde{\mathbf{p}}}_{j,j2j} {\mathbf{J}}_{j,\omega } } \\ \end{array} } \right. $$

(A8)

$$ {\dot{\mathbf{\omega }}}_{j} = \left\{ {\begin{array}{*{20}l} {{\dot{\mathbf{\omega }}}_{b} , j = 0} \\ {{\dot{\mathbf{\omega }}}_{j - 1} + \ddot{\alpha }_{i} {\mathbf{x}}_{j - 1} } \\ { + \dot{\alpha }_{i} {\tilde{\mathbf{\omega }}}_{j - 1} {\mathbf{x}}_{j - 1} + \ddot{\beta }_{i} {\mathbf{z}}_{j} + \dot{\beta }_{i} {\tilde{\mathbf{\omega }}}_{j} {\mathbf{z}}_{j} , j > 0} \\ \end{array} } \right. $$

(A9)

$$ {\dot{\mathbf{v}}}_{j,com} = {\dot{\mathbf{v}}}_{j,jnt} + {\dot{\mathbf{v}}}_{j,j2c} $$

(A10)

$$ {\dot{\mathbf{v}}}_{j,jnt} = \left\{ {\begin{array}{*{20}l} {{\dot{\mathbf{v}}}_{b} + \widetilde{{{\dot{\mathbf{\omega }}}}}_{b} {\mathbf{p}}_{b2t} + {\tilde{\mathbf{\omega }}}_{b}^{2} {\mathbf{p}}_{b2t} , j = 1} \\ {{\dot{\mathbf{v}}}_{j - 1,jnt} + {\dot{\mathbf{v}}}_{j - 1,j2j} , j > 1} \\ \end{array} } \right. $$

(A11)

$$ \left\{ {\begin{array}{*{20}c} {{\dot{\mathbf{v}}}_{j,j2c} = \widetilde{{{\dot{\mathbf{\omega }}}}}_{j} {\mathbf{p}}_{j,j2c} + {\tilde{\mathbf{\omega }}}_{j}^{2} {\mathbf{p}}_{j,j2c} } \\ {{\dot{\mathbf{v}}}_{j,j2j} = \widetilde{{{\dot{\mathbf{\omega }}}}}_{j} {\mathbf{p}}_{j,j2j} + {\tilde{\mathbf{\omega }}}_{j}^{2} {\mathbf{p}}_{j,j2j} } \\ \end{array} } \right. $$

(A12)

$$ {\mathbf{I}}_{j,at} = {\mathbf{R}}_{j}\,\,^{j} {\mathbf{I}}_{j,at} {\mathbf{R}}_{j}^{T} $$

(A13)

$$ {\mathbf{I}}_{b} = {\mathbf{R}}_{b} {}_{ }^{b} {\mathbf{I}}_{b} {\mathbf{R}}_{b}^{T} $$

(A14)

Appendix B: Continuum tail kinematics

The detailed expression of matrix ${\mathbf{E}}_{i,v}$ is given as follows where ${\mathrm{c}}_{\theta }=\mathrm{cos}{\theta }_{i}$, ${\mathrm{s}}_{\theta }=\mathrm{sin}{\theta }_{i}$, ${\mathrm{c}}_{2\theta }=\mathrm{cos}2{\theta }_{i}$, ${\mathrm{s}}_{2\theta }=\mathrm{sin}2{\theta }_{i}$. Since ${\mathbf{E}}_{i,v}$ is symmetric, only the upper triangle elements are listed.

$${\mathbf{E}}_{i,v}\left(\mathrm{1,1}\right)=1$$

$${\mathbf{E}}_{i,v}\left(\mathrm{1,2}\right)=(1-{\mathrm{c}}_{\theta })/{\theta }_{i}$$

$${\mathbf{E}}_{i,v}\left(\mathrm{1,3}\right)=(-1+{\mathrm{c}}_{\theta }+{\theta }_{i}{\mathrm{s}}_{\theta })/{\theta }_{i}^{2}$$

$${\mathbf{E}}_{i,v}\left(\mathrm{1,4}\right)=-{\mathrm{s}}_{\theta }/{\theta }_{i}$$

$${\mathbf{E}}_{i,v}\left(\mathrm{1,5}\right)=(-{\theta }_{i}{\mathrm{c}}_{\theta }+{\mathrm{s}}_{\theta })/{\theta }_{i}^{2}$$

$${\mathbf{E}}_{i,v}\left(\mathrm{2,2}\right)=1/2-{\mathrm{s}}_{2\theta }/(4{\theta }_{i})$$

$${\mathbf{E}}_{i,v}\left(\mathrm{2,3}\right)=(-2{\theta }_{i}{\mathrm{c}}_{2\theta }+{\mathrm{s}}_{2\theta })/(8{\theta }_{i}^{2})$$

$${\mathbf{E}}_{i,v}\left(\mathrm{2,4}\right)=({\mathrm{c}}_{2\theta }-1)/(4{\theta }_{i})$$

$${\mathbf{E}}_{i,v}\left(\mathrm{2,5}\right)=(1-{\mathrm{c}}_{2\theta }-2{\theta }_{i}{\mathrm{s}}_{2\theta }+2{\theta }_{i}^{2})/(8{\theta }_{i}^{2})$$

$${\mathbf{E}}_{i,v}\left(\mathrm{3,3}\right)=(4{\theta }_{i}^{3}+6{\theta }_{i}{\mathrm{c}}_{2\theta }+(6{\theta }_{i}^{2}-3){\mathrm{s}}_{2\theta })/(24{\theta }_{i}^{3})$$

$${\mathbf{E}}_{i,v}\left(\mathrm{3,4}\right)={\mathbf{E}}_{i,v}\left(\mathrm{2,5}\right)-1/2$$

$${\mathbf{E}}_{i,v}\left(\mathrm{3,5}\right)=(-1+(1-2{\theta }_{i}^{2}){\mathrm{c}}_{2\theta }+2{\theta }_{i}{\mathrm{s}}_{2\theta })/(8{\theta }_{i}^{3})$$

$${\mathbf{E}}_{i,v}\left(\mathrm{4,4}\right)=1-{\mathbf{E}}_{i,v}\left(\mathrm{2,2}\right)$$

$${\mathbf{E}}_{i,v}\left(\mathrm{4,5}\right)=-{\mathbf{E}}_{i,v}\left(\mathrm{2,3}\right)$$

$${\mathbf{E}}_{i,v}\left(\mathrm{5,5}\right)=(4{\theta }_{i}^{3}-6{\theta }_{i}{\mathrm{c}}_{2\theta }+(3-6{\theta }_{i}^{2}){\mathrm{s}}_{2\theta })/(24{\theta }_{i}^{3})$$

The elements in matrix ${\mathbf{Q}}_{i,v}$ are given as:

$${\mathbf{Q}}_{i,v}\left(\mathrm{1,1}\right)=1$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{2,1}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{1,2}\right)=(1-{\mathrm{c}}_{\theta })/{\theta }_{i}$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{3,1}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{1,4}\right)=(-1+{\mathrm{c}}_{\theta }+{\theta }_{i}{\mathrm{s}}_{\theta })/{\theta }_{i}^{2}$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{4,1}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{1,6}\right)=-{\mathrm{s}}_{\theta }/{\theta }_{i}$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{5,1}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{1,8}\right)=(-{\theta }_{i}{\mathrm{c}}_{\theta }+{\mathrm{s}}_{\theta })/{\theta }_{i}^{2}$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{2,2}\right)=1-{\mathbf{Q}}_{i,v}\left(\mathrm{4,6}\right)=1/2-{\mathrm{s}}_{2\theta }/(4{\theta }_{i})$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{3,2}\right)=(-2{\theta }_{i}{\mathrm{c}}_{2\theta }+{\mathrm{s}}_{2\theta })/(8{\theta }_{i}^{2})$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{3,2}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{2,4}\right)=-{\mathbf{Q}}_{i,v}\left(\mathrm{5,6}\right)=-{\mathbf{Q}}_{i,v}\left(\mathrm{4,8}\right)$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{4,2}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{2,6}\right)=({\mathrm{c}}_{2\theta }-1)/(4{\theta }_{i})$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{5,2}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{2,8}\right)=(1-{\mathrm{c}}_{2\theta }-2{\theta }_{i}{\mathrm{s}}_{2\theta }+2{\theta }_{i}^{2})/(8{\theta }_{i}^{2})$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{3,4}\right)=(4{\theta }_{i}^{3}+6{\theta }_{i}{\mathrm{c}}_{2\theta }+(6{\theta }_{i}^{2}-3){\mathrm{s}}_{2\theta })/(24{\theta }_{i}^{3})$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{3,4}\right)=-{\mathbf{Q}}_{i,v}\left(\mathrm{4,9}\right)$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{4,4}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{3,6}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{5,2}\right)-1/2$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{5,4}\right)=(-1+(1-2{\theta }_{i}^{2}){\mathrm{c}}_{2\theta }+2{\theta }_{i}{\mathrm{s}}_{2\theta })/(8{\theta }_{i}^{3})$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{5,4}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{4,5}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{3,8}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{2,9}\right)$$

$${\mathbf{Q}}_{i,v}\left(:,3\right)=2{\mathbf{Q}}_{i,v}\left(:,4\right)$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{1,5}\right)=(2+\left({\theta }_{i}^{2}-2\right){\mathrm{c}}_{\theta }-2{\theta }_{i}{\mathrm{s}}_{\theta })/{\theta }_{i}^{3}$$

$${\mathbf{Q}}_{i,v}\left(2,5\right)=-{\mathbf{Q}}_{i,v}\left(\mathrm{5,8}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{3,4}\right)-1/3$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{3,5}\right)=(2{\theta }_{i}(2{\theta }_{i}^{2}-3){\mathrm{c}}_{2\theta }-(6{\theta }_{i}^{2}-3){\mathrm{s}}_{2\theta })/(16{\theta }_{i}^{4})$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{3,5}\right)=-{\mathbf{Q}}_{i,v}\left(\mathrm{5,9}\right)$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{5,5}\right)=(3+2{\theta }_{i}\left(2{\theta }_{i}^{2}-3\right){\mathrm{s}}_{2\theta }+(6{\theta }_{i}^{2}-3){\mathrm{c}}_{2\theta }-2{\theta }_{i}^{4})/(16{\theta }_{i}^{4})$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{5,5}\right)={\mathbf{Q}}_{i,v}\left(\mathrm{3,9}\right)-1/4$$

$${\mathbf{Q}}_{i,v}\left(:,7\right)=2{\mathbf{Q}}_{i,v}\left(:,8\right)$$

$${\mathbf{Q}}_{i,v}\left(\mathrm{1,9}\right)=(\left({\theta }_{i}^{2}-2\right){\mathrm{s}}_{\theta }+2{\theta }_{i}{\mathrm{c}}_{\theta })/{\theta }_{i}^{3}$$

The block-wise matrix multiplication notation “$\circ$” is introduced in Sect. 2.3 to express the sum of scalar multiplications (linear combination of vectors). To better present its operations, an example is given here.

$$\mathbf{X}=[\begin{array}{cc}{\mathbf{A}}_{3\times 2}& {\mathbf{B}}_{3\times 2}\end{array}]$$

$$\mathbf{Y}=[\begin{array}{cc}\mathrm{a}& b\end{array}]$$

where the subscripts of $\mathbf{A}$, $\mathbf{B}$ denote their dimensions. $a$ and $b$ are scalars. Then $\mathbf{X}\circ {\mathbf{Y}}^{T}$ is evaluated as

$$\mathbf{X}\circ {\mathbf{Y}}^{T}=\left[\begin{array}{cc}{\mathbf{A}}_{3\times 2}& {\mathbf{B}}_{3\times 2}\end{array}\right]\circ {\left[\begin{array}{cc}a& b\end{array}\right]}^{T}=a{\mathbf{A}}_{3\times 2}+b{\mathbf{B}}_{3\times 2}$$

It has the similar transpose property as matrix multiplication.

$${(\mathbf{X}\circ {\mathbf{Y}}^{T})}^{T}=\mathbf{Y}\circ {\mathbf{X}}^{T}$$