Navigating the Wrench-Feasible C-Space of Cable-Driven Hexapods

Abstract

Motion paths of cable-driven hexapods must carefully be planned to ensure that the lengths and tensions of all cables remain within acceptable limits, for a given wrench applied to the platform. The cables cannot go slack—to keep the control of the platform—nor excessively tight—to prevent cable breakage—even in the presence of bounded perturbations of the wrench. This paper proposes a path planning method that accommodates such constraints simultaneously. Given two configurations of the platform, the method attempts to connect them through a path that, at any point, allows the cables to counteract any wrench lying inside a predefined uncertainty region. The resulting C-space is placed in correspondence with a smooth manifold, which allows defining a continuation strategy to search this space systematically from one configuration, until the second configuration is found, or path non-existence is proved by exhaustion of the search. The approach is illustrated on the NIST Robocrane hexapod, but it remains applicable to general cable-driven hexapods, either to navigate their full six-dimensional C-space, or any of its slices.

Appendix: Properties of \({\fancyscript{M}}\)

This appendix is devoted to the proof of two properties that are essential in order to apply the continuation strategy described in Sect. 4. The first one is the non-nullity of $s_i$, $t_i$, $g_i$ and $v_{i,i}$, and the second one is the smoothness of the six-dimensional manifold ${\fancyscript{M}}$. A by-product of the second property is the fact that $\varvec{J}(\varvec{q})$ is non-singular for all $\varvec{q}\in {\fancyscript{C}}$.

From Eqs. (6–8) it follows directly that $s_i$, $t_i$ and $g_i$ can never be zero on ${\fancyscript{M}}$. The same property for $v_{i,i}$ can be proved by contradiction. Let us assume that $v_{i,i}=0$ for some $i$. If we consider Eq. (5), then, by replacing $\varvec{B}^i \varvec{v}_i = \varvec{0}$ into $\varvec{v}_i^{{\mathsf T }}\varvec{B} \; \varvec{v}_i =1$, we obtain the dot product of two vectors: $\varvec{v}_i^{{\mathsf T }}$, with $v_{i,i}=0$, and the vector $\varvec{B}\varvec{v}_i$, whose components are all zero except that in position $i$. The result of this dot product is $0$, which contradicts Eq. (5), as it should be $1$. As a result, the set ${\fancyscript{M}}^+$ and its complement ${\fancyscript{M}}\setminus {\fancyscript{M}}^+$ are disconnected.

Let us now prove the smoothness of ${\fancyscript{M}}$. If we can verify that $\varvec{F}(\varvec{x})$ is a differentiable function with full rank differential $\varvec{F}_{\varvec{x}}$, then the smoothness of ${\fancyscript{M}}$ will follow from the implicit function theorem. By construction, all functions intervening in $\varvec{F}(\varvec{x})$ are differentiable all over ${\fancyscript{M}}$, and the differential matrix $\varvec{F}_{\varvec{x}}$ can be expressed in the following block-triangular form after re-organizing some equations and variables

where empty blocks represent zero-matrices and asterisks indicate non-zero blocks.

Due to the triangular structure of $\varvec{F}_{\varvec{x}}$ it suffices to verify that the five blocks in the diagonal are full-rank in order to prove the smoothness of ${\fancyscript{M}}$. The first block is

which is the differential matrix of the system ${\varvec{\varPhi }}(\varvec{y})=\varvec{0}$ formed by Eqs. (1–3) and (8) with respect to $\varvec{y}=(\varvec{\tau },\varvec{p},\varvec{R},\varvec{u}_i,\rho _i,g_i)$. Here $\varvec{L}$ and $\varvec{G}$ are $6\times 6$ diagonal matrices with diagonal elements $\rho _i$ and $(\rho _i - \underline{\rho _i}) \cdot (\overline{\rho _i} - \rho _i)$, respectively. To see that ${\varvec{\varPhi }}_{\varvec{y}}$ is full rank, observe that its last four block-columns comprise a non-singular square submatrix of maximum size, as its diagonal elements do not vanish over ${\fancyscript{M}}$ by virtue of Eq. (8) and the fact that $\underline{\rho _i}>0$.

The remaining four diagonal blocks of $\varvec{F}_{\varvec{x}}$ are the differential matrices of Eqs. (4–7) with respect to the variables $\varvec{f}_0$, $\varvec{v}_i$, $s_i$ and $t_i$, respectively, where the blocks $\varvec{S}$ and $\varvec{T}$ are $6\times 6$ diagonal matrices with elements $f_{0,i} - v_{i,i} -\underline{f_i}$ and $\overline{f_i}-f_{0,i} - v_{i,i}$, respectively. The screw Jacobian $\varvec{J}(\varvec{q})$ can be shown to be full rank over ${\fancyscript{M}}$ by contradiction. Indeed, if $\varvec{J}(\varvec{q}_s)$ were rank deficient for some $\varvec{q}_s$, then so would be $\varvec{B}$, and therefore $\ker \varvec{B}$ would contain non-zero vectors. In such case, for some $i$ all solutions of $\varvec{B}^i\varvec{v}_i=0$ would satisfy $\varvec{v}_i\in \ker \varvec{B}$ and, thus, it would be $\varvec{v}_i^{{\mathsf T }}\varvec{B} \varvec{v}_i =0$, which contradicts Eq. (5) and, hence, $\varvec{J}(\varvec{q})$ cannot be rank deficient over ${\fancyscript{M}}$. The $6\times 6$ block matrices involving $\varvec{B}$ and $\varvec{B}^i$ can only be rank deficient if $v_{i,i}=0$, but this can never happen as we have already seen. All these blocks are therefore full rank over ${\fancyscript{M}}$. Finally, it is clear that $\varvec{S}$ and $\varvec{T}$ are also full rank over ${\fancyscript{M}}$, since their diagonal elements never vanish due to Eqs. (6–7), and this completes the proof of the smoothness of ${\fancyscript{M}}$ and, in particular, that of ${\fancyscript{M}}^+$.

It is worth mentioning that not only ${\fancyscript{M}}$ and ${\fancyscript{M}}^+$ are smooth, but also any slice taken as a combination of the angular and position parameters, $\varvec{\tau }$ and $\varvec{p}$. Indeed, taking any of these slices implies only the removal of some columns amongst the first two blocks of ${\varvec{\varPhi }}_{\varvec{y}}$, which does not change the global rank of the differential $\varvec{F}_{\varvec{x}}$ corresponding to the considered slice.