# ZMP Support Areas for Multi-contact Mobility Under Frictional Constraints

## Abstract

We propose a method for checking and enforcing multi-contact stability based on
the Zero-tilting Moment Point (ZMP). The key to our development is the
generalization of ZMP *support areas* to take into account (a) frictional
constraints and (b) multiple non-coplanar contacts. We introduce and
investigate two kinds of ZMP support areas. First, we characterize and provide
a fast geometric construction for the support area generated by valid contact
forces, with no other constraint on the robot motion. We call this set the
*full support area*. Next, we consider the control of humanoid robots using the
Linear Pendulum Mode (LPM). We observe that the constraints stemming from the
LPM induce a shrinking of the support area, even for walking on horizontal
floors. We propose an algorithm to compute the new area, which we call
*pendular support area*. We show that, in the LPM, having the ZMP in the
pendular support area is a necessary *and sufficient* condition for contact
stability. Based on these developments, we implement a whole-body controller
and generate feasible multi-contact motions where an HRP-4 humanoid locomotes
in challenging multi-contact scenarios.

## Video

## Content

Paper | |

Source code | |

Slides (opens in new window/tab for online reading) | |

10.1109/TRO.2016.2623338 |

## BibTeX

```
@article{caron2016tro,
title = {ZMP Support Areas for Multi-contact Mobility Under Frictional Constraints},
author = {Caron, St{\'e}phane and Pham, Quang-Cuong and Nakamura, Yoshihiko},
journal = {IEEE Transactions on Robotics},
year = {2017},
month = feb,
volume = {33},
number = {1},
pages = {67--80},
publisher = {IEEE},
doi = {10.1109/TRO.2016.2623338},
}
```

## Discussion

**In Equation (5), the wrench coordinates** \(\boldsymbol{w}^c_{O}\) **are
taken with respect to the origin of the world frame. Why don't you rather take
this wrench at the CoM, as is usually done?**

Wrench coordinates are indeed taken at the origin of the world frame (or any other fixed point, for what matters). Taking screw coordinates at the origin of the world frame is typical of spatial vector algebra, which was formalized by Roy Featherstone.

The main reason for working with the CWC at the origin is that it only depends on the stance (set of contacts), while the CWC taken at the CoM would also depend on CoM coordinates. Actually, once you have the former, it is straightforward to compute the latter using a simple dual transformation formula, as we described in Section III of the following paper.

**What was the reason for introducing the cross product by**
\(\boldsymbol{n}\) **on both sides of equation (50)? Is it just here to
expand the cross product and isolate** \(\bfp_G\) **?**

A core assumption of this paper is that dynamics are decoupled into 2D plane dynamics (involving horizontal CoM and ZMP motions) and 4D complementary dynamics. Taking the cross product by \(\bfn\) is a convenient way to project onto the plane, as the projection of any vector \(\bfv\) onto a plane of normal \(\bfn\) can be written \(\bfn \times \bfv \times \bfn\) (the cross product is not associative in general but in this particular case it is).

**What was the need for adding equation (51)? Aren't equations (49) and (50)
sufficient to characterize static equilibrium?**

Rather than being added, equation (51) is already present in Newton and Euler equations (

i.e.the dynamics of the floating base) and we are not ignoring it. Specifically, we decompose the 6D floating base dynamics into:

- 2D plane dynamics,
e.g.equation (50)- complementary 3D linear dynamics,
e.g.equation (49)- complementary 1D angular dynamics along the vertical axis,
e.g.equation (51)In total we model the full dynamics, while leaving out equation (51) would be an approximation of these dynamics.

**When computing the static-equilibrium polygon, why is** \(\bfn\) **chosen
to be vertical? You mention earlier in the paper that** \(\bfn\) **can be "a
fixed unit space vector, not necessarily aligned with the gravity vector". How
is it determined or what does it represent in general?**

In the case of the CoM polygon, it is convenient but not necessary to choose a vertical \(\bfn\). You can take a non-vertical one: it will give you the same polygon but in a tilted plane.

In general, \(\bfn\) represents the plane in which you are controlling your CoM. Because the CoM is constrained to lie in this plane, you should choose it according to the motions to be realized: for instance, if the robot is climbing stairs, the plane should have the same inclination as the staircase.