Why ankle torque minimization in wrench distribution?

[shbang91]

Hello Dr.Caron,

Thank you for your effort in maintaining this nice repo!
Sorry if this question is duplicated one in this repo.

I have a question regarding the ankle torque minimization (Eq.(13)) in II-C. Contact Wrench Distribution on your stair climbing stabilization paper.

First of all, what is the benefit of minimizing ankle torque minimization (Eq.(13)) over minimizing the contact wrench in each sole center frame (lc_W_left, rc_W_right)? (I believe the position of the sole center frame(lc, rc) is in the bottom center of the foot which directly contacts with the ground while the ankle frame is located in the same planer surface offset along x and y axis of the sole center frame based on the discussion in here. Please correct me if I am wrong).

Secondly, looking at the

lipm_walking_controller/src/Stabilizer.cpp

Line 617 in 34f5d1e

A_lankle.diagonal() << 1., 1., 1e-4, 1e-3, 1e-3, 1e-4;

, I wonder why you weighted more on the tau_x and tau_y. Is this because you don't want to generate tilting motions in the x and y local ankle frame (By minimizing the external wrench applied on the ankle joint through Eq.(13), you would expect the ankle joint to execute small torque command)? There are also f_x and f_y which contribute to the ankle torque but you only penalized them with a smaller weight than tau_x and tau_y. Is this a common heuristic or only for HRP-4 system?

Finally, do the sole center frame and ankle frame share the same orientation in your definition?

Thank you,
Seung Hyeon

[stephane-caron]

First of all, what is the benefit of minimizing ankle torque minimization (Eq.(13)) over minimizing the contact wrench in each sole center frame (lc_W_left, rc_W_right)?

Ankle torque minimization is motivated by the design of the soles of HRP robots, which is inherited from Honda's humanoid robot design:

(Source: slides 4 and 5 in this presentation.)

The mechanical flexibility below the ankles is here to reduce the rigidity of the contact interaction with the environment. This means protecting the force sensor from hard impacts, but also importantly reducing the control frequency required for foot admittance control.

One downside of this flexibility is that our sensing and admittance control laws are based on the assumption that its deflection is small. (To be precise, the deflection is the orientation of the sole frame with respect to the ankle frame.) The larger the deflection, the worse our control becomes. That's why most works dealing with HRP robots aim to keep this deflection small.

That's also the motivation for projects like SpringEstimator which observe the deflection directly. In this controller, we assume that there exists a linear underlying model $$\Delta_{spring} = \frac{1}{K} \tau_{ankle},$$ and minimize the ankle torque $\tau_{ankle}$ as a proxy to minimize the deflection $\Delta_{spring}$.

[Saeed-Mansouri]

The mechanical flexibility below the ankles is here to reduce the rigidity of the contact interaction with the environment. This means protecting the force sensor from hard impacts but also importantly reducing the control frequency required for foot admittance control.

Can you explain more about the control frequency that you mentioned here?

[stephane-caron]

Let's consider both our linear flexibility model and foot admittance control:

The closed-loop behavior mentioned in this slide is for a continuous system, but in practice we have a discretized one: at every step $k$, we set a new

$$\dot{\theta}[k] = A_{cop} (\tau^d - \tau[k])$$

to the underlying joint velocity controller. Let's assume $\theta_e=0$ and $\tau^d=0$ to make formulas simpler (it has no effect on the stability argument). Integrating to the next time step yields:

$$\tau[k + 1] = (1 - A_{cop} K_e \delta t) \tau[k]$$

Stability of this system requires $|1 - A_{cop} K_e \delta t | < 1$, that is, $0 < A_{cop} K_e \delta t < 2$. The left-hand side inequality is no problem: $K_e > 0$ by definition (stiffness coefficient), $A_{cop} > 0$ by design, $\delta t > 0$ is the sampling period. The meaningful part is the other side:

$$A_{cop} \delta t < \frac{2}{K_e}$$

This model shows how a more rigid contact (that is, a larger stiffness $K_e$) requires us to either lower the sampling period $\delta t$, i.e. control at higher frequency, or switch to a lower force-control admittance $A_{cop}$. To go back to the original point:

The mechanical flexibility below the ankles is here to reduce the rigidity of the contact interaction with the environment [which in turn reduces the] the control frequency required for foot admittance control.

One role of the flexibility is to reduce $K_e$, thus adding room for us to increase either $\delta t$ or $A_{cop}$.

📢: I will discuss this topic next Monday at the Humanoids 2022 Tutorial on Challenge-driven Learning of Humanoid Robot Control in Virtual Environments. Feel free to join e.g. using the Zoom link.

[Saeed-Mansouri]

Thanks @stephane-caron.
It was so instructive. I will join your seminar.
I always had this question: Is it desirable to design a system ( $\delta t$ and $K_{e}$ ) to provide a larger force-control admittance $A_{cop}$?

[stephane-caron]

Following up on the tutorial this morning:

• Slides
• Discussion that followed

One role of the flexibility is to reduce , thus adding room for us to increase either $\delta t$ or $A_{cop}$.

Disclaimer: I'm sharing thoughts iteratively here, this is not a question I have thought all the way through 😉

I think it is not the value $A_{cop}$ in itself that we want to look at, but rather either of its products $A_{cop} \delta t$ or $A_{cop} K_e$ with the other two quantities. For example, looking at the former: if we double $\delta t$ and halve $A_{cop}$, we get the same outcome $\tau[k + 1]$ after a discrete control step. As to the latter, from the continuous-time dynamics $\dot{\tau} = A_{cop} K_e (\tau^d - \tau)$ it gives us the characteristic time $T = (A_{cop} K_e)^{-1}$ of the closed-loop force control performance.

[stephane-caron]

Secondly, looking at the lipm_walking_controller/src/Stabilizer.cpp, I wonder why you weighted more on the taux and tauy. Is this because you don't want to generate tilting motions in the x and y local ankle frame (By minimizing the external wrench applied on the ankle joint through Eq.(13), you would expect the ankle joint to execute small torque command)?

The coordinates $\tau_{ankle,x/y}$, in full: "the moment of ground contact forces around the local $x$ and $y$ axes of the (projected) ankle frame", correspond to the center of pressure via the cross-product formula:

$$ p_{CoP} = p_{ankle} + \frac{\tau_{ankle} \times e_z}{f \cdot e_z}, $$

where $e_z$ is the normal vector of the projected ankle frame (orthogonal to the sole surface). The motivation for weighing those two coordinates more than the third one $(\tau_{ankle,z})$ is to give priority to center-of-pressure control over yaw-moment control. There are at least two reasons why CoP control is more important:

1. Centers of pressure under each contact contribute directly to the net ZMP, which is what we aim to regulate in the end.
2. Centers of pressure contribute indirectly to the yaw-moment friction condition:

$$ |\tau_z - \tau_z^*| \leq \mu f_z d_{CoP}, $$

where $d_{CoP}$ is the distance between the CoP and the edge of the contact surface, which we try to keep as large as possible (by minimizing ankle torques, i.e. keeping each CoP below its corresponding ankle). The larger $d_{CoP}$ is, the less the robot has to pay attention to $\tau_{ankle,z}$ as yaw slippage becomes less likely.

[stephane-caron]

Finally, do the sole center frame and ankle frame share the same orientation in your definition?

Yes, the sole center frame and projected ankle frame of a foot share the same orientation.