A spacecraft model

Imagine you have a flying vehicle modeled as a single rigid body in a gravity field with some number of force "thrusters" attached. We'll describe the pose of the body in the world frame, ${}^WX^B$, parameterized by its position $(x,z)$ and orientation $\theta$, and take the center of mass to be at the origin of this body frame. The vector from the center of mass to thruster $i$ in the world frame is given by ${}^Bp^{B_i}_W$, yielding the equations of motion: \begin{align} m\ddot{x} =& \sum_i f^{B_i}_{i, W_x}, \label{eq:spacecraft_x} \\ m\ddot{z} =& \sum_i f^{B_i}_{i,W_z} - mg,\label{eq:spacecraft_z}\\ I\ddot\theta =& \sum_i \left[ {}^Bp^{B_i}_W(\theta) \times f^{B_i}_{i,W} \right]_y, \label{eq:spacecraft_theta}\end{align} where I've used $f^{B_i}_{i,W_x}$ to represent the $x$-component of the force applied to $B$ at the location of thruster $i$, expressed in the world frame, and we take the $y$-axis component of the cross product on the last line. You might compare this to the equations of motion for the planar quadrotor; the only difference is that I've written the force inputs here in world coordinates instead of body coordinates.

Our goal is to move the spacecraft to a desired position/orientation and to keep it there. If we take the inputs to be $f^{B_i}_{i,W}$, then the translational dynamics are affine (linear plus a constant term). The rotational dynamics are nonlinear assuming that the positions of the thrusters rotate with the spacecraft. But for small angles, we can stabilize a stabilizable fixed point using linearization plus a change of coordinates then LQR, or even plan/track a desired trajectory using time-varying LQR. If I were to add additional linear constraints, for instance constraining $f_{min} \le f^{B_i}_{i,W} \le f_{max}$, then I can still use linear model-predictive control (MPC) to plan and stabilize a desired motion. By most accounts, this is a relatively easy control problem. (If we do consider the nonlinear dynamics in $\theta$, then we will resort to nonlinear optimization, but this still feels fairly managable).

Now imagine that you have just a small number of thrusters (let's say two) each with severe input constraints. To make things more interesting, let us say that you're allowed to move the thrusters, so ${}^B\text{v}^{B_i} = \frac{d}{dt} {}^Bp^{B_i}$ becomes an additional control input, but with some important limitations: you have to turn the thrusters off when they are moving (e.g. $f^{B_i}_i = 0$ or ${}^B \text{v}^{B_i} = 0$) and there is a limit to how quickly you can move the thrusters $|{}^B\text{v}^{B_i}| \le \text{v}_{max}$. This problem is now a lot more difficult, due especially to the fact that constraints of the form "$x=0$ or $y=0$"" are non-convex.

I find this a very useful thought exercise; somehow our controller needs to make an effectively discrete decision to turn off a thruster and move it. The framework of optimal control should support this - you are sacrificing a short-term loss of control authority for the long-term benefit of having the thruster positioned where you would like, but we haven't developed tools yet that deal well with this discrete plus continuous decision making. We'll need to address that here.

Unfortunately, although the problem is already difficult, we need a few more constraints to make it relevant for legged robots. Now let's say additionally, that the thrusters can only be turned on in certain locations (in the world frame), as cartooned here:

Even if the individual regions are convex, the union of these regions need not form a convex set. Furthermore, these locations are often expressed in the world frame, not the robot frame, e.g. $${\bf A}\, {}^W p^{B_i}_W(\theta) \le {\bf b}.$$

Robots with (massless) legs

In my view, the spacecraft with thrusters problem above is a central component of the walking problem. If we consider a walking robot as our single rigid body, $B$, with massless legs, then the feet are exactly like movable thrusters. As above, they are highly constrained - they can only produce force when they are touching the ground, and (typically) they can only produce forces in certain directions, for instance as described by a "friction cone" (you can push on the ground but not pull on the ground, and with Coulomb friction the forces tangential to the ground must be smaller than the forces normal to the ground, as described by a coefficient of friction, e.g. $f^{B_i}_{i,C_z} \ge 0, |f^{B_i}_{i,C_x}| < \mu f^{B_i}_{i,C_z}$) where $C$ is the contact frame.

The constraints on where you can put your feet / thrusters will depend on the kinematics of your leg, and the speed at which you can move them will depend on the full dynamics of the leg -- these are difficult constraints to deal with. But the actual dynamics of the rigid body, $B$, are actually still just the spacecraft dynamics, and are still simple!

The special case of flat terrain

While not the only important case, it is extremely common for our robots to be walking over flat, or nearly flat terrain. In this situation, even if the robot is touching the ground in a number of places (e.g., two heels and two toes), the constraints on the center of mass dynamics can be summarized very efficiently.

First, on flat terrain $f^{B_i}_{i,W_z}$ represents the normal component of the force contact point $i$. If we assume that the robot can only push on the ground (and not pull), then this implies $$\forall i, f^{B_i}_{i,W_z} \ge 0 \Rightarrow \sum_i f^{B_i}_{i,W_z} \ge 0 \Rightarrow \ddot{z} \ge -g.$$ In other words, if I cannot pull on the ground, then my center of mass cannot accelerate towards the ground faster than gravity.

Furthermore, if we use a Coulomb model of friction on the ground, with friction coefficient $\mu$, then $$\forall i, |f^{B_i}_{i,W_x}| \le \mu f^{B_i}_{i,W_z} \Rightarrow \sum_i |f^{B_i}_{i,W_x}| \le \mu \sum_i f^{B_i}_{i,W_z} \Rightarrow |\ddot{x}| \le \mu (\ddot{z} + g).$$ For instance, if I keep my center of mass height at a constant velocity, then $\ddot{z}=0$ and $|\ddot{x}| \le \mu g$; this is a nice reminder of just how important friction is if you want to be able to move around in the world.

Even better, let us define the "center of pressure" (CoP) as the point on the ground where $$\begin{bmatrix} x_{cp} \\ z_{cp}\end{bmatrix} \equiv \,{}^Wp^{R_{cp}}_W = \frac{\sum_i {}^Wp^{B_i}_{W}\, f^{B_i}_{W_z}}{\sum_i f^{B_i}_{W_z}}.$$ Since all ${}^Wp^{B_i}_{W_z}$ are equal on flat terrain, $z_{cp}$ is just the height of the terrain. It turns out that the center of pressure is a "zero-moment point" (ZMP) -- a property that we will demonstrate below -- and a moment-balance equation (see below) gives us a very important relationship between the location of the CoP and the dynamics of the CoM: \begin{equation} (m\ddot{z} + mg) (x_{cp} - x) = (z_{cp} - z) m\ddot{x} - I\ddot\theta. \label{eq:cop} \end{equation} If $\ddot{z} = \ddot{\theta} = 0$, or more generally if the rate of change in vertical and angular momentum about the center of mass is small, then the result is the famous "ZMP equations": \begin{equation} \ddot{x} = g \frac{(x - x_{cp})}{(z - z_{cp})}. \label{eq:zmp} \end{equation} So the location of the center of pressure completely determines the acceleration of the center of mass, and vice versa! What's more, when $z - z_{cp}$ is a constant (the center of mass remains at a constant height relative to the ground) or following a predetermined schedule, the remaining relationship is affine -- a property that we can exploit in a number of ways.

As an example, we can easily relate constraints on the CoP to constraints on $\ddot{x}$. It is easy to verify from the definition that the CoP must live inside the convex hull of the ground contact points. Therefore, if we use the $\ddot{z}=\ddot\theta=0$ strategy, then this directly implies strict bounds on the possible accelerations of the center of mass given the locations of the ground contacts.

An aside: Zero-moment point derivation

The zero-moment point (ZMP) is discussed very frequently in the current literature on legged robots. It also has an unfortunate tendency to be surrounded by some confusion; a number of people have defined the ZMP is slightly different ways (see e.g. Goswami99 for a summary). Therefore, it makes sense to provide a simple derivation here.

First let us recall that for rigid body systems I can always summarize the contributions from many external forces as a single wrench (force and torque) on the object -- this is simply because the $F_i$ terms enter our equations of motion linearly. Specifically, the position $p^A$ of an arbitrary point $A$:

I can re-write the equations of motion as \begin{align*} m\ddot{x} =& \sum_i f^{B_i}_{W_x} = f^A_{net,W_x},\\ m\ddot{z} =& \sum_i f^{B_i}_{W_z} - mg = f^A_{net,W_z} - mg,\\ I\ddot\theta =& \sum_i \left[ {}^Bp^{B_i}_W \times f^{B_i}_W \right]_y = [{}^Bp^A_W \times f^A_{net,W}]_y + \tau^A_{net,W}, \end{align*} where $f^A_{net} = \sum_i f^{B_i}$ and the value of $\tau^A_{net}$ depends on the position, $p^A$. For some choices of $p^A$, we can make $\tau^A_{net}=0$. We'll call these points $\text{ZMP}$, for which we have \begin{equation} \left[ {}^Bp^{\text{ZMP}}_W \times f^{\text{ZMP}}_{net,W} \right]_y = {}^Bp^{\text{ZMP}}_{W_z} f^{\text{ZMP}}_{net,W_x} - {}^Bp^{\text{ZMP}}_{W_x} f^{\text{ZMP}}_{net,W_z} = \sum_i \left[ {}^Bp^{B_i}_W \times f^{B_i}_W \right]_y. \label{eq:zmp_def} \end{equation} We can see that whenever we have ground reaction forces ($f^{\text{ZMP}}_{net,W_z} \ne 0$) there is an entire line of solutions, ${}^Bp^{\text{ZMP}}_{W_x} = a\, {}^Bp^{\text{ZMP}}_{W_z} + b$, including one that will intercept the terrain. For walking robots, it is this point on the ground from which the external wrench can be described by a single force vector (and no moment) that is the famous "zero-moment point" (ZMP).

We can rewrite the equations of motion in terms of the ZMP. Substituting the equations of motion into equation \ref{eq:zmp_def}, to replace the individual terms of $f^{\text{ZMP}}_{net}$, we have \[ {}^Bp^{\text{ZMP}}_{W_z} m\ddot{x} - {}^Bp^{\text{ZMP}}_{W_x} (m\ddot{z} + mg) = I \ddot{\theta}. \] But this is precisely the equation \ref{eq:cop} presented above.

Furthermore, since \[\sum_i \left[ {}^Bp^{B_i}_W \times f^{B_i}_W \right]_y = \sum_i \left( {}^Bp^{B_i}_{W_z} f^{B_i}_{W_x} - {}^Bp^{B_i}_{W_x} f^{B_i}_{W_z} \right), \] and for flat terrain we have \[ \sum_i {}^Bp^{B_i}_{W_z} f^{B_i}_{W_x} = {}^Bp^{\text{ZMP}}_{W_z} \sum_i f^{B_i}_{W_x} = {}^Bp^\text{ZMP}_{W_z} f^\text{ZMP}_{net, W_x}, \] then we can see that this ZMP is exactly the CoP: \[ {}^Bp^\text{ZMP}_{W_x} = \frac{\sum_i {}^Bp^{B_i}_{W_x} f^{B_i}_{W_z}}{\sum_i f^{B_i}_{W_z} }. \]

In three dimensions, we solve for the point on the ground where $\tau^\text{ZMP}_{net,W_y} = \tau^\text{ZMP}_{net,W_x} = 0$, but allow $\tau^\text{ZMP}_{net,W_z} \ne 0$ to extract the analogous equations in the $y$-axis: \[ {}^Bp^{\text{ZMP}}_{W_z} m\ddot{y} - {}^Bp^{\text{ZMP}}_{W_y} (m\ddot{z} + mg) = I \ddot{\theta}. \]

A note about impact dynamics

In the previous chapter we devoted relatively a lot of attention to dynamics of impact, characterized for instance by a guard that resets dynamics in a hybrid dynamical system. In those models we used impulsive ground-reaction forces (these forces are instantaneously infinite, but doing finite work) in order to explain the discontinuous change in velocity required to avoid penetration with the ground. This story can be extended naturally to the dynamics of the center of mass.

For an articulated robot, though, there are a number of possible strategies for the legs which can affect the dynamics of the center of mass. For example, if the robot hits the ground with a stiff leg like the rimless wheel, then the angular momentum about the point of collision will be conserved, but any momentum going into the ground will be lost. However, if the robot has a spring leg and a massless toe like the SLIP model, then no energy need be lost.

One strategy that many highly-articulated legged robots use is to keep the vertical component of their center of mass at a constant velocity, $$\dot{z} = c \quad \Rightarrow \quad \ddot{z} = 0,$$ and minimize their angular momentum about the center of mass (here $I\dot\theta=0 \Rightarrow I\ddot\theta = 0$). Using this strategy, if the swinging foot lands roughly below the center of mass, then even with a stiff leg there is no energy dissipated in the collision - all of the momentum is conserved. This the common approach/justification for most of the humanoid literature ignoring the impact dynamics when reasoning about the centroidal dynamics.

Heuristic footstep planning

Coming soon. For a description of our approach with Atlas, see Deits14a+Kuindersma14.

Planning trajectories for the center of mass

The implementation that we used in the DARPA Robotics Challenge is described in Tedrake15 and is available in Drake as ZmpPlanner.

The ZMP Planner

I've coded up a simple example calling the planner with nominal footstep positions, generating a nominal zmp trajectory as a simple trajectory moving between the foot centers, and then calling ZmpPlanner to generate the center of mass trajectory.

Please take it for a spin! In particular, I recommend changing the footstep timing to see the impact on the center of mass trajectory.

Note that the ZmpPlanner also produces a feedback policy (on the center of mass position and velocity) in addition to the nominal plan.

From a CoM plan to a whole-body plan

Spatial momentum

Let's generalize our notation for a single rigid body a bit more before we make things more complex.

In the simple model above, we chose to use the generalized positions $\bq = [\theta, x, z]$, where $x$ and $z$ were taken to be the position of the center of mass, ${}^Wp^{B_{cm}}$. Let's observe that this particular choice of generalized velocity actually corresponds to an important physical quantity -- the spatial velocity (the concatenation of the angular velocity and the translational velocity). Using the full monogram notation, we can write the spatial velocity of the center of mass of body B, relative to the world frame, as $${}^WV^{B_{cm}} = \begin{bmatrix} {}^W\dot\theta^B \\ {}^W\dot{p}^{B_{cm}} \end{bmatrix}.$$ We can similarly define a spatial force, $F^{B_{cm}}_i$, by concentating the moment and forces, and spatial inertia of the body taken around the center of mass: $$M^{B/B_{cm}} = \begin{bmatrix} I & 0 & 0\\ 0 & m & 0 \\ 0 & 0 & m \end{bmatrix},$$ and the spatial momentum vector (the concatenation of angular momentum + linear/translational momentum) as $${}^W L^{B/B_{cm}}(\dot\bq) = \begin{bmatrix} I \dot\theta \\ m \dot{x} \\ m \dot{z} \end{bmatrix} = M^{B/B_{cm}} {}^WV^{B_{cm}}.$$ Now recognize that the left-hand side of equations (\ref{eq:spacecraft_x}-\ref{eq:spacecraft_theta}) can be written simply as \begin{equation}{}^W\dot{L}^{B/B_{cm}}(\ddot\bq) = \sum_i F^{B_{cm}}_i,\label{eq:Ldot}\end{equation} the rate of change of the spatial momentum is just the sum of the spatial forces applied applied at $B_{cm}$.

Generalization to multibody

We can write the configuration-dependent center of mass of the entire robot, $R,$ as the (weighted) sum of the center of masses of the individual links: $$\begin{bmatrix} x_{cm} \\ z_{cm} \end{bmatrix} \equiv {}^Wp^{R_{cm}}(\bq) = \frac{\sum_i {}^W p^{B_{i,cm}}(\bq) \, m_i}{\sum_i m_i} = \frac{\sum_i {}^WX^{B_i}(\bq) \,{}^B_i p^{B_{i,cm}} \, m_i}{\sum_i m_i}.$$

One can define the spatial momentum of the entire robot, $R,$ about this point. It is now configuration dependent, but takes a particularly simple form: $${}^W L^{R/R_{cm}}(\bq, \dot\bq) = {\bf A}(\bq)\dot\bq,$$ it can be computed in Drake for a MultibodyPlant using the CalcSpatialMomentumInWorldAboutPoint() method. The matrix ${\bf A}(\bq)$ is known as the centroidal momentum matrixOrin13. The rate of change of this whole-body spatial momentum is still given by the sum of the forces (once the spatial tranforms are properly applied): \begin{equation}{}^W\dot{L}^{R/R_{cm}}(\bq, \dot\bq, \ddot\bq) = {\bf A}(\bq)\ddot{\bq} + \dot{\bf A}(\bq, \dot\bq)\dot\bq = \sum_i F^{B_{cm}}_i.\label{eq:centroidal} \end{equation}

I suspect you already know this intuitively. Imagine you are an astronaut experiencing (effectively) zero-gravity. If you are not applying external forces, then your center of mass acceleration will be zero, and your total angular momentum will be conserved. Equation (\ref{eq:centroidal}) simply generalizes to the case where forces are still applied.

Let me make one important point very clear: this is not an approximation. This equations exactly describe the motion of the center of mass (and angular momentum about the center of mass). The simplification here is not because we are approximating the dynamics, but rather because we have not yet captured all of the constraints we might need to put on the forces we can produce with our limbs (due to kinematic constraints) and our actuators (due to e.g. torque limits).