Multi-Body Dynamics
Deriving the equations of motion (an example)
The equations of motion for a standard robot can be derived using the method of Lagrange. Using $T$ as the total kinetic energy of the system, and $U$ as the total potential energy of the system, $L = T-U$, and $Q_i$ as the generalized force corresponding to $q_i$, the Lagrangian dynamic equations are: \begin{equation} \frac{d}{dt}\pd{L}{\dot{q}_i} - \pd{L}{q_i} = Q_i.\end{equation} You can think of them as a generalization of Newton's equations. For a particle, we have $T=\frac{1}{2}m \dot{x}^2,$ so $\frac{d}{dt}\pd{L}{\dot{x}} = m\ddot{x},$ and $\pd{L}{x} = -\pd{U}{x} = f$ amounting to $f=ma$. But the Lagrangian derivation works in generalized coordinate systems and even for constrained motion.
If you are not comfortable with these equations, then any good book
chapter on rigid body mechanics can bring you up to speed -- try
For completeness, I've included a derivation of the Lagrangian from the principle of stationary action below.
Simple Double Pendulum
Consider the simple double pendulum with torque actuation at both
joints and all of the mass concentrated in two points (for simplicity).
Using $\bq = [\theta_1,\theta_2]^T$, and ${\bf p}_1,{\bf p}_2$ to denote
the locations of $m_1,m_2$, respectively, the kinematics of this system
are:
\begin{eqnarray*}
{\bf p}_1 =& l_1\begin{bmatrix} s_1 \\ - c_1 \end{bmatrix}, &{\bf p}_2 =
{\bf p}_1 + l_2\begin{bmatrix} s_{1+2} \\ - c_{1+2} \end{bmatrix} \\
\dot{{\bf p}}_1 =& l_1 \dot{q}_1\begin{bmatrix} c_1 \\ s_1 \end{bmatrix},
&\dot{{\bf p}}_2 = \dot{{\bf p}}_1 + l_2 (\dot{q}_1+\dot{q}_2) \begin{bmatrix} c_{1+2} \\ s_{1+2} \end{bmatrix}
\end{eqnarray*}
Note that $s_1$ is shorthand for $\sin(q_1)$, $c_{1+2}$ is shorthand for
$\cos(q_1+q_2)$, etc. From this we can write the kinetic and potential
energy:
\begin{align*}
T =& \frac{1}{2} m_1 \dot{\bf p}_1^T \dot{\bf p}_1 + \frac{1}{2} m_2
\dot{\bf p}_2^T \dot{\bf p}_2 \\
=& \frac{1}{2}(m_1 + m_2) l_1^2 \dot{q}_1^2 + \frac{1}{2} m_2 l_2^2 (\dot{q}_1 + \dot{q}_2)^2 + m_2 l_1 l_2 \dot{q}_1 (\dot{q}_1 + \dot{q}_2) c_2 \\
U =& m_1 g y_1 + m_2 g y_2 = -(m_1+m_2) g l_1 c_1 - m_2 g l_2 c_{1+2}
\end{align*}
Taking the partial derivatives $\pd{T}{q_i}$, $\pd{T}{\dot{q}_i}$, and
$\pd{U}{q_i}$ ($\pd{U}{\dot{q}_i}$ terms are always zero), then
$\frac{d}{dt}\pd{T}{\dot{q}_i}$, and plugging them into the Lagrangian,
reveals the equations of motion:
\begin{align*}
(m_1 + m_2) l_1^2 \ddot{q}_1& + m_2 l_2^2 (\ddot{q}_1 + \ddot{q}_2) + m_2 l_1 l_2 (2 \ddot{q}_1 + \ddot{q}_2) c_2 \\
&- m_2 l_1 l_2 (2 \dot{q}_1 + \dot{q}_2) \dot{q}_2 s_2 + (m_1 + m_2) l_1 g s_1 + m_2 g l_2 s_{1+2} = \tau_1 \\
m_2 l_2^2 (\ddot{q}_1 + \ddot{q}_2)& + m_2 l_1 l_2 \ddot{q}_1 c_2 + m_2 l_1 l_2
\dot{q}_1^2 s_2 + m_2 g l_2 s_{1+2} = \tau_2
\end{align*}
As we saw in chapter 1, numerically integrating (and animating) these
equations in
The Manipulator Equations
If you crank through the Lagrangian dynamics for a few simple robotic
manipulators, you will begin to see a pattern emerge - the resulting
equations of motion all have a characteristic form. For example, the
kinetic energy of your robot can always be written in the form:
\begin{equation} T = \frac{1}{2} \dot{\bq}^T \bM(\bq)
\dot{\bq},\end{equation} where $\bM$ is the state-dependent inertia matrix
(aka mass matrix). This observation affords some insight into general
manipulator dynamics - for example we know that ${\bf M}$ is always
positive definite, and symmetric
Continuing our abstractions, we find that the equations of motion of a
general robotic manipulator (without kinematic loops) take the form
\begin{equation}{\bf M}(\bq)\ddot{\bq} + \bC(\bq,\dot{\bq})\dot{\bq} =
\btau_g(\bq) + {\bf B}\bu, \label{eq:manip} \end{equation} where $\bq$ is
the joint position vector, ${\bf M}$ is the inertia matrix, $\bC$ captures
Coriolis forces, and $\btau_g$ is the gravity vector. The matrix $\bB$
maps control inputs $\bu$ into generalized forces. Note that we pair
$\bM\ddot\bq + \bC\dot\bq$ on the left side because "... the equations of
motion depend on the choice of coordinates $\bq$. For this reason neither
$\bM\ddot\bq$ nor $\bC\dot\bq$ individually should be thought of as a
generalized force; only their sum is a force"
Manipulator Equation form of the Simple Double Pendulum
The equations of motion from Example 1 can be written compactly as: \begin{align*} \bM(\bq) =& \begin{bmatrix} (m_1 + m_2)l_1^2 + m_2 l_2^2 + 2 m_2 l_1l_2 c_2 & m_2 l_2^2 + m_2 l_1 l_2 c_2 \\ m_2 l_2^2 + m_2 l_1 l_2 c_2 & m_2 l_2^2 \end{bmatrix} \\ \bC(\bq,\dot\bq) =& \begin{bmatrix} 0 & -m_2 l_1 l_2 (2\dot{q}_1 + \dot{q}_2)s_2 \\ m_2 l_1 l_2 \dot{q}_1 s_2 & 0 \end{bmatrix} \\ \btau_g(\bq) =& -g \begin{bmatrix} (m_1 + m_2) l_1 s_1 + m_2 l_2 s_{1+2} \\ m_2 l_2 s_{1+2} \end{bmatrix} , \quad \bB = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{align*} Note that this choice of the $\bC$ matrix was not unique. The manipulator equations are very general, but they do define some
important characteristics. For example, $\ddot{\bq}$ is (state-dependent)
linearly related to the control input, $\bu$. This observation justifies
the control-affine form of the dynamics assumed throughout the notes. There
are many other important properties that might prove useful. For instance,
we know that the $i$ element of $\bC\dot{q}$ is given by
Note that we have chosen to use the notation of second-order systems
(with $\dot{\bq}$ and $\ddot{\bq}$ appearing in the equations) throughout
this book. Although I believe it provides more clarity, there is an
important limitation to this notation: it is impossible to describe 3D
rotations in "minimal coordinates" using this notation without introducing
kinematic singularities (like the famous "gimbal lock"). For instance, a
common singularity-free choice for representing a 3D rotation is the unit
quaternion, described by 4 real values (plus a norm constraint). However we
can still represent the rotational velocity without singularities using just
3 real values. This means that the length of our velocity vector is no
longer the same as the length of our position vector. For this reason, you
will see that most of the software in
Recursive Dynamics Algorithms
The equations of motions for our machines get complicated quickly.
Fortunately, for robots with a tree-link kinematic structure, there are
very efficient and natural recursive algorithms for generating these
equations of motion. For a detailed reference on these methods, see
Hamiltonian Mechanics
In some cases, it might be preferable to use the Hamiltonian formulation of the dynamics, with state variables $\bq$ and $\bp$ (instead of $\dot{\bq}$), where $\bp = \pd{L}{\dot{\bq}}^T = \bM \dot\bq$ results in the dynamics: \begin{align*}\bM(\bq) \dot{\bq} =& \bp \\ \dot{\bp} =& {\bf c}_H(\bq,\bp) + \tau_g(\bq) + \bB\bu, \qquad \text{where}\quad {\bf c}_H(\bq,\bp) = \frac{1}{2} \left(\pd{ \left[ \dot{\bq}^T \bM(\bq) \dot{\bq}\right]}{\bq}\right)^T.\end{align*}
Bilateral Position Constraints
If our robot has closed-kinematic chains, for instance those that arise from a four-bar linkage, then we need a little more. The Lagrangian machinery above assumes "minimal coordinates"; if our state vector $\bq$ contains all of the links in the kinematic chain, then we do not have a minimal parameterization -- each kinematic loop adds (at least) one constraint so should remove (at least) one degree of freedom. Although some constraints can be solved away, the more general solution is to use the Lagrangian to derive the dynamics of the unconstrained system (a kinematic tree without the closed-loop constraints), then add additional generalized forces that ensure that the constraint is always satisfied.
Consider the constraint equation \begin{equation}\bh(\bq) = 0.\end{equation} For the case of the kinematic closed-chain, this can be the kinematic constraint that the location of one end of the chain is equal to the location of the other end of the chain. The equations of motion can be written \begin{equation}\bM({\bq})\ddot{\bq} + \bC(\bq,\dot{\bq})\dot\bq = \btau_g(\bq) + \bB\bu + \bH^T(\bq) \blambda,\label{eq:constrained_manip}\end{equation} where $\bH(\bq) = \pd\bh{\bq}$ and ${\blambda}$ is the constraint force. Let's use the shorthand \begin{equation} \bM({\bq})\ddot{\bq} = \btau(q,\dot{q},u) + \bH^T(\bq) \blambda. \label{eq:manip_short} \end{equation}
Using \begin{gather}\dot\bh = \bH\dot\bq,\\ \ddot\bh = \bH \ddot\bq + \dot\bH \dot\bq, \label{eq:ddoth} \end{gather} we can solve for $\blambda$, by observing that when the constraint is imposed, $\bh=0$ and therefore $\dot\bh=0$ and $\ddot\bh=0$. Inserting the dynamics (\ref{eq:manip_short}) into (\ref{eq:ddoth}) yields \begin{equation}\blambda =- (\bH \bM^{-1} \bH^T)^+ (\bH \bM^{-1} \btau + \dot\bH\dot\bq).\label{eq:lambda_from_h}\end{equation} The $^+$ notation refers to a Moore-Penrose pseudo-inverse. In many cases, this matrix will be full rank (for instance, in the case of multiple independent four-bar linkages) and the traditional inverse could have been used. When the matrix drops rank (multiple solutions), then the pseudo-inverse will select the solution with the smallest constraint forces in the least-squares sense.
To combat numerical "constraint drift", one might like to add a restoring force in the event that the constraint is not satisfied to numerical precision. To accomplish this, rather than solving for $\ddot\bh = 0$ as above, we can instead solve for \begin{equation}\ddot\bh = \bH\ddot\bq + \dot\bH\dot\bq = -2\alpha \dot\bh - \alpha^2 \bh,\end{equation} where $\alpha>0$ is a stiffness parameter. This is known as Baumgarte's stabilization technique, implemented here with a single parameter to provide a critically-damped response. Carrying this through yields \begin{equation} \blambda =- (\bH \bM^{-1} \bH^T)^+ (\bH \bM^{-1} \btau + (\dot{\bH} + 2\alpha \bH)\dot{\bq} + \alpha^2 \bh). \end{equation}
There is an important optimization-based derivation/interpretation of
these equations, which we will build on below, using Gauss's
Principle of Least Constraint. This principle says that we can
derive the constrained dynamics as : \begin{align} \min_\ddot{\bq} \quad
& \frac{1}{2} (\ddot{\bq} - \ddot{\bq}_{uc})^T \bM (\ddot{\bq} -
\ddot{\bq}_{uc}), \label{eq:least_constraint} \\ \subjto \quad &
\ddot{\bh}(\bq,\dot\bq,\ddot\bq) = 0 = \bH \ddot{\bq} + \dot\bH \dot{\bq},
\nonumber \end{align} where $\ddot{\bq}_{uc} = \bM^{-1}\tau$ is the
"unconstrained acceleration"
Bilateral Velocity Constraints
Consider the constraint equation \begin{equation}\bh_v(\bq,\dot{\bq}) = 0,\end{equation} where $\pd{\bh_v}{\dot{\bq}} \ne 0.$ These are less common, but arise when, for instance, a joint is driven through a prescribed motion. Here, the manipulator equations are given by \begin{equation}\bM\ddot{\bq} = \btau + \pd{\bh_v}{\dot{\bq}}^T \blambda.\end{equation} Using \begin{equation} \dot\bh_v = \pd{\bh_v}{\bq} \dot{\bq} + \pd{\bh_v}{\dot{\bq}}\ddot{\bq},\end{equation} setting $\dot\bh_v = 0$ yields \begin{equation}\blambda = - \left( \pd{\bh_v}{\dot{\bq}} \bM^{-1} \pd{\bh_v}{\dot{\bq}} \right)^+ \left[ \pd{\bh_v}{\dot{\bq}} \bM^{-1} \btau + \pd{\bh_v}{\bq} \dot{\bq} \right].\end{equation}
Again, to combat constraint drift, we can ask instead for $\dot\bh_v = -\alpha \bh_v$, which yields \begin{equation}\blambda = - \left( \pd{\bh_v}{\dot{\bq}} \bM^{-1} \pd{\bh_v}{\dot{\bq}} \right)^+ \left[ \pd{\bh_v}{\dot{\bq}} \bM^{-1} \btau + \pd{\bh_v}{\bq} \dot{\bq} + \alpha \bh_v \right].\end{equation}
The Dynamics of Contact
The dynamics of multibody systems that make and break contact are closely related to the dynamics of constrained systems, but tend to be much more complex. In the simplest form, you can think of non-penetration as an inequality constraint: the signed distance between collision bodies must be non-negative. But, as we have seen in the chapters on walking, the transitions when these constraints become active correspond to collisions, and for systems with momentum they require some care. We'll also see that frictional contact adds its own challenges.
There are three main approaches used to modeling contact with "rigid" body systems: 1) rigid contact approximated with stiff compliant contact, 2) hybrid models with collision event detection, impulsive reset maps, and continuous (constrained) dynamics between collision events, and 3) rigid contact approximated with time-averaged forces (impulses) in a time-stepping scheme. Each modeling approach has advantages/disadvantages for different applications.
Before we begin, there is a bit of notation that we will use throughout. Let $\phi(\bq)$ indicate the relative (signed) distance between two rigid bodies. For rigid contact, we would like to enforce the unilateral constraint: \begin{equation} \phi(\bq) \geq 0. \end{equation} We'll use ${\bf n} = \pd{\phi}{\bq}$ to denote the "contact normal", as well as ${\bf t}_1$ and ${\bf t}_2$ as a basis for the tangents at the contact (orthonormal vectors in the Cartesian space, projected into joint space), all represented as row vectors in joint coordinates.
We will also find it useful to assemble the contact normal and tangent vectors into a single matrix, $\bJ$, that we'll call the contact Jacobian: $$\bJ(\bq) = \begin{bmatrix} {\bf n}\\ {\bf t}_1\\ {\bf t}_2 \end{bmatrix}.$$ As written, $\bJ\bv$ gives the relative velocity of the closest points in the contact coordinates; it can be also be extended with three more rows to output a full spatial velocity (e.g. when modeling torsional friction). The generalized force produced by the contact is given by $\bJ^T \lambda$, where $\lambda = [f_n, f_{t1}, f_{t2}]^T:$ \begin{equation} \bM(\bq) \dot{\bv} + \bC(\bq,\bv)\bv = \btau_g(\bq) + \bB \bu + \bJ^T(\bq)\lambda. \end{equation}
Compliant Contact Models
Most compliant contact models are conceptually straight-forward: we
will implement contact forces using a stiff spring (and
damper
Coulomb friction is described by two parameters -- $\mu_{static}$ and $\mu_{dynamic}$ -- which are the coefficients of static and dynamic friction. When the contact tangential velocity (given by ${\bf t}\bv$) is zero, then friction will produce whatever force is necessary to resist motion, up to the threshold $\mu_{static} f_n.$ But once this threshold is exceeding, we have slip (non-zero contact tangential velocity); during slip friction will produce a constant drag force $|f_t| = \mu_{dynamic} f_n$ in the direction opposing the motion. This behaviour is not a simple function of the current state, but we can approximate it with a continuous function as pictured below.
With these two laws, we can recover the contact forces as relatively simple functions of the current state. However, the devil is in the details. There are a few features of this approach that can make it numerically unstable. If you've ever been working in a robotics simulator and watched your robot take a step only to explode out of the ground and go flying through the air, you know what I mean.
In order to tightly approximate the (nearly) rigid contact that most
robots make with the world, the stiffness of the contact "springs" must be
quite high. For instance, I might want my 180kg humanoid robot model to
penetrate into the ground no more than 1mm during steady-state standing.
The challenge with adding stiff springs to our model is that this results
in stiff
differential equations (it is not a coincidence that the word
stiff is the common term for this in both springs and ODEs). As a
result, the best implementations of compliant contact for simulation make
use of special-purpose numerical integration recipes (e.g.
But there is another serious numerical challenge with the basic implementation of this model. Computing the signed-distance function, $\phi(\bq)$, when it is non-negative is straightforward, but robustly computing the "penetration depth" once two bodies are in collision is not. Consider the case of use $\bphi(\bq)$ to represent the maximum penetration depth of, say, two boxes that are contacting each other. With compliant contact, we must have some penetration to produce any contact force. But the direction of the normal vector, ${\bf n} = \pd{\bphi}{\bq},$ can easily change discontinuously as the point of maximal penetration moves, as I've tried to illustrate here:
If you really think about this example, it actually even more of the foibles of trying to even define the contact points and normals consistently. It seems reasonable to define the point on body B as the point of maximal penetration into body A, but notice that as I've drawn it, that isn't actually unique! The corresponding point on body A should probably be the point on the surface with the minimal distance from the max penetration point (other tempting choices would likely cause the distance to be discontinuous at the onset of penetration). But this whole strategy is asymmetric; why shouldn't I have picked the vector going with maximal penetration into body B? My point is simply that these cases exist, and that even our best software implementations get pretty unsatisfying when you look into the details. And in practice, it's a lot to expect the collision engine to give consistent and smooth contact points out in every situation.
Some of the advantages of this approach include (1) the fact that it is easy to implement, at least for simple geometries, (2) by virtue of being a continuous-time model it can be simulated with error-controlled integrators, and (3) the tightness of the approximation of "rigid" contact can be controlled through relatively intuitive parameters. However, the numerical challenges of dealing with penetration are very real, and they motivate our other two approaches that attempt to more strictly enforce the non-penetration constraints.
Rigid Contact with Event Detection
Impulsive Collisions
The collision event is described by the zero-crossings (from positive to negative) of $\phi$. Let us start by assuming frictionless collisions, allowing us to write \begin{equation}\bM\ddot{\bq} = \btau + \lambda {\bf n}^T,\end{equation} where ${\bf n}$ is the "contact normal" we defined above and $\lambda \ge 0$ is now a (scalar) impulsive force that is well-defined when integrated over the time of the collision (denoted $t_c^-$ to $t_c^+$). Integrate both sides of the equation over that (instantaneous) interval: \begin{align*}\int_{t_c^-}^{t_c^+} dt\left[\bM \ddot{\bq} \right] = \int_{t_c^-}^{t_c^+} dt \left[ \btau + \lambda {\bf n}^T \right] \end{align*} Since $\bM$, $\btau$, and ${\bf n}$ are constants over this interval, we are left with $$\bM\dot{\bq}^+ - \bM\dot{\bq}^- = {\bf n}^T \int_{t_c^-}^{t_c^+} \lambda dt,$$ where $\dot{\bq}^+$ is short-hand for $\dot{\bq}(t_c^+)$. Multiplying both sides by ${\bf n} \bM^{-1}$, we have $$ {\bf n} \dot{\bq}^+ - {\bf n}\dot{\bq}^- = {\bf n} \bM^{-1} {\bf n}^T \int_{t_c^-}^{t_c^+} \lambda dt.$$ After the collision, we have $\dot\phi^+ = -e \dot\phi^-$, with $0 \le e \le 1$ denoting the coefficient of restitution, yielding: $$ {\bf n} \dot{\bq}^+ - {\bf n}\dot{\bq}^- = -(1+e){\bf n}\dot{\bq}^-,$$ and therefore $$\int_{t_c^-}^{t_c^+} \lambda dt = - (1+e)\left[ {\bf n} \bM^{-1} {\bf n}^T \right]^\# {\bf n} \dot{\bq}^-.$$ I've used the notation $A^\#$ here to denote the pseudoinverse of $A$ (I would normally write $A^+,$ but have changed it for this section to avoid confusion). Substituting this back in above results in \begin{equation}\dot{\bq}^+ = \left[ \bI - (1+e)\bM^{-1} {\bf n}^T \left[{\bf n} \bM^{-1} {\bf n}^T \right]^\# {\bf n} \right] \dot{\bq}^-.\end{equation}
We can add friction at the contact. Rigid bodies will almost always
experience contact at only a point, so we typically ignore torsional
friction
A spinning ball bouncing on the ground.
Imagine a ball (a hollow-sphere) in the plane with mass $m$, radius $r$. The configuration is given by $q = \begin{bmatrix} x, z, \theta \end{bmatrix}^T.$ The equations of motion are $$\bM(\bq)\ddot\bq = \begin{bmatrix} m & 0 & 0 \\ 0 & m & 0 \\ 0 & 0 & \frac{2}{3}mr^2 \end{bmatrix} \ddot\bq = \begin{bmatrix} 0 \\ -g \\ 0 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ 0 & r \end{bmatrix} \begin{bmatrix} \lambda_z \\ \lambda_x \end{bmatrix} = \tau_g + \bJ^T {\bf \lambda},$$ where ${\bf \lambda}$ are the contact forces (which are zero except during the impulsive collision). Given a coefficient of restitution $e$ and a collision with a horizontal ground, the post-impact velocities are: $$ \dot{\bq}^+ = \begin{bmatrix} \frac{3}{5} & 0 & -\frac{2}{5} r \\ 0 & - e & 0 \\ -\frac{3}{5r} & 0 & \frac{2}{5}\end{bmatrix}\dot{\bq}^-.$$
Putting it all together
We can put the bilateral constraint equations and the impulsive collision equations together to implement a hybrid model for unilateral constraints of the form. Let us generalize \begin{equation}\bphi(\bq) \ge 0,\end{equation} to be the vector of all pairwise (signed) distance functions between rigid bodies; this vector describes all possible contacts. At every moment, some subset of the contacts are active, and can be treated as a bilateral constraint ($\bphi_i=0$). The guards of the hybrid model are when an inactive constraint becomes active ($\bphi_i>0 \rightarrow \bphi_i=0$), or when an active constraint becomes inactive ($\blambda_i>0 \rightarrow \blambda_i=0$ and $\dot\phi_i > 0$). Note that collision events will (almost always) only result in new active constraints when $e=0$, e.g. we have pure inelastic collisions, because elastic collisions will rarely permit sustained contact.
If the contact involves Coulomb friction, then the transitions between sticking and sliding can be modeled as additional hybrid guards.
Time-stepping Approximations for Rigid Contact
So far we have seen two different approaches to enforcing the inequality constraints of contact, $\phi(\bq) \ge 0$ and the friction cone. First we introduced compliant contact which effectively enforces non-penetration with a stiff spring. Then we discussed event detection as a means to switch between different models which treat the active constraints as equality constraints. But, perhaps surprisingly, one of the most popular and scalable approaches is something different: it involves formulating a mathematical program that can solve the inequality constraints directly on each time step of the simulation. These algorithms may be more expensive to compute on each time step, but they allow for stable simulations with potentially much larger and more consistent time steps.
Complementarity formulations
What class of mathematical program due we need to simulate contact?
The discrete nature of contact suggests that we might need some form of
combinatorial optimization. Indeed, the most common transcription is to
a Linear Complementarity Problem (LCP)
Rather than dive into the full transcription, which has many terms and can be relatively difficult to parse, let me start with two very simple examples.
Time-stepping LCP: Normal force
Consider our favorite simple mass being actuated by a horizontal force (with the double integrator dynamics), but this time we will add a wall that will prevent the cart position from taking negative values: our non-penetration constraint is $q \ge 0$. Physically, this constraint is implemented by a normal (horizontal) force, $f$, yielding the equations of motion: $$m\ddot{q} = u + f.$$
$f$ is defined as the force required to enforce the non-penetration constraint; certainly the following are true: $f \ge 0$ and $q \cdot f = 0$. $q \cdot f = 0$ is the "complementarity constraint", and you can read it here as "either q is zero or force is zero" (or both); it is our "no force at a distance" constraint, and it is clearly non-convex. It turns out that satisfying these constraints, plus $q \ge 0$, is sufficient to fully define $f$.
To define the LCP, we first discretize time, by approximating \begin{gather*}q[n+1] = q[n] + h v[n+1], \\ v[n+1] = v[n] + \frac{h}{m}(u[n] + f[n]).\end{gather*} This is almost the standard Euler approximation, but note the use of $v[n+1]$ in the right-hand side of the first equation -- this is actually a semi-implicit Euler approximation, and this choice is essential in the derivation.
Given $h, q[n], v[n],$ and $u[n]$, we can solve for $f[n]$ and $q[n+1]$ simultaneously, by solving the following LCP: \begin{gather*} q[n+1] = \left[ q[n] + h v[n] + \frac{h^2}{m} u[n] \right] + \frac{h^2}{m} f[n] \\ q[n+1] \ge 0, \quad f[n] \ge 0, \quad q[n+1]\cdot f[n] = 0. \end{gather*} Take a moment and convince yourself that this fits into the LCP prescription given above.
Note: Please don't confuse this visualization with the more common
visualization of the solution space for an LCP (in two or more
dimensions) in terms of "complementary cones"
Perhaps it's also helpful to plot the solution, $q[n+1], f[n]$ as a function of $q[n], v[n]$. I've done it here for $m=1, h=0.1, u[n]=0$:
In the (time-stepping, "impulse-velocity") LCP formulation, we write the contact problem in it's combinatorial form. In the simple example above, the complementarity constraints force any solution to lie on either the positive x-axis ($f[n] \ge 0$) or the positive y-axis ($q[n+1] \ge 0$). The equality constraint is simply a line that will intersect with this constraint manifold at the solution. But in this frictionless case, it is important to realize that these are simply the optimality conditions of a convex optimization problem: the discrete-time version of Gauss's principle that we used above. Using $\bv'$ as shorthand for $\bv[n+1]$, and replacing $\ddot{\bq} = \frac{\bv' - \bv}{h}$ in Eq \eqref{eq:least_constraint} and scaling the objective by $h^2$ we have: \begin{align*} \min_{\bv'} \quad & \frac{1}{2} \left(\bv' - \bv - h\bM^{-1}\btau\right)^T \bM \left(\bv' - \bv - h\bM^{-1}\btau\right) \\ \subjto \quad & \frac{1}{h}\phi(\bq') = \frac{1}{h} \phi(\bq + h\bv') \approx \frac{1}{h}\phi(\bq) + {\bf n}\bv' \ge 0. \end{align*} The objective is even cleaner/more intuitive if we denote the next step velocity that would have occurred before the contact impulse is applied as $\bv^- = \bv + h\bM^{-1}\btau$: \begin{align*} \min_{\bv'} \quad & \frac{1}{2} \left(\bv' - \bv^-\right)^T \bM \left(\bv' - \bv^-\right) \\ \subjto \quad & \frac{1}{h}\phi(\bq') \ge 0. \end{align*} The LCP for the frictionless contact dynamics is precisely the optimality conditions of this convex (because $\bM \succeq 0$) quadratic program, and once again the contact impulse, ${\bf f} \ge 0$, plays the role of the Lagrange multiplier (with units $N\cdot s$).
So why do we talk so much about LCPs instead of QPs? Well LCPs can also represent a wider class of problems, which is what we arrive at with the standard transcription of Coulomb friction. In the limit of infinite friction, then we could add an additional constraint that the tangential velocity at each contact point was equal to zero (but these equation may not always have a feasible solution). Once we admit limits on the magnitude of the friction forces, the non-convexity of the disjunctive form rear's it's ugly head.
Time-stepping LCP: Coulomb Friction
We can use LCP to find a feasible solution with Coulomb friction, but it requires some gymnastics with slack variables to make it work. For this case in particular, I believe a very simple example is best. Let's take our brick and remove the wall and the wheels (so we now have friction with the ground).
The dynamics are the same as our previous example, $$m\ddot{q} = u
+ f,$$ but this time I'm using $f$ for the friction force which is
inside the friction cone if $\dot{q} = 0$ and on the boundary of the
friction cone if $\dot{q} \ne 0;$ this is known as the principle of
maximum dissipation
Now, to write the friction constraints as an LCP, we must introduce some slack variables. First, we'll break the force into a positive component and a negative component: $f[n] = f^+ - f^-.$ And we will introduce one more variable $v_{abs}$ which will be non-zero if the velocity is non-zero (in either direction). Now we can write the LCP: \begin{align*} \find_{v_{abs}, f^+, f^-} \quad \subjto && \\ 0 \le v_{abs} + v[n+1] \quad &\perp& f^+ \ge 0, \\ 0 \le v_{abs} - v[n+1] \quad &\perp& f^- \ge 0, \\ 0 \le \mu mg - f^+ - f^- \quad &\perp& v_{abs} \ge 0,\end{align*} where each instance of $v[n+1]$ we write out with $$v[n+1] = v[n] + \frac{h}{m}(u + f^+ - f^-).$$ It's enough to make your head spin! But if you work it out, all of the constraints we want are there. For example, for $v[n+1] > 0$, then we must have $f^+=0$, $v_{abs} = v[n+1]$, and $f^- = \mu mg$. When $v[n+1] = 0$, we can have $v_{abs} = 0$, $f^+ - f^- \le \mu mg$, and those forces must add up to make $v[n+1] = 0$.
We can put it all together and write an LCP with both normal forces
and friction forces, related by Coulomb friction (using a polyhedral
approximation of the friction cone in 3d)
Anitescu's convex formulation
In
But be careful! Although the primal solution was convex, and the
dual problems are always convex, the objective here can be positive
semi-definite. This isn't a mistake --
When the tangential velocity is zero, this constraint is tight; for sliding contact the relaxation effectively causes the contact to "hydroplane" up out of contact, because $\phi(\bq') \ge h\mu {\bf d}_i\bv'.$ It seems like a quite reasonable approximation to me, especially for small $h$!
Let's continue on and write the dual program. To make the notation a little better, the us stack the Jacobian vectors into $\bJ_\beta$, such that the $i$th row is ${\bf n} + \mu{\bf d_i}$, so that we have $$\pd{L}{\bv'}^T = \bM(\bv' - \bv) - h\btau - \bJ_\beta^T \beta = 0.$$ Substituting this back into the Lagrangian give us the dual program: $$\min_{\beta \ge 0} \frac{1}{2} \beta^T \bJ_\beta \bM^{-1} \bJ_\beta^T \beta + \frac{1}{h}\phi(\bq) \sum_i \beta_i.$$
One final note: I've written just one contact point here to simplify
the notation, but of course we repeat the constraint(s) for each contact
point;
Todorov's convex formulation
According to
How should we interpret this? If you expand the linear terms, you will see that this is almost exactly the dual form we get from the position equality constraints formulation, including the Baumgarte stabilization. It's interesting to think about the implications of this formulation -- like the Anitescu relaxation, it is possible to get some "force at a distance" because we have not in any way expressed that $\lambda = 0$ when $\phi(\bq) > 0.$ In Anitescu, it happened in a velocity-dependent boundary layer; here it could happen if you are "coming in hot" -- moving towards contact with enough relative velocity that the stabilization terms want to slow you down.
In dual form, it is natural to consider the full conic description of the friction cone: $$\mathcal{FC}(\bq) = \left\{{\bf \lambda} = [f_n, f_{t1}, f_{t2}]^T \middle| f_n \ge 0, \sqrt{f^2_{t1}+f^2_{t2}} \le \mu f_n \right\}.$$ The resulting dual optimization is has a quadratic objective and second-order cone constraints (SOCP).
There are a number of other important relaxations that happen in
MuJoCo. To address the positive indefiniteness in the objective,
Relaxed complementarity-free formulation: Normal force
What equations do we get if we simulate the simple cart colliding with a wall example that we used above?
Following
What remains is to compute $f[n]$, and here is where things get
interesting. Motivated by the idea of "softening the Gauss principle"
(e.g. the principle of least constraint
Beyond Point Contact
Coming soon... Box-on-plane example. Multi-contact.
Also a single point force cannot capture effects like torsional friction, and performs badly in some very simple cases (imaging a box sitting on a table). As a result, many of the most effective algorithms for contact restrict themselves to very limited/simplified geometry. For instance, one can place "point contacts" (zero-radius spheres) in the four corners of a robot's foot; in this case adding forces at exactly those four points as they penetrate some other body/mesh can give more consistent contact force locations/directions. A surprising number of simulators, especially for legged robots, do this.
In practice, most collision detection algorithms will return a list of "collision point pairs" given the list of bodies (with one point pair per pair of bodies in collision, or more if they are using the aforementioned "multi-contact" heuristics), and our contact force model simply attaches springs between these pairs. Given a point-pair, $p_A$ and $p_B$, both in world coordinates, ...
Hydroelastic model in drake
Principle of Stationary Action
The principle of least action -- really the principle of stationary action -- is the most compact form of the classical laws of physics. This simple rule (it can be written in a single line) summarizes everything! Not only the principles of classical mechanics, but electromagnetism, general relativity, quantum mechanics, everything known about chemistry -- right down to the ultimate known constituents of matter, elementary particles.Sounds important!
I find a lot of presentations of the principle of stationary action derivation of the Lagrangian unnecessarily confusing. Here's my version, in case it helps.
Consider a trajectory $\bq(t)$ defined over $t \in [t_0, t_1]$. The principle of stationary action states that this trajectory is a valid solution to the differential equation governing our system if it represents a stationary point of the "action", defined as $$A = \int_{t_0}^{t_1} L(\bq,\dot{\bq})dt.$$ What does it mean for a trajectory to be a stationary point? Using the calculus of variations, we think of an infinitesimal variation of this trajectory: $\bq(t) + {\bf \epsilon}(t),$ where ${\bf \epsilon}(t)$ is an arbitrary differentiable function that is zero at $t_0$ and $t_1$. That this variation should not change the action means $\delta A = 0$ for any small ${\bf \epsilon}(t)$: \begin{align*} \delta A =& A(\bq(t) + {\bf \epsilon}(t)) - A(\bq(t)) \\ =& \int_{t_0}^{t_1} L\left(\bq(t)+{\bf \epsilon}(t),\, \dot\bq(t) + \dot{\bf \epsilon}(t)\right) dt - \int_{t_0}^{t_1} L\left(\bq(t),\,\dot\bq(t)\right)dt.\end{align*} We can expand the term in the first integral: $$L\left(\bq(t)+{\bf \epsilon}(t),\, \dot\bq(t) + \dot{\bf \epsilon}(t)\right) = L\left(\bq(t),\,\dot\bq(t)\right) + \pd{L}{\bq(t)}{\bf \epsilon}(t) + \pd{L}{\dot\bq(t)}\dot{\bf \epsilon}(t).$$ Substituting this back in and simplifying leaves: $$\delta A = \int_{t_0}^{t_1} \left[\pd{L}{\bq(t)}{\bf \epsilon}(t) + \pd{L}{\dot\bq(t)}\dot{\bf \epsilon}(t)\right]dt.$$ Now use integration by parts on the second term: $$\int_{t_0}^{t_1} \pd{L}{\dot\bq(t)}\dot{\bf \epsilon}(t)dt = \left[\pd{L}{\dot\bq(t)} {\bf \epsilon}(t) \right]_{t_0}^{t_1} - \int_{t_0}^{t_1} \frac{d}{dt} \pd{L}{\dot\bq(t)} {\bf \epsilon}(t) dt,$$ and observe that the first term in the right-hand side is zero since ${\bf \epsilon}(t_0) = {\bf \epsilon}(t_1) = 0$. This leaves $$\delta A = \int_{t_0}^{t_1} \left[\pd{L}{\bq} - \frac{d}{dt}\pd{L}{\dot\bq}\right]{\bf \epsilon}(t) dt = 0.$$ Since this must integrate to zero for any ${\bf \epsilon}(t)$, we must have $$\pd{L}{\bq} - \frac{d}{dt}\pd{L}{\dot\bq} = 0,$$ which can be multiplied by negative one to obtain the familiar form of the (unforced) Lagrangian equations of motion. The forced version follows from the variation $$\delta A = \int_{t_0}^{t_1} L(\bq(t),\dot\bq(t))dt + \int_{t_0}^{t_1} \bQ^T(t){\bf \epsilon}(t) dt = 0.$$