Holonomic constraints and stabilization

Learning outcomes

Derive velocity and acceleration constraints.
Solve a multiplier system with consistent signs.
Explain drift and Baumgarte stabilization.

Prerequisites: Linear algebra, ODEs, and earlier lectures in this sequence.

Constraint levels

A holonomic constraint is $\phi(q)=0$ . For simple Euclidean coordinates, differentiating gives $Jv=0$ and $Ja+\dot Jv=0$ , where $J=\partial\phi/\partial q$ . A constrained mechanical system satisfies

Ma=\tau-c+J^\top f.

The multiplier f enforces the acceleration constraint. Using this plus-sign force convention gives

\begin{bmatrix}M&-J^\top\\J&0\end{bmatrix} \begin{bmatrix}a\\f\end{bmatrix} =\begin{bmatrix}\tau-c\\-\dot Jv\end{bmatrix}.

Changing the definition of multiplier sign changes the off-diagonal block. Keep the equations and implementation together.

Worked particle on a circle

Let $\phi(p)=\frac12(p^\top p-\ell^2)$ , so $J=p^\top$ and $\dot Jv=v^\top v$ . At $p=(1,0)$ , $v=(0,2)$ , $\ell=1$ , and no external force, the constraint requires $p^\top a=-4$ . A unit mass therefore has centripetal acceleration $a=(-4,0)$ .

A velocity can be tangent while the acceleration has a radial component. Setting radial acceleration to zero would immediately leave the circle.

Constraint drift

An acceleration-level solve does not by itself eliminate discrete position and velocity errors. Even if $Ja+\dot Jv=0$ holds at each computed state, numerical integration can accumulate $\phi\ne0$ .

Baumgarte stabilization replaces the target with

Ja+\dot Jv=-2\alpha Jv-\beta^2\phi.

The ideal constraint error then obeys

\ddot\phi+2\alpha\dot\phi+\beta^2\phi=0

. Large gains may make the numerical problem stiff. Projection methods are another approach, but projection changes momentum or energy unless designed carefully.

Practice

Practice 5 simulates the circle-constrained particle under gravity with and without stabilization. It reports maximum position-constraint error and the acceleration residual. Compare several timesteps before choosing gains.

Use consistent initial conditions: the position must satisfy $\phi=0$ and the velocity must satisfy $Jv=0$ . Starting off the manifold tests recovery, a different question from drift under consistent initialization.

Exercise and answer guide

If the same constraint row is duplicated, can you still invert $JM^{-1}J^\top$ ? Answer: generally no; the rows are dependent. Acceleration can remain well-defined while multipliers are nonunique. Use a rank-aware solve or remove redundant constraints, and inspect residuals rather than assuming a successful solve.

The next lecture interprets the acceleration correction as a minimum-change principle, then explains why MuJoCo's soft constraints should not be identified with this exact rigid equality solve MIT multibody dynamics.