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A penalty approximation of unilateral contact allows a small modeled overlap and uses a stiff restoring force instead of enforcing an exact rigid nonpenetration constraint. With signed gap $d$, positive in separation, a simple nonadhesive law isAn open or just-touching contact has zero force in this spring law; a sustained compressive load requires negative $d$. Thus it approximates unilateral contact rather than satisfying its rigid gap condition exactly. Tedrake's primary compliant-contact treatment uses stiff springs and damping to resist penetration and explains the resulting numerical stiffness.^[ref]

For an original gravity-supported example, a $2\mathrm{kg}$ mass with no other vertical load has $N=mg=19.62\mathrm{N}$. With $k=2\times 10^5\mathrm{N}{\mathrm{m}}^{-1}$, its static overlap is $mg/k=98.1\mu \mathrm{m}$. Increasing $k$ by a factor of 100 reduces that overlap to $0.981\mu \mathrm{m}$, but also makes the contact dynamics faster.

While the contact stays compressed, linearization about equilibrium givesThe normal contact penalty time scale is therefore proportional to $\sqrt{m/k}$. For the first stiffness above, ${\omega }_c\approx 316.2\mathrm{rad}{\mathrm{s}}^{-1}$; after increasing stiffness it is ten times larger. At fixed numerical step size, a smaller overlap error need not mean a more accurate trajectory.

For example, the Forward Euler method applied to this undamped oscillator has amplification factors $1\pm i{\omega }_c\Delta t$, whose magnitudes exceed one for every positive step. Merely shrinking the step does not make that method asymptotically stable for indefinite undamped oscillation, although it can reduce error on a finite interval. A suitable integration method, damping model and convergence check must be chosen. The source notes specifically identify stiff differential equations as a challenge for compliant contact simulation.^[ref]

A penalty stiffness is a chosen model parameter, not automatically the actual material stiffness. Impact dissipation and tangential friction still need their own constitutive choices; a purely elastic normal spring does not imply a perfectly inelastic collision.

A support polygon is the convex hull of a body's supporting contact positions, projected onto a horizontal plane. For a rigid body resting on that plane in a uniform gravitational field, with gravity as the only applied load besides support reactions, it describes which horizontal center of mass positions can admit static equilibrium. Widely separated feet can support a center of mass above empty space between them: the footprint's convex hull, rather than the material under every point, matters.^[ref]

The geometry follows directly from force and torque balance. Let the contact positions be $(x_i,y_i,0)$ and their upward normal forces be $N_i\ge 0$. The inequality expresses unilateral contact: an ordinary unbonded support can push the body upward but cannot pull it downward. For total mass $m>0$, gravitational acceleration $g>0$ and horizontal center of mass projection $c=(x_c,y_c)$, balance givesDivide by $mg$ and write ${\lambda }_i=N_i/(mg)$. ThenThus $c$ must be a convex combination of the contacts, which is exactly membership in their convex hull. Conversely, any such convex combination supplies nonnegative reactions $N_i=mg{\lambda }_i$ satisfying this model's balance equations. The center of pressure of these vertical contact forces coincides with $c$. This proves existence of allowable reactions; it does not determine every reaction uniquely when there are more contacts than independent balance equations.

For an original three-foot example, take contact positions $(0,0)$, $(1,0)$ and $(0,1)$ in metres. The support polygon is the triangle described by $x\ge 0$, $y\ge 0$, $x+y\le 1$. A $10\mathrm{kg}$ body with $c=(0.30,0.20)\mathrm{m}$ requires the convex combination coefficientsWith $g=9.81m/{\mathrm{s}}^2$, the corresponding normal forces are $49.05$, $29.43$ and $19.62\mathrm{N}$. Moving the projection to $(0.70,0.50)\mathrm{m}$ would require ${\lambda }_1=-0.20$, a downward reaction at the first foot. No set of nonnegative normal forces can hold that configuration at rest under the stated loads.

A simple test uses a rigid plate on three small supports, a movable known mass, and three load cells measuring the upward reactions. Include the plate's own mass and center of mass when calculating the combined projection. As that projection approaches the edge $x+y=1$, the first support's reading approaches zero. Beyond the edge, maintaining the same contact configuration would require the forbidden downward reaction; the observed motion or changed contact set marks the failure of the original static equilibrium. This is a proposed apparatus test of the balance equations, not reported experimental data.

The distance from an interior projection to the polygon's nearest edge is a static tipping margin; here it is $\min (0.30,0.20,0.50/\sqrt{2})=0.20\mathrm{m}$. A boundary projection has zero margin: some supports unload, and an arbitrarily small outward shift can make balance impossible. A strictly interior projection gives a positive geometric margin against small horizontal shifts under the same model. It is not a general certificate of dynamical stability for a flexible or articulated system.

This explains a constraint in a quadrupedal robot's gait. Four feet at the corners of a square give a square support polygon. Lifting the corner $(1,1)$ leaves the triangle $x\ge 0$, $y\ge 0$, $x+y\le 1$. A center of mass projection at $(0.60,0.60)$ was inside the square but is outside that triangle, so the body must shift or use dynamics to lift that foot without losing the assumed balance. The contact set must be updated when computing the support polygon.

The assumptions matter when applying the test. Nonhorizontal contacts require the full balance equations and admissible contact forces, including Coulomb friction; a projected hull alone need not decide feasibility on irregular terrain.^[ref] Additional horizontal loads also change the balance problem. For example, a horizontal force $F$ applied at center of mass height $h$ produces an overturning torque $Fh$ about the relevant edge. If the inward horizontal distance to that edge is $d$, incipient tipping of a rigid body satisfies $Fh=mgd$. Whether the body slides first depends on static friction. The gravity-only support polygon contains no information about $h$ or that additional load.

Unilateral contact permits bodies to separate while preventing them from passing through each other. In an ideal rigid, nonadhesive model, the supporting normal contact force can push but cannot pull. It is therefore different from a permanently attached support: an upward pull may lift a body away instead of producing a downward support reaction.

Let $d$ be the signed normal gap, positive when the bodies are separated, and $N$ the outward compressive normal contact force. The nonpenetration constraint and absence of adhesion give the unilateral contact complementarity conditionsAn open gap requires $N=0$, while $N>0$ requires a closed gap. The pair $d=N=0$ is also allowed: touching does not by itself imply a loaded contact.^[ref] Tangential friction requires an additional relation such as Coulomb friction; it is not specified by the normal-gap conditions.

The equation of motion is still needed to determine the reaction. Consider a block of mass $m$ on a fixed horizontal floor, initially at rest at $d=0$, with an upward applied force $u$ through its center of mass. With upward positive, Newton's second law isIf the contact remains closed, $\ddot{d}=0$ would require $N=mg-u$. That is feasible for $u\le mg$. For $u>mg$, a negative normal reaction would violate unilateral contact, so the contact opens with $N=0$ and $\ddot{d}=(u-mg)/m>0$. At a closed contact with zero normal velocity, the appropriate acceleration-level contact complementarity iswhich yieldsThese acceleration conditions are being applied at a closed, initially stationary contact; they are not a rule forbidding downward acceleration during separated free flight.

For an original numerical example, let $m=2\mathrm{kg}$ and $g=9.81m/{\mathrm{s}}^2$. Pulling upward with $u=5\mathrm{N}$ leaves a normal reaction of $14.62\mathrm{N}$. At $u=25\mathrm{N}$, the reaction is zero and the initial upward acceleration is $2.69m/{\mathrm{s}}^2$. At $u=19.62\mathrm{N}$, both the reaction and the initial acceleration vanish, while touching remains possible.

A simple apparatus test places the block on a calibrated load cell and applies the upward pull through a spring balance, keeping the pull vertical through the center of mass to avoid an extra overturning torque. The support reading should decrease with the pull while contact persists, then vanish at lift-off. Recording the gap as well distinguishes an unloaded touch from a genuinely open contact. Real support elastic deformation and adhesion must be accounted for before treating that apparatus as the ideal rigid, nonadhesive model.

A penalty approximation of unilateral contact replaces perfect rigidity by a chosen stiff response, for exampleHere a small negative gap represents an allowed modeled overlap relative to undeformed reference surfaces. It is not an exact solution of $d\ge 0$. For the $14.62\mathrm{N}$ supported load above and chosen penalty stiffness $k=10^{5}N/m$, equilibrium requires $d=-0.1462\mathrm{mm}$. Increasing $k$ reduces that overlap at the same load. The penalty method and the ideal constraint are different approximations of the contact response.

Finally, impact requires a rule for the velocity change. The position-level inequalities alone do not decide whether an approaching body sticks or rebounds. A coefficient of restitution or another specified impact law determines that behavior, and the impulsive contact force is summarized by its impulse. Tedrake's particular time-stepping formulation identifies its block collisions as perfectly inelastic; that is a feature of the stated model, not a consequence of the three gap/reaction inequalities alone.^[ref]