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The center of pressure summarizes where a distributed normal pressure acts. On a planar contact region $A$ in $z=0$, take a nonnegative pressure field $p(x,y)$ and positive total normal force $N$. Its center of pressure is the pressure-weighted positionIt is a continuous weighted mean, rather than necessarily the point of largest pressure. For discrete normal forces $N_i\ge 0$ at $(x_i,y_i)$, replace the integrals by sums. The resulting convex combination places the center of pressure inside the contacts' convex hull, or support polygon.^[ref]

This definition preserves the torque as well as the resultant normal force. About the origin in the contact plane,A single vertical force $(0,0,N)$ applied at $(x_P,y_P,0)$ gives the same torque. Thus its line of action passes through the center of pressure. Conversely, force-and-torque measurements reconstruct the point throughThe origin and sign convention are part of these equations; moving the moment reference above the contact plane requires the corresponding torque translation.

For an original discrete example, four vertical reactions of $100$, $200$, $300$ and $200\mathrm{N}$ act at the successive corners $(0,0)$, $(0.40,0)$, $(0.40,0.30)$ and $(0,0.30)$ in metres. Their total normal force is $800\mathrm{N}$, with $M_x=150\mathrm{N}\mathrm{m}$ and $M_y=-200\mathrm{N}\mathrm{m}$. Hence the center of pressure is $(0.250,0.1875)\mathrm{m}$. A rigid plate on four load cells provides a simple force platform implementation: calibrate the force channels and positions, then test known vertical loads at known locations. Subtracting the unloaded plate's own force and moment removes its contribution when reconstructing the added load. This is a proposed measurement test, not experimental data.

A continuous example shows the inverse problem hidden by this useful compression. On a rectangle of length $L$ and width $B$, put $u=x/L$ and choose profiles independent of $y$:Both profiles are positive on $0\le u\le 1$. The added polynomial has zero integral and zero first weighted integral:Thus both profiles give $N=p_{0}LB$ and the same center of pressure $(0.60L,B/2)$, while their peak pressures are $1.6p_0$ and $1.9p_0$. These are pressure distributions with identical resultant force and center of pressure. For $L=0.40\mathrm{m}$, $B=0.30\mathrm{m}$ and $N=800\mathrm{N}$, both give $(0.240,0.150)\mathrm{m}$ despite different local loads. A force platform alone therefore does not identify every local pressure.

Two limits matter. If $N=0$, the normal-pressure center of pressure is undefined. At fixed $N$, an absolute moment error $\delta M_y$ changes the reconstructed $x_P$ by $-\delta M_y/N$, so fixed absolute measurement uncertainty is amplified as a contact unloads. Also, tangential contact forces can leave a residual contact torque at the center of pressure: with full resultant $\mathbf{F}$,Selecting the normal-pressure centroid does not force this yaw torque to vanish.

Under the gravity-only, resting assumptions of support polygon, the center of pressure coincides with the horizontal center of mass projection. During motion it need not. For a planar model driven only by gravity and ground contact forces, at fixed center of mass height $h$ and with constant angular momentum about that centre, force and torque balance give $x_P=x_{\mathrm{CM}}-(h/g){\ddot{x}}_{\mathrm{CM}}$. The acceleration term explains why the center of pressure is a contact-force measurement rather than a general substitute for center of mass position.^[ref]

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.

Tipping of a rigid body begins when a supported configuration can no longer balance its applied forces and torques with admissible contact forces, and rotation about a support edge becomes possible. For gravity alone on a horizontal surface, the center of mass projection must lie in the support polygon. A boundary projection marks incipient unloading of contacts; it is not a general guarantee about subsequent dynamics. The unilateral-contact and force/torque conditions are explicit in the primary robotics treatment.^[ref]

A planar horizontal pull gives a useful distinction between tipping and sliding. Let the gravitational weight be $mg$, the inward distance of the center of mass projection from the candidate tipping edge be $d>0$, and the pull's application height be $h>0$. Taking torque about that edge gives the limiting balanceThe contact forces elsewhere on the base then unload. With coefficient of static friction $\mu$, the horizontal sliding limit in this same model is $F_{\mathrm{slide}}=\mu mg$. Tipping can occur first only if $d/h<\mu$; the geometric support polygon by itself does not decide this competition.

For an original example, choose $m=20\mathrm{kg}$, $d=0.15\mathrm{m}$, $h=0.60\mathrm{m}$, and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. The tipping threshold is $49.05\mathrm{N}$. With $\mu =0.40$, sliding would require $78.48\mathrm{N}$, so tipping is the first predicted loss of this planar static configuration. With $\mu =0.15$, sliding starts at $29.43\mathrm{N}$ instead. Equal thresholds require a coupled contact analysis rather than a strict choice between the two labels.

A bench comparison can use a rigid box, a horizontal force gauge at a known height, and a slow force increase while observing the first loss of full-base contact or translation. Include the box's actual combined center of mass, not just its visible geometric centre. The static tipping margin measures the gravity-only geometric reserve; a pull, anchoring, changing contact set, or nonhorizontal ground changes the governing balance problem. The full friction constraints must then be retained.^[ref]

A trot in a quadrupedal robot coordinates diagonal leg pairs: front-left with hind-right, and front-right with hind-left. The two pairs alternate their stance phases and swing phases. The pairing and relative timing define the pattern; the fraction of each cycle spent in contact can vary.

For an original periodic model, take cycle period $T$ and normalized cycle phase $u=t/T(\mathrm{mod}1)$. Let the front-left/hind-right pair have phase zero and the other pair have phase $1/2$. Each foot remains in stance phase for fraction $\beta$ of the cycle. At $\beta =1/2$, one diagonal pair contacts while the other swings, apart from the idealized transition instants.

The contact-time consequences follow directly. For $\beta <1/2$, the diagonal-pair contacts do not overlap, and the fraction of the cycle with no foot contact isFor $\beta >1/2$, the intervals overlap, and the fraction with all four feet in contact is $2\beta -1$. Thus $\beta =0.4$ gives 20% no-contact time, while $\beta =0.6$ gives 20% four-foot overlap. The average number of contacting feet is $4\beta$ in either case, but an average does not describe the instantaneous support.

For stipulated $T=0.5\mathrm{s}$ and $\beta =0.6$, each foot's stance lasts $0.3\mathrm{s}$ and swing lasts $0.2\mathrm{s}$. Each pair overlaps the other for $0.05\mathrm{s}$ at each exchange, totaling $0.1\mathrm{s}$ per cycle. These are constructed timings, not measurements of a particular machine.

Two ideal diagonal point contacts have a support polygon reduced to a line segment. Static balance would require the projected center of mass to lie on that segment under the usual horizontal-plane assumptions, leaving no area margin to lateral disturbance. Dynamic motion additionally involves contact forces, angular momentum, friction and actuator limits. The contact and equilibrium models in the primary robotics text explain why a geometric schedule alone is insufficient.^[ref]

A proposed robot trial records all foot-contact events, body motion and ground contact forces while commanding the diagonal pattern. It checks whether the planned duty fraction and phase offsets are realized and whether motion remains feasible. Contact timing identifies the gait; the force and motion records test its execution.

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]