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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]

The center of pressure locates the resultant of a distributed fluid pressure load relative to its moment. On a planar surface carrying parallel pressure forces, the first-moment ratio in center of pressure applies. On an airfoil, pressure variation and its resultant can change with the flow state, so the point is not a permanently fixed geometric centre. NASA's primary account explicitly distinguishes a moving center of pressure from retaining a force and pitching moment about an aerodynamic center.^[ref]

For an original hydrostatic plate example, take a vertical rectangle of width $w$, extending from the free surface to depth $H$ in a constant-density fluid. If the same atmospheric pressure acts on its other face, the net gauge pressure at depth $y$ is $p(y)=\rho gy$. This hydrostatic relation follows from vertical force balance.^[ref] The resultant and its depth areFor $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $w=1\mathrm{m}$ and $H=2\mathrm{m}$, the net force is $19.62\mathrm{kN}$ at depth $1.333\mathrm{m}$. The area centroid is only one metre below the surface; the increasing pressure shifts the load's centroid downward.

A tank demonstration can infer the resultant line from the force and hinge moment on a small plate while changing the water depth. The plate's own gravitational moment and the pressure on both faces must be included. This tests a specified pressure-load model, not a general claim that the point always lies below every geometric centre.

For aerodynamic net loads, differences of pressures on opposite faces can have both signs. The convex-hull property proved for nonnegative contact pressure then need not apply, and a small resultant with a finite moment makes its location ill-conditioned. A curved surface or a general wrench may also require a residual couple rather than a single force that removes every component of torque. The resultant convention and moment reference are therefore part of the definition.

Pressure distributions with identical resultant force and center of pressure demonstrate that a pressure centroid is not a reconstruction of a pressure field. A normal-load measurement and its two first spatial moments constrain onlyA perturbation $\delta p$ preserving all three integrals leaves the resultant and center of pressure unchanged. It is physically admissible only when the perturbed field remains nonnegative under the nonadhesive pressure model. The polynomial profiles in the parent article give one smooth example; the following construction shows that even a peak-pressure bound is missing.

Take a rectangular region of length $L$ and width $B$, centered about its midpoint, and fix a normal load $N>0$. A uniform field has pressure $N/(LB)$ and centroid $(L/2,B/2)$. Instead place the entire load uniformly on a smaller centered square of side $a\le \min (L,B)$, with zero pressure outside that square. Its pressure isSymmetry fixes the same centroid, and integration still gives total force $N$. Yet its peak is $N/a^2$, which has no upper bound as $a$ shrinks. Smooth symmetric concentrated profiles can provide the same conclusion if a continuous pressure field is required.

For an original numerical example, let $L=0.40\mathrm{m}$, $B=0.20\mathrm{m}$, and $N=800\mathrm{N}$. Uniform pressure is $10\mathrm{kPa}$. Concentrating the same load on a centered $0.05\mathrm{m}$ square gives $320\mathrm{kPa}$ while preserving the measured point $(0.20,0.10)\mathrm{m}$. Both satisfy the stated nonnegative-field moment constraints.

Consequently a force/moment sensor cannot infer local peaks from the centroid alone. Local contact pressure from additional spatial measurements needs more data or justified material/contact assumptions, such as a constitutive model and known geometry. The source's normal-force centroid definition explains exactly which resultant information was retained.^[ref] The examples are inverse-problem constructions; they are not claims that arbitrary pressure maps can be realized by the same elastic surfaces.

A residual contact torque at the center of pressure is the moment left after translating a planar contact's resultant force to its normal-pressure centroid. Let the resultant wrench at an origin in $z=0$ be force $F=(F_x,F_y,N)$ and moment $M=(M_x,M_y,M_z)$, with $N>0$. Put $r_P=(x_P,y_P,0)$, where $x_P=-M_y/N$ and $y_P=M_x/N$. Translation givesThe two horizontal moment components vanish, but the yaw component need not. The zero-moment-point construction for flat robot contact removes moments perpendicular to the surface normal; it does not assert that every component is zero.^[ref]

A useful invariant makes the limitation exact. Since $F\cdot (r_P\times F)=0$,If this scalar is nonzero, no change of the moment origin can eliminate the entire moment vector. Replacing the whole contact by only one force at the center of pressure would discard that information.

For an original two-contact example, put equal normal loads of $100\mathrm{N}$ at $(+0.10,0)$ and $(-0.10,0)$ metres. Add tangential forces $(0,+15,0)\mathrm{N}$ at the positive position and $(0,-15,0)\mathrm{N}$ at the negative position. The total force is $(0,0,200)\mathrm{N}$, the pressure centroid is the origin, and the opposing tangential forces give a pure yaw couple of $3\mathrm{N}\mathrm{m}$. A friction coefficient of at least $0.15$ permits the stipulated tangential magnitudes at these loads.

A normal-load-only reconstruction would identify the correct centroid while missing the couple. A sensor measuring all force and moment components can distinguish the two situations. The origin, surface orientation and axis signs must be retained when comparing measurements. Near zero normal load, the centroid reconstruction itself becomes singular, independently of whether a tangential moment remains.

Spatial pressure-map resolution below a sensor cell concerns which variations of a contact pressure field can actually be distinguished by spatial measurements. A smaller display pixel does not imply a smaller independently measured contact region. The response of a pressure sensor, its sampling positions and any imposed reconstruction assumptions must be specified.

For an original ideal array model, cell $A_j$ of known area reports an area-average pressure,It determines that cell’s load $F_j=|A_j|y_j$, but not every value inside it. For a more general linear response, $y_j=\int K_j(x,y)p(x,y)dA$ with known sensitivity $K_j$. The null space of a spatial pressure measurement consists of perturbations $\delta p$ whose weighted integrals vanish for all measured channels. Nonnegative fields that differ by such a perturbation remain indistinguishable in that model.

Consider square cells of side $a$ aligned with $x=ja$, and compare fields independent of $y$:Every cell contains one full cosine period, so both fields give every channel the identical value $p_0$. Yet $p_2$ varies between $0.5p_0$ and $1.5p_0$. With $a=10\mathrm{mm}$ and $p_0=100\mathrm{kPa}$, a perfectly uniform field and a field spanning $50$–$150\mathrm{kPa}$ produce the same ideal map. On an aligned finite rectangular array, the added cosine also has zero first moment within each cell, so even the global resultant and center of pressure agree. This is an inverse-problem construction, not a claim that both profiles arise from the same elastic contact geometry.

Cell averaging has a calculable spatial response. For a one-dimensional centered interval of width $a$, averaging a complex wave $e^{2\pi i\nu x}$ multiplies its value at the interval center byThe first zero at $\nu =1/a$ explains the erased cosine above. Sampling introduces further ambiguity; even translating this ideal array cannot recover a component erased by its aperture at every position.

A regularity assumption can supply a conditional bound. If $p$ has a validated spatial Lipschitz constant $L_p$ over a square cell, its center value differs from its area average by at mostEvery point in the cell is at most half the diagonal from its center, so integrating the Lipschitz inequality proves this bound. For $a=10\mathrm{mm}$ and $L_p=2kPa/mm$, it gives about $14.14\mathrm{kPa}$. The array readings alone do not establish that slope bound.

Interpolation chooses a finer representation from the measured values. SciPy explicitly describes its array zoom operation as spline interpolation; it does not describe collecting new pressure data.^[ref] A proposed apparatus test scans a known small indenter across a sensor array at controlled total load, comparing independent load measurements with channel responses. The indenter’s actual contact area, load spreading, sensor cross-coupling and repeatability must be measured before inferring a spatial response. Additional smaller-aperture measurements or justified mechanical constraints can improve identifiability; a smoother image alone supplies neither.

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.