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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]

Local contact pressure from additional spatial measurements requires more information than total normal force and two first moments. One explicit reconstruction model partitions the contact into cells of known areas $A_i$ and centroids $(x_i,y_i)$, assuming uniform pressure $p_i$ within each cell. Writing $F_i=A_i{p}_i$ givesThese are three linear measurements of the cell loads. They need not determine every $F_i$, as the nonuniqueness argument in pressure distributions with identical resultant force and center of pressure shows. The primary centroid definition specifies precisely which force/moment information is retained.^[ref]

For an original four-cell example, use equal $0.01{\mathrm{m}}^2$ cells centered at successive positions $(-a,-a),(a,-a),(a,a),(-a,a)$ with $a=0.05\mathrm{m}$. Loads $(10,20,30,40)\mathrm{N}$ give $N=100\mathrm{N}$ and centroid $(0,0.020)\mathrm{m}$. Adding $s(1,-1,1,-1)$ to these loads changes neither the sum nor either first moment. Nonnegative loads permit $-10\le s\le 20\mathrm{N}$. The alternative $(15,15,35,35)\mathrm{N}$ therefore gives the same global measurements.

An independently calibrated measurement of the first cell load resolves this one missing degree of freedom under the four-cell model. If it reads $10\mathrm{N}$, the global moment equations recover the original other three loads. Their cell pressures are $(1,2,3,4)\mathrm{kPa}$. Measuring all four cells instead provides redundant data for checking total load and the assumed geometry.

The local sensor output must actually represent a known cell-integrated force, or have a validated response model; a point sample cannot automatically be treated as an area average. Calibration errors, cell cross-coupling and imperfect coverage become additional measurement inputs. Spatial pressure-map resolution below a sensor cell remains limited even when every cell average is known: unmeasured within-cell peaks are not recovered by simply drawing a finer image. Constitutive or smoothness assumptions can add information, but must be identified as assumptions rather than extra measurements.

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.