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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]

In terrestrial robot locomotion, the center of pressure is the ground point summarizing the normal load shared by the active contacts. With contact positions $(x_i,y_i)$ and nonnegative normal reactions $N_i$, its coordinates are the weighted averages $x_P={\sum }_i{N}_i{x}_i/{\sum }_i{N}_i$ and $y_P={\sum }_i{N}_i{y}_i/{\sum }_i{N}_i$, provided the total normal force is positive. It lies in the current support polygon, but is a contact-force quantity rather than the robot's center of mass position.^[ref]

Force-sensor coordinates matter when reconstructing this point. Suppose a sensor origin lies height $h$ above a horizontal contact plane, with parallel coordinate axes. If it reports force $(F_x,F_y,F_z)$ and moment $(M_x,M_y,M_z)$, the plane point is located at $(x_P,y_P,-h)$ relative to that origin. Moment translation givesThe ground-plane transformation of a measured contact wrench must therefore precede interpreting moment/force ratios as a contact location.

For an original sensor example, take $h=0.05\mathrm{m}$, force $(30,-20,600)\mathrm{N}$, and measured roll/pitch moments $(17,-49.5)\mathrm{N}\mathrm{m}$. The reconstructed point is $(0.080,0.030)\mathrm{m}$. Ignoring the origin height would instead give $(0.0825,0.02833)\mathrm{m}$. Those millimetre differences are coordinate effects, not a change in the physical pressure distribution.

During motion, the contact point need not equal the horizontal projection of the center of mass. The source's constant-height, negligible-angular-momentum-rate model relates their difference to horizontal acceleration; those assumptions must be established before using that simplified relation.^[ref] A robot-foot experiment can compare known applied loads with the sensor's translated force/moment measurements. When $F_z$ approaches zero, reconstruction becomes sensitive to moment error; at zero normal load the normal-pressure point is undefined. A residual yaw couple may remain, as explained in residual contact torque at the center of pressure.

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.