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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 principle of transmissibility states that moving a force along its line of action preserves its external mechanical effect on an ideal rigid body. The force vector is unchanged, and its torque about every reference point is unchanged. The statement concerns external rigid-body motion or equilibrium; it does not assert unchanged internal deformation.

For an original proof, a nonzero force $\mathbf{F}$ is applied at $\mathbf{r}$. Move it to $\mathbf{r}+s\hat{\mathbf{F}}$, where $\hat{\mathbf{F}}=\mathbf{F}/\Vert \mathbf{F}\Vert$ and $s$ is a signed distance. The new moment isThe last cross product vanishes because its vectors are parallel. This is the same moment geometry used in the primary torque definition.^[ref]

For a numerical example, let $\mathbf{F}=(0,10,0)\mathrm{N}$ act at $(0.30,0.20,0)\mathrm{m}$. Its moment is $(0,0,3)\mathrm{N}\mathrm{m}$. Moving the application point to $(0.30,0.80,0)$ preserves that moment because the displacement follows the force direction. Moving it instead to $(0.50,0.20,0)$ gives $(0,0,5)\mathrm{N}\mathrm{m}$. The second move needs an additional compensating couple to preserve the original wrench.

A one-dimensional elastic counterexample shows the rigid-body limitation. Consider a uniform bar along $0\le x\le L$, fixed at the left end, with a rightward axial force $F$ applied to a cross-section at $x=a$. In a small-strain axial model, internal force is $F$ on $0<x<a$ and zero on $a<x<L$. Its right-end displacement iswhere $E$ is Young's modulus and $A$ the section area. Applying the same force at $a=L$ instead gives $FL/(EA)$, although both applied forces share one line of action and give the same external resultant and moment.

With stipulated $F=100\mathrm{N}$, $L=1\mathrm{m}$, $E=200\mathrm{GPa}$ and $A=1\mathrm{m}{\mathrm{m}}^2$, the end displacement is $0.25\mathrm{mm}$ for $a=L/2$ and $0.50\mathrm{mm}$ for $a=L$. The linear stress–strain relation underlying that calculation is described in the primary elasticity section.^[ref]

A proposed comparison therefore measures support reaction and body deformation separately when changing load position along a common line. Equal reactions and moments can coexist with different local stress and displacement. The external free-body diagram may be equivalent while the internal loading model differs.