= Center of pressure
{title2=$\mathbf c_P=\int_A(x,y)p\,dA/\int_Ap\,dA$}
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 position
$$
N=\int_A p\,dA>0,\qquad
\boxed{x_P=\frac{\int_A xp\,dA}{N},\quad y_P=\frac{\int_A yp\,dA}{N}.}
$$
It is a continuous <weighted mean>, rather than necessarily the point of largest <pressure>. For discrete <normal forces> $N_i\ge0$ 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>.<Underactuated Robotics by Russ Tedrake>[Highly-articulated Legged Robots, Center of pressure (CoP) and Zero-moment point (ZMP), The special case of flat terrain]{ref}
This definition preserves the <torque> as well as the resultant normal <force>. About the origin in the contact plane,
$$
\mathbf M_O=\int_A (x,y,0)\times(0,0,p)\,dA
=\left(\int_A yp\,dA,-\int_A xp\,dA,0\right).
$$
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 through
$$
\boxed{x_P=-M_y/N,\qquad y_P=M_x/N.}
$$
The 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 $M_x=150\,\mathrm{N\,m}$ and $M_y=-200\,\mathrm{N\,m}$. Hence the <center of pressure> is \b[$(0.250,0.1875)\,\mathrm m$]. 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 $y$:
$$
p_1=p_0[1+0.6(2u-1)],\qquad
p_2=p_1+0.3p_0(6u^2-6u+1).
$$
Both profiles are positive on $0\le u\le1$. The added polynomial has zero integral and zero first weighted integral:
$$
\int_0^1(6u^2-6u+1)\,du=0,\qquad
\int_0^1u(6u^2-6u+1)\,du=0.
$$
Thus both profiles give $N=p_0LB$ 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>.
\Image[/center-of-pressure-nonuniqueness.png]
{title=Different <pressure> profiles can give the same <center of pressure> and total <normal force>}
{description=Original positive pressure profiles on a rectangular contact. Both have normalized integral one and normalized first weighted integral 0.60, but their peak pressures differ. The vertical line marks their shared pressure-weighted position, not a pressure maximum.}
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 $\mathbf F$,
$$
M_z^P=M_z-x_PF_y+y_PF_x.
$$
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 $x_P=x_{\rm CM}-(h/g)\ddot x_{\rm CM}$. The acceleration term explains why the <center of pressure> is a contact-force measurement rather than a general substitute for <center of mass> position.<Underactuated Robotics by Russ Tedrake>[Highly-articulated Legged Robots, equations cop and zmp and their stated constant-height/angular-momentum assumptions]{ref}
Back to article page
