Consider a rigid body in contact with the ground plane. Let the ground plane normal be $n$. Let various contact forces $F_i$ act on the rigid body at positions $r_i$. The center of pressure is then defined as $$x_c = \frac{\sum_i (F_i \cdot n) r_i}{\sum_i F_i \cdot n}$$ where the generalization from finitely many forces to a force field should be obvious. We generally assume that the normal component of the contact forces, $F_i \cdot n$, is greater or equal to 0 (i.e. that the contact forces are non-sticky). This implies that, for $$\alpha_i := \frac{F_i \cdot n}{\sum_j F_j \cdot n}$$ we have $0 \le \alpha_i \le 1$. The center of pressure, as defined above, is thus a convex sum and must lie within the convex hull of the contact points $r_i$.

The value of the center of pressure concept comes from the fact that we can calculate it without having to know the contact points $r_i$ and the corresponding forces $F_i$. Let us assume we are given only the total force and torque resulting from the contact, \begin{eqnarray} F &=& \sum_i F_i\\ T &=& \sum_i r_i \times F_i \end{eqnarray} To calculate the center of pressure from these vectors, we note that \begin{eqnarray} n \times T &=& n \times \left( \sum_i r_i \times F_i \right) = \sum_i (n \times (r_i \times F_i))\\ &=& \sum_i \left( r_i (n \cdot F_i) - F_i (n \cdot r_i) \right) \end{eqnarray} As the contact points all lie in the plane, $n \cdot r_i$ is equal for all $i$. Let $n \cdot r_i =: h$. Then $$n \times T = \sum_i r_i (n \cdot F_i) - h \sum_i F_i = \sum_i r_i (n \cdot F_i) - h F$$ Thus $$x_c = \frac{n \times T + h F}{F \cdot n}$$