# Link points on a bicubic bezier patch

A bicubic bezier patch is defined by 16 control points.
Given two points both lying on the patch boundaries, I think that if you link the two points you will end up with a cubic bezier curve in 3D. Is that true ? If yes, how can I find the two middle control points of this bezier curve ?

Let's denote the bicubic Bezier patch by $S(u,v)$, with $0 \le u \le 1$, $0 \le v \le 1$. Suppose that the two given points are at parameter locations $(u_0, v_0)$ and $(u_1, v_1)$. Given the situation you described, we can assume that $v_0 = 0$ and $v_1 = 1$. If $u_0 = u_1$, then the curve that "joins" the two given locations is indeed a Bezier cubic curve. This curve is called an "isoparametric curve", or sometimes just an "iso-curve", for short. The "iso" prefix is because the points on the curve are produced by holding one parameter value fixed ($u=u_0$ in our case), and varying the other one ($v$).
The equation of the patch is $$S(u,v) = \sum_{i=0}^3 \sum_{j=0}^3 b_i(u)b_j(v)\mathbf {P}_{ij}$$ where the $b_i$ and $b_j$ are the cubic Bernstein polynomials. So, if we fix $u=u_0$, as above, then we get an iso-curve that is a function of the parameter $v$: $$S(u_0,v) = \sum_{j=0}^3 b_j(v) \left\{\sum_{i=0}^3 b_i(u_0)\mathbf {P}_{ij} \right\}$$ This is a cubic Bezier curve, and its control points $\mathbf Q_0$, $\mathbf Q_1$, $\mathbf Q_2$, $\mathbf Q_3$ are given by: $$\mathbf Q_j = \sum_{i=0}^3 b_i(u_0)\mathbf {P}_{ij} \quad (j = 0,1,2,3)$$ The control points $\mathbf Q_0$ and $\mathbf Q_3$ will lie on the edges of the patch, as you know; the other two will not.
The red dots are the points $\mathbf Q_0$, $\mathbf Q_1$, $\mathbf Q_2$, $\mathbf Q_3$, and the picture shows how they are calculated.