This type of surface patch id defined by its four boundary curves.

Coons decided to search for an expression of the surface that satisfies,

i) It is symmetric in u and v and

ii) It is an interpolation of $P(u, 0)$ and $P(u, 1)$,in one direction and of $P(0, v)$ and $P(1, v)$ in the other direction.

The first step is to construct two lofted surfaces from the two sets of opposite boundary curves.

A. It includes areas 1,2 and 3

$P_{a}(u, v)=(1-v) P(u, 0)+v P(u, 1)$

B. It includes areas 4,2 and 5

$P_{b}(u, v)=(1-u) P(0, v)+v P(1, v)$

The second step is to tentatively attempt to create the final surface. To find the equation of required surface area we have to add above equation and subtract bilinear surface defined by points $P_{00}, P_{10}, P_{11} \ and \ P_{01}(\text { area } 2)$

C. $P_{a b}(u, v)=(1-v)(1-u) P_{00}+(1-v) P_{10}+v(1-u) P_{01}+u v P_{11}$

Note that $P_{q}$ and $P_{b}$ are lofted surface, whereas is bilinear surface, the final expression is,

$P(u, v)=P_{a}(u, v)+P_{b}(u, v)-P_{a b}(u, v)$

$\quad \quad \quad = \left[\begin{array}{ll}{1-v} & {v}\end{array}\right]\left[\begin{array}{l}{P(u, 0)} \\ {P(u, 1)}\end{array}\right]+\left[\begin{array}{ll}{1-u} & {u}\end{array}\right]\left[\begin{array}{l}{P(0, v)} \\ {P(1, v)}\end{array}\right]-\left[\begin{array}{cc}{1-u} & {u}\end{array}\right]\left[\begin{array}{cc}{P_{00}} & {P_{01}} \\ {P_{10}} & {P_{11}}\end{array}\right]\left[\begin{array}{c}{1-v} \\ {v}\end{array}\right]$

$P(u, v)=\left[\begin{array}{ccc}{1-u} & {u} & {1}\end{array}\right]\left[\begin{array}{ccc}{-P_{00}} & {-P_{01}} & {P(0, v)} \\ {-P_{10}} & {-P_{11}} & {P(1, v)} \\ {P(u, 0)} & {P(u, 1)} & {0}\end{array}\right]\left[\begin{array}{c}{1-v} \\ {v} \\ {1}\end{array}\right]$

Q. Four bounding curves for coon's path are given by,

$P(u, 0)=(\cos \pi u, 0, \sin \pi u)$

$P(u, 1)=(\cos \pi u, \sin \pi u, 0)$

$P(0, v)=(1,0,0)$

$P(1, v)=(-1,0,0)$

Find: $P_{x}(u, v)$

Solution:

$P(u, 0)=(\cos \pi u, 0, \sin \pi u)$

$\therefore P_{00}=P(0,0)=(1,0,0)$

$\therefore P_{10}=P(1,0)=(-1,0,0)$

$P(u, 1)=(\cos \pi u, \sin \pi u, 0)$

$\therefore P_{01}=P(0,1)=(1,0,0)$

$\therefore P_{11}=P(1,1)=(-1,0,0)$

$P(0, v)=(1,0,0)$

$P(1, v)=(-1,0,0)$

$P(u, v)=\left[\begin{array}{ccc}{1-u} & {u} & {1}\end{array}\right]\left[\begin{array}{ccc}{-P_{00}} & {-P_{01}} & {P(0, v)} \\ {-P_{10}} & {-P_{11}} & {P(1, v)} \\ {P(u, 0)} & {P(u, 1)} & {0}\end{array}\right]\left[\begin{array}{c}{1-v} \\ {v} \\ {1}\end{array}\right]$

$P_{x}(u, v)=\left[\begin{array}{ccc}{1-u} & {u} & {1}\end{array}\right]\left[\begin{array}{ccc}{-1} & {-1} & {1} \\ {1} & {1} & {-1} \\ {\cos \pi u} & {\cos \pi u} & {0}\end{array}\right]\left[\begin{array}{c}{1-v} \\ {v} \\ {1}\end{array}\right]$

$=[(1-u)+u+\cos \pi u (1-u)+u+\cos \pi u \quad(1-u)-u]\left[\begin{array}{c}{1-v} \\ {v} \\ {1}\end{array}\right]$

$=[(1-v)+((u-1)+u)+\cos \pi u+v((u-1)+u+\cos \pi u)+1-2 u]$

$=[(1-v)+(2 u-1+\cos \pi u)+v(2 u-1+\cos \pi u)+1-2 u]$

$= P_{x}(u, v)=[\cos \pi u]$