我知道怎么做bicubic patch来自16 control points（即4x4网格），如B-spline，Catmull-Rom，Besier ...
但是，我希望在这些点上由4个控制点（p00,p01,p10,p11）和4个法线（n00,n01,n10,n11）描述的类似于Hermite spline的2D / 3D类似。
我尝试对它进行编程（首先沿u进行2x 1D插值，然后沿v插值结果）。
当我意识到我不知道如何使用表面法线而不是一维Hermite样条中使用的导数时，我陷入了困境。
如果我天真地插入法线而不是衍生物，我会得到无意义的结果。请参阅此处的C ++代码（我只参考了相关部分）：
// basis function
template <class TYPE>
inline void spline_hermite_basis( TYPE x, TYPE& c0, TYPE& c1, TYPE& d0, TYPE& d1 ){
TYPE x2 = x*x;
TYPE x3 = x*x2;
c0 = 2*x3 - 3*x2 + 1;
c1 = -2*x3 + 3*x2 ;
d0 = x3 - 2*x2 + x;
d1 = x3 - x2 ;
};
// derivative of basis functions
template <class TYPE>
inline void dspline_hermite_basis( TYPE x, TYPE& c0, TYPE& c1, TYPE& d0, TYPE& d1 ){
TYPE x2 = x*x;
c0 = 6*x2 - 6*x ;
c1 = -6*x2 + 6*x ;
d0 = 3*x2 - 4*x + 1;
d1 = 3*x2 - 2*x ;
};
void cubicPatch_point( double u, double v,
const Vec3d& p00, const Vec3d& p01, const Vec3d& p10, const Vec3d& p11,
const Vec3d& n00, const Vec3d& n01, const Vec3d& n10, const Vec3d& n11,
Vec3d& p, Vec3d& n
){
double cc0,cc1,cd0,cd1;
double dc0,dc1,dd0,dd1;
// interpolation along u
spline_hermite_basis<double>( u, cc0, cc1, cd0, cd1 );
dspline_hermite_basis<double>( u, dc0, dc1, dd0, dd1 );
Vec3d p0u,p1u,n0u,n1u;
p0u.set_mul( p00, cc0 ); p0u.add_mul( p01, cc1 ); p0u.add_mul( n00, cd0 ); p0u.add_mul( n01, cd1 ); // p0u = cc0*p00 + cc1*p01 + cd0*n00 + cd1*n01;
n0u.set_mul( p00, dc0 ); n0u.add_mul( p01, dc1 ); n0u.add_mul( n00, dd0 ); n0u.add_mul( n01, dd1 ); // n0u = dc0*p00 + dc1*p01 + dd0*n00 + dd1*n01;
p1u.set_mul( p10, cc0 ); p1u.add_mul( p11, cc1 ); p1u.add_mul( n10, cd0 ); p1u.add_mul( n11, cd1 ); // p1u = cc0*p10 + cc1*p11 + cd0*n10 + cd1*n11;
n1u.set_mul( p10, dc0 ); n1u.add_mul( p11, dc1 ); n1u.add_mul( n10, dd0 ); n1u.add_mul( n11, dd1 ); // n1u = dc0*p10 + dc1*p11 + dd0*n10 + dd1*n11;
// interpolation along v
spline_hermite_basis<double>( v, cc0, cc1, cd0, cd1 );
dspline_hermite_basis<double>( v, dc0, dc1, dd0, dd1 );
p.set_mul( p0u, cc0 ); p.add_mul( p1u, cc1 ); p.add_mul( n0u, cd0 ); p.add_mul( n1u, cd1 ); // p = cc0*p0u + cc1*p1u + cd0*n0u + cd1*n1u;
n.set_mul( p0u, dc0 ); n.add_mul( p1u, dc1 ); n.add_mul( n0u, dd0 ); n.add_mul( n1u, dd1 ); // n = dc0*p0u + dc1*p1u + dd0*n0u + dd1*n1u;
}