bezier曲面中的自动内边

Question

通常，Bezier曲面作为双三次曲面，有16个控制点。然而，在3 3dsMax中，内部边缘可以被隐藏起来以避免编辑和计算自动 (这是它们的默认状态)。这只剩下12个控制点，这使得编辑变得更简单。

XNA的Primitives3D样本项目代码(简化)：

void CreatePatchVertices(Vector3[] patch, int tessellation)
{
    Debug.Assert(patch.Length == 16);

    for (int i = 0; i <= tessellation; i++)
    {
        float ti = (float)i / tessellation;

        for (int j = 0; j <= tessellation; j++)
        {
            float tj = (float)j / tessellation;

            // Perform four horizontal bezier interpolations
            // between the control points of this patch.
            Vector3 p1 = Bezier(patch[0], patch[1], patch[2], patch[3], ti);
            Vector3 p2 = Bezier(patch[4], patch[5], patch[6], patch[7], ti);
            Vector3 p3 = Bezier(patch[8], patch[9], patch[10], patch[11], ti);
            Vector3 p4 = Bezier(patch[12], patch[13], patch[14], patch[15], ti);

            // Perform a vertical interpolation between the results of the
            // previous horizontal interpolations, to compute the position.
            Vector3 position = Bezier(p1, p2, p3, p4, tj);

            // Perform another four bezier interpolations between the control
            // points, but this time vertically rather than horizontally.
            Vector3 q1 = Bezier(patch[0], patch[4], patch[8], patch[12], tj);
            Vector3 q2 = Bezier(patch[1], patch[5], patch[9], patch[13], tj);
            Vector3 q3 = Bezier(patch[2], patch[6], patch[10], patch[14], tj);
            Vector3 q4 = Bezier(patch[3], patch[7], patch[11], patch[15], tj);

            // Compute vertical and horizontal tangent vectors.
            Vector3 tangentA = BezierTangent(p1, p2, p3, p4, tj);
            Vector3 tangentB = BezierTangent(q1, q2, q3, q4, ti);

            // Cross the two tangent vectors to compute the normal.
            Vector3 normal = Vector3.Cross(tangentA, tangentB);
            normal.Normalize();

            // Create the vertex.
            AddVertex(position, normal);
        }
    }
}

在此示例中，如何计算矢量5、6、9和10 (patch[5]等)自动的，就像3 3dsMax那样？

Answer 1

看起来3ds Max构建了一个平行四边形来获取内部点。

patch[5] = patch[0] + (patch[1] - patch[0]) + (patch[4] - patch[0]);
patch[5] = patch[1] + patch[4] - patch[0];

以及整个功能：

void CreatePatchVertices(Vector3[] s, int tessellation, bool isMirrored)
{
    Debug.Assert(s.Length == 16);

    for (int i = 0; i <= tessellation; i++)
    {
        float ti = (float)i / tessellation;

        for (int j = 0; j <= tessellation; j++)
        {
            float tj = (float)j / tessellation;

            // Compute automatic interior edges.
            s[5] = s[1] + s[4] - s[0];
            s[6] = s[2] + s[7] - s[3];
            s[9] = s[8] + s[13] - s[12];
            s[10] = s[11] + s[14] - s[15];

            // Perform four horizontal bezier interpolations
            // between the control points of this patch.
            Vector3 p1 = Bezier(s[0], s[1], s[2], s[3], ti);
            Vector3 p2 = Bezier(s[4], s[5], s[6], s[7], ti);
            Vector3 p3 = Bezier(s[8], s[9], s[10], s[11], ti);
            Vector3 p4 = Bezier(s[12], s[13], s[14], s[15], ti);

            // Perform a vertical interpolation between the results of the
            // previous horizontal interpolations, to compute the position.
            Vector3 position = Bezier(p1, p2, p3, p4, tj);

            // Perform another four bezier interpolations between the control
            // points, but this time vertically rather than horizontally.
            Vector3 q1 = Bezier(s[0], s[4], s[8], s[12], tj);
            Vector3 q2 = Bezier(s[1], s[5], s[9], s[13], tj);
            Vector3 q3 = Bezier(s[2], s[6], s[10], s[14], tj);
            Vector3 q4 = Bezier(s[3], s[7], s[11], s[15], tj);

            // Compute vertical and horizontal tangent vectors.
            Vector3 tangentA = BezierTangent(p1, p2, p3, p4, tj);
            Vector3 tangentB = BezierTangent(q1, q2, q3, q4, ti);

            // Cross the two tangent vectors to compute the normal.
            Vector3 normal = Vector3.Cross(tangentA, tangentB);
            normal.Normalize();

            // Create the vertex.
            AddVertex(position, normal);
        }
    }
}

在3ds Max和Primitives3D示例中，具有自动内部点的茶壶原语就是这样的：

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Answer 2

这个答案必须是多余的，但我想提醒您注意另一种构造贝塞尔曲面的方法，称为https://en.wikipedia.org/wiki/Coons_patch。Coons曲面只使用边界曲线，向双三次曲面的转换是非常直接的。

我建议这样做的原因是，它是为了精确地外推汽车设计中典型的边界特征线，并自动计算内部点(对于双三次bezier贴片)或镶嵌。

给出了bu0、bu1、bv0、bv1四条边界曲线；

bu0, bu1  // boundaries in u direction u <- [0, 1]
bv0, bv1  // in v direction  v <- [0, 1]
// bu0, bu1, bv0, bv1 are arrays of bezier coefficients

coons曲面上的内点和点可以通过分别求u和v方向的直纹曲面和u和v的双线性插值来求出。

// In matrix form, so that you can reduce it all together

                      ⎡        ⎤
ruled_u = [ (1-u) u ] ⎜ bv0(v) ⎟
                      ⎜ bv1(v) ⎟
                      ⎣        ⎦ 

// Similarly
                      ⎡        ⎤
ruled_v = [ (1-v) v ] ⎜ bu0(u) ⎟
                      ⎜ bu1(u) ⎟
                      ⎣        ⎦ 

// Assuming cubic bezier curves as boundary, if you select appropriate control point 
// instead of bv0(u) etc, i.e. bv0[1] for u = 1/3 and bv0[2] for u = 2/3, we can convert
// a coons patch into a bicubic patch, instead of evaluating the point on surface directly

                          ⎡               ⎤ ⎡         ⎤
bilinear_uv = [ (1-u) u ] ⎜ bu0(0) bu0(1) ⎟ ⎜ (1 - v) ⎟
                          ⎜ bu1(0) bu1(1) ⎟ ⎜    v    ⎟
                          ⎣               ⎦ ⎣         ⎦

// Then the 'interior' point at u and v is
Coons (u, v) = ruled_u + ruled_v - bilinear_uv

在coons补丁中，内部控制点可以计算为

Coons(1/3, 1/3)
Coons(1/3, 2/3)
Coons(2/3, 1/3)
Coons(2/3, 2/3)

在上述方程中，只选择1/3的bv01而不是bv0(1/3)等。

在Coons贴片中，插值比线性插值边界控制点更自然。在我的经验中也更“正确”。不仅在美学上，而且在参试后也是如此。

最近，我不得不在LCH颜色空间中生成逼近sRGB色域的bezier曲面。它只是一个转换和扭曲的RGB立方体。我所拥有的只是斑块的边界。仅使用线性插值时，内部值是错误的。最后，coons补丁给出的误差最小。

这里是从一个集合coons补丁中生成的双三次曲面的示例。

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