获取多面体的表面区域(3D对象)

Question

我有一个3D表面,(想想xy平面).飞机可以倾斜.(想想一条斜坡路).

鉴于三维坐标的列表定义面(Point3D1X,Point3D1Y,Point3D1Z,Point3D12X,Point3D2Y,Point3D2Z,Point3D3X,Point3D3Y,Point3D3Z,等),如何计算表面的面积是多少？

请注意,我的问题类似于在2D平面中查找区域.在2D平面中,我们有一个定义多边形的点列表,使用这个点列表我们可以找到多边形的面积.现在假设所有这些点都具有z这样的值,即它们在3D中升高以形成表面.我的问题是如何找到3D表面的区域？

Answer 1

既然你说它是一个多面体,堆叠器的链接(http://softsurfer.com/Archive/algorithm_0101/algorithm_0101.htm)适用.

这是我对您的情况的C代码的近似C#转换:

// NOTE: The original code contained the following notice:
// ---------------------------------------
// Copyright 2000 softSurfer, 2012 Dan Sunday
// This code may be freely used and modified for any purpose
// providing that this copyright notice is included with it.
// iSurfer.org makes no warranty for this code, and cannot be held
// liable for any real or imagined damage resulting from its use.
// Users of this code must verify correctness for their application.
// ---------------------------------------
// area3D_Polygon(): computes the area of a 3D planar polygon
//    Input:  int n = the number of vertices in the polygon
//            Point[] V = an array of n+2 vertices in a plane
//                       with V[n]=V[0] and V[n+1]=V[1]
//            Point N = unit normal vector of the polygon's plane
//    Return: the (float) area of the polygon
static float
area3D_Polygon( int n, Point3D[] V, Point3D N )
{
    float area = 0;
    float an, ax, ay, az;  // abs value of normal and its coords
    int   coord;           // coord to ignore: 1=x, 2=y, 3=z
    int   i, j, k;         // loop indices

    // select largest abs coordinate to ignore for projection
    ax = (N.x>0 ? N.x : -N.x);     // abs x-coord
    ay = (N.y>0 ? N.y : -N.y);     // abs y-coord
    az = (N.z>0 ? N.z : -N.z);     // abs z-coord

    coord = 3;                     // ignore z-coord
    if (ax > ay) {
        if (ax > az) coord = 1;    // ignore x-coord
    }
    else if (ay > az) coord = 2;   // ignore y-coord

    // compute area of the 2D projection
    for (i=1, j=2, k=0; i<=n; i++, j++, k++)
        switch (coord) {
        case 1:
            area += (V[i].y * (V[j].z - V[k].z));
            continue;
        case 2:
            area += (V[i].x * (V[j].z - V[k].z));
            continue;
        case 3:
            area += (V[i].x * (V[j].y - V[k].y));
            continue;
        }

    // scale to get area before projection
    an = Math.Sqrt( ax*ax + ay*ay + az*az);  // length of normal vector
    switch (coord) {
    case 1:
        area *= (an / (2*ax));
        break;
    case 2:
        area *= (an / (2*ay));
        break;
    case 3:
        area *= (an / (2*az));
        break;
    }
    return area;
}

Answer 2

我赞成了一些 我认为正确的答案。但我认为最简单的方法 - 无论是 2D 还是 3D，都是使用以下公式：

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area = sum(V(i+1) \xc3\x97 V(i))/2;\n

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向量十字\xc3\x97在哪里。

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执行此操作的代码是：

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    public double Area(List<Point3D> PtList)\n    {\n\n        int nPts = PtList.Count;\n        Point3D a;\n        int j = 0;\n\n        for (int i = 0; i < nPts; ++i)\n        {\n            j = (i + 1) % nPts;\n            a += Point3D.Cross(PtList[i], PtList[j]);\n        }\n        a /= 2;\n        return Point3D.Distance(a,default(Point3D));\n    }\n\n    public static Point3D Cross(Point3D v0, Point3D v1)\n    {\n        return new Point3D(v0.Y * v1.Z - v0.Z * v1.Y,\n            v0.Z * v1.X - v0.X * v1.Z,\n            v0.X * v1.Y - v0.Y * v1.X);\n    }\n

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请注意，该解决方案不依赖于 x 平面的投影，我认为这很笨拙。

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你怎么认为？

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