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Get the surface area of a polyhedron (3D object)

I have a 3D surface, (think about the xy plane). The plane can be slanted. (think about a slope road).

Given a list of 3D coordinates that define the surface(Point3D1X, Point3D1Y, Point3D1Z, Point3D12X, Point3D2Y, Point3D2Z, Point3D3X, Point3D3Y, Point3D3Z, and so on), how to calculate the area of the surface?

Note that my question here is analogous to finding area in 2D plane. In 2D plane we have a list of points that defines a polygon, and using this list of points we can find the area of the polygon. Now assuming that all these points have z values in such a way that they are elevated in 3D to form a surface. My question is how to find the area of that 3D surface?

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Graviton Avatar asked Feb 28 '10 09:02

Graviton


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1 Answers

Since you say it's a polyhedron, stacker's link (http://softsurfer.com/Archive/algorithm_0101/algorithm_0101.htm) is applicable.

Here's my approximate C# translation of the C code for your situation:

// 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;
}
like image 127
Gabe Avatar answered Sep 17 '22 07:09

Gabe