我有一个3D表面(想想xy平面)。飞机可以倾斜。(想一想坡路)。
给定定义曲面的3D坐标列表(Point3D1X
、Point3D1Y
、Point3D1Z
、Point3D12X
、Point3D2Y
、Point3D2Z
、Point3D3X
、Point3D3Y
、Point3D3Z
等),如何计算曲面的面积?
请注意,我在这里的问题类似于在2D平面中寻找面积。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
值,以使它们在3D中升高以形成曲面。我的问题是如何找到3D曲面的面积?
发布于 2010-02-28 18:48:52
我支持a few answers,我认为它是正确的。但我认为最简单的方法--不管是2D还是3D,都是使用下面的公式:
area = sum(V(i+1) × V(i))/2;
其中×
是vector cross。
执行此操作的代码为:
public double Area(List<Point3D> PtList)
{
int nPts = PtList.Count;
Point3D a;
int j = 0;
for (int i = 0; i < nPts; ++i)
{
j = (i + 1) % nPts;
a += Point3D.Cross(PtList[i], PtList[j]);
}
a /= 2;
return Point3D.Distance(a,default(Point3D));
}
public static Point3D Cross(Point3D v0, Point3D v1)
{
return new Point3D(v0.Y * v1.Z - v0.Z * v1.Y,
v0.Z * v1.X - v0.X * v1.Z,
v0.X * v1.Y - v0.Y * v1.X);
}
请注意,解决方案不依赖于对x平面的投影,我认为这很笨拙。
你认为如何?
发布于 2010-02-28 17:46:18
既然你说它是一个多面体,那么stacker的链接(http://softsurfer.com/Archive/algorithm_0101/algorithm_0101.htm)也是适用的。
下面是我根据您的情况对C代码进行的大致C#翻译:
// 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;
}
发布于 2010-02-28 17:46:56
你是说三维平面多边形的面积吗?
https://stackoverflow.com/questions/2350604
复制相似问题