Get the surface area of a polyhedron (3D object)
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;
}
Do you mean Area of 3D planar polygons?
- http://softsurfer.com/Archive/algorithm_0101/algorithm_0101.htm
- http://local.wasp.uwa.edu.au/~pbourke/geometry/area3d/
- http://thebuildingcoder.typepad.com/blog/2008/12/3d-polygon-areas.html
I upvoted a few answers which I think are correct. But I think the simplest way to do it-- regardless of whether it's in 2D or 3D, is to use the following formula:
area = sum(V(i+1) × V(i))/2;
Where ×
is the vector cross.
The code to do this is:
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);
}
Note that the solution doesn't depend on projection to x-plane, which I think is clunky.
What do you think?