Polygons and meshes

In what follows are various notes and algorithms dealing with polygons and meshes.

Surface (polygonal) Simplification

Written by Paul Bourke
July 1997

The following describes a method for reducing the number of polygons making up a surface representation while still attempting to retain the fundamental form of the surface. The applications for this are obvious if performance improvements are being sought for rendering and/or interactive environments.

The basic approach is to repeatedly remove the polygons with the shortest shared edge. Two polygons are removed on each step, the vertices of the remaining polygons are shifted to the midpoint of the shorted edge.

The following example illustrates the technique for a facet representation of a sphere. The initial sphere has 4000 facets, on each iteration the number of polygons is reduced by 1000. The initial sphere is obviously inefficient, to begin with, there are regions with much more detail than others (eg: the poles). If a smooth shaded rendering is being used then the model with 1000 facets is probably just as good as the original with 4 times the number of facets.

As expected, for severe reductions in the number of polygons the surface undergoes a smoothing and loss of detail. This is readily illustrated for a gridded cube.

Careful consideration needs to be given to the edges of non-manifold surfaces. A straightforward implementation will slowly eat away at the edge when the shortest facet edge is on the edge of the surface.

And finally, applied to a model of the human cortex.

Clipping Polygonal Facets
with an Arbitrary Plane

Written by Paul Bourke
February 1997

This note along with the source code at the end clips a 3 vertex facet by an arbitrary plane. The facet is described by its 3 vertices, the clipping plane is specified by its normal and one point on the plane. The clipping side of the plane is taken to be that one containing the normal, in the diagram below this is the right side of the clipping plane.

The standard equation of a plane is

A x + B y + C z + D = 0

where (A,B,C) is the unit normal. The value of D is determined by substituting in the known point (Px,Py,Pz) on the plane, namely

D = - (A Px + B Py + C Pz)

For a vertex (Qx,Qy,Qz) the expression

side(Q) = A Qx + B Qy + C Qz + D

can be used to determine which side of the plane the vertex lies on. If it is positive the point lies on the same side as the normal, if negative it lies on the other side, if zero it lies on the plane.

After determining if an edge intersects the cutting plane it is necessary to calculate the intersection point as that will become a vertex of the clipped facet. Let the edge be between two points P0 and P1, the equation of the points along the line segment

P = P0 + u(P1 - P0)

where u lies between 0 and 1. Substituting this into the expression for the plane

A (P0x + u (P1x - P0x)) + B (P0y + u (P1y - P0y)) + C (P0z + u (P1z - P0z)) + D = 0

Solving for u

u = - side(P0) / (side(P1) - side(P0))

Substituting this into the equation of the line gives the actual intersection point.

There are 8 different cases to consider, they are illustrated in the table below and appear in order in the source code. The top two cases are when all the points are on either side of the clipping plane. The three cases on the left are when there is only one vertex on the clipping side of the plane, the three cases on the right occur when there are two vertices on the clipping side of the plane.


  • When there is only one point on the clipping side the resulting facet has 4 vertices, if the underlying database only deals with 3 vertex facets then this can be bisected using a number of methods.

  • When the facets to be clipped have more than 3 vertices then they can be subdivided first and each segment clipped individually.

  • The algorithm as presented has no divide by zero problems

  • The source below does the clipping in place, the order in which the vertices are calculated is important for the 3 cases when there is one point on the clipping side of the plane.

C Source

As an example and testing exercise, the following shows a 2 dimensional Gaussian the the result from three clipping planes.

Surface Relaxation and Smoothing

Written by Paul Bourke
January 1997

This document describes a method of smoothing a general surface described by a collection of planar facets. The facets need not be evenly spaced nor lie on a grid, they can form an arbitrary 3D surface. The method can be applied iteratively to smooth the surface to an arbitrary degree. In general the method attempts to preserve gross features and it tends to result in equal size facets (normally a desirable characteristic). This general technique is often known as surface relaxation.

Consider a point Pi surrounded by n vertices Pj making up the facets sharing vertex Pi.

Then the point Pi is perturbed as follows:

As an example consider the top figure below, it is a 2D Gaussian with noise added to each x,y,z component of each vertex. The subsequent smoothed versions are the result of iteratively applying the technique 1,2,3 and 6 times respectively.

The following is a more "real life" example, in particular, the facets are not arranged on a simple grid and are of varying sizes. The geometry is a 2cm slice from a model of the human cortex. The original surface on the left has clear vertical ribs which arose as a result of digitising scanned images which were slightly misaligned. The surface on the right has one iteration of relaxation applied to it. The vertical artifacts are clearly reduced.

The following shows a different sectioning of the same model through two relaxation iterations. The pairs of images are the same except that the facet edges are drawn on the surface on the right. In both cases the surface has been rendered using flat shading, that is, without smoothing across facets.

Original surface

One degree of relaxation

Two degrees of relaxation

And finally the whole surface

Geometric Crumpling

Written by Paul Bourke
January 1994

In most rendering applications rough surfaces can be simulated with what are known as texture maps or more precisely bump maps. These are images which perturb the surface normal during the rendering process creating the appearance of a rough surface or raised and lowered structure such as grouting in a tiled floor.

In some instances it is desirable to actually represent the surface roughness geometrically. For example, many textures don't retain their desired effect when viewed from very close. Some effects are also hard to generate using bump maps, for example, the extent of the deformation is limited.

There are many ways of modifying the geometry depending on the exact effect. One effect is that of crumpling, what you might do to a piece of paper. In the following example, each facet is split into a 3x3 grid. Each vertex is then displaced by a random amount (Gaussian distribution). Of course for each displacement all facets which share that vertex are also modified.

The above shows a Menger sponge before and after the crumpling exercise.

There are a few things to note
* This whole process is inefficient to say to least. In this case there has almost been a 10 fold increase in the geometric content.
* 4 point facets are in general no longer coplanar even if they were in the original model, triangulating these further increases the geometric content.
* The extent of the displacement is limited if it is undesirable for facets to intersect.

Splitting facets

Written by Paul Bourke
November 1991

The following discusses a number of ways of splitting up facet based computer models into smaller facets. There are a number of reasons for wanting to do this, some include:

  • In order to add detail to the surface, for example, fractal spatial subdivision
  • For surface relaxation algorithms
  • To eliminate narrow angled facets which cause problems for some rendering and texture algorithms
Method Additional Geometry Comments
Increases the number of facets by 3 Perhaps the most common splitting technique. Doesn't result in finer internal angles.
Doubles the number of facets Simplest method, bisection. Normally the longest edge or the widest internal angle would be bisected. While it reduces long edges it also tends to produce narrow internal angles
2 additional facets The centroid becomes the new vertex. If the facet is part of a height field the centroid height would normally be the average of the 3 original vertex heights.
Results in an identical but smaller triangular facet and three new 4 vertex facets. Uncommon
Yields three 4 vertex facets. The centroid is usually used as the mid point.
Gives a smaller 3 point facet and a new 4 point facet. This is a common first approach for long thin facets, the cut is made along the two longest edges
Subdivide a 4 point facet into two 4 point facets Simple example of a more general repeated bisection of a facet. The longest opposite pair of edges are split if equal size facets are desirable.
Results in a small 4 point facet and four new triangular facets. Uncommon
Bisection into two triangular facets The simplest splitting of a 4 point facet. Most commonly used by rendering program to ensure all facets are planar. Results in facets with small internal angles.
Gives three triangular facets. The centroid is normally used. An improved technique to simple bisection for smoothing height fields.

Facets, planes, normals, rendering

Written by Paul Bourke
November 1992

The usual way of representing a bounded planar surface (facet) in computer graphic applications is as a sequence of vertices. Shading algorithms and ray-tracing generally requires knowledge of the normal to the facet, this is calculated by taking the cross product two of the edge vectors of the facet. The angle the normal makes with the light source vector determines the degree of shading of the facet. In particular, if the normal points towards the light source then the surface is brightly illuminated, if it points away from the light source then the surface is in shadow.

A common problem arises because the vertices need then to be specified in a specific order. A common convention is they need to be ordered such that the normal points outwards. This assumes there is an "outer" and "inner" ie: that the object is closed.

The usual way of calculating the angle between the normal and light source vector involves taking the cross product between these two vectors giving the cosine of the angle between them. Some rendering packages simple use the absolute value of this angle thus shading the back side the same as the side facing the light source.

Another technique which is common when you don't have control over the source of the facets nor the rendering package is to double up each facet, the duplicate having its vertices in the reverse order. These two sided facets can obviously make up objects which are not closed.

The basic problem arises because facets/planes don't exist in real life, all planar surfaces have a finite thickness. This is similar to the issue of representing lines in ray-tracing packages....they must be turned into objects with finite thickness such as cylinders. Creating planes with a finite thickness has been recognised for a long time in Architectural modelling where creating walls from infinitely thin planes leads to all sorts of problems not encountered if their true thickness is used.

Polygon types

Written by Paul Bourke
January 1993

There are a number of categories of polygons in common usage in computer modelling and graphics. The particular polygon type being used can have a dramatic effect on the complexity of many rendering and editing algorithms. For example, an algorithm for splitting a polygon into a number of 3 vertex facets is trivial for convex polygons and quite problematic for polygons with holes.

While most of the discussion will be with regard to polygons as bounded planes in 3D, the same ideas apply to polygons in 2D.

Some of the more frequently used categories will be listed and discussed below.

3 vertex facet

This is the simplest type of polygon. Perhaps one of the most important characteristics is the points lie on a plane, as such it is often the most fundamental primitive for 3D rendering applications which expect a single unambiguous normal for the whole polygonal region.

Rectangular facet

Rectangular or 4 vertex polygons are generated from gridded datasets and polygonal approximations of 2D surfaces. Many applications where such rectangular facets arise naturally don't ensure the vertices are coplanar, fortunately it is trivial to turn such a polygon into triangular and hence planar polygons.

Convex polygon

This is the simplest type of polygon with more than 3 or 4 vertices.

Concave polygon

This is the most general type of "simple" polygon, that is, without holes.

Complex polygon

Polygons with holes are defined in a number of ways. The holes can be "tagged" as such, a common method is to define the vertices of the holes in a order different from that of the solid parts. For example the solid parts might be defined clockwise about the normal, the holes anticlockwise about the normal.

Such polygons can be turned into concave polygons by introducing 2 coincident edges, between the solid and hole polygons. For multiple holes and concave solid pieces the coincident edges need to be taken between appropriate vertices to avoid making an overlapping polygon.

Intersecting polygon

These somewhat perverse polygons normally get lumped in with complex polygons with holes. The example on the right is the most straight forward case, situations where the polygon covers part of itself are generally avoided and are not handled consistently by rendering engines.

Determining whether or not a polygon (2D) has its vertices ordered clockwise or counterclockwise

Written by Paul Bourke
March 1998

The following describes a method for determining whether or not a polygon has its vertices ordered clockwise or anticlockwise for both convex and concave polygons. A polygon will be assumed to be described by N vertices, ordered

(x0,y0), (x1,y1), (x2,y2), . . . (xn-1,yn-1)

A simple test of vertex ordering for convex polygons is based on considerations of the cross product between adjacent edges. If the cross product is positive then it rises above the plane (z axis up out of the plane) and if negative then the cross product is into the plane.

cross product = ((xi - xi-1),(yi - yi-1)) x ((xi+1 - xi),(yi+1 - yi))

= (xi - xi-1) * (yi+1 - yi) - (yi - yi-1) * (xi+1 - xi)

A positive cross product means we have a counterclockwise polygon.

To determine the vertex ordering for concave polygons one can use a result from the calculation of polygon areas, where the area is given by

If the above expression is positive then the polygon is ordered counter clockwise otherwise if it is negative then the polygon vertices are ordered clockwise.

Test for concave/convex polygon

For a convex polygon all the cross products of adjacent edges will be the same sign, a concave polygon will have a mixture of cross product signs.

Source Code

Example and test program for testing whether a polygon is convex or concave. For MS, contributed by G. Adam Stanislav.
Source Code by the author

Clipping a line segment to a complex polygon

Written by Paul Bourke
August 1997
The following describes a procedure for clipping a line segment to a complex polygon. Complex polygon refers to both concave polygons and polygons with holes.

Whether the requirement is to retain that portion of the line within the polygon or remove the portion of the polygon within the polygon the concept is the same.

Consider a parametric expression for the line segment between two points P1 and P2.

P = P1 + mu (P2 - P1) where 0 <= mu <= 1

Then all the points of intersection of this line segment with edges of the polygon can be calculated.

Arrange these points of intersection by increasing values of mu along the line. These points form pairs of edges alternatively inside and outside the polygon.

The only remaining ambiguity is whether the first point P1 of the line segment (mu = 0) is within or outside the polygon, see here for details.

Calculating the area of a 3D polygon

Written by Paul Bourke
April 2000

Area of a triangular facet

This simply stems from the definition of the cross product.

A = || (P1 - P0) x (P2 - P0) || / 2

Area of a quad facet (assume planar)

This is somewhat more interesting, it is left as an exercise to the reader that the quad formed by connecting the 4 midpoints of the edges is a parallelogram and further that the area of the quad is half the area of this parallelogram. For more information see Pierre Varignon who is credited with discovering this around 1730.

A = || (P2 - P0) x (P3 - P1) || / 2

Area of an arbitrary planar polygon

This general case is somewhat more difficult to derive. One approach is Stokes theorem, another is to decompose the polygon into triangles or quads. In the following N is the normal to the plane on which the polygon lies.

Determining if a point lies on the interior of a polygon

Written by Paul Bourke
November 1987

Solution 1 (2D)

The following is a simple solution to the problem often encountered in computer graphics, determining whether or not a point (x,y) lies inside or outside a 2D polygonally bounded plane. This is necessary for example in applications such as polygon filling on raster devices, hatching in drafting software, and determining the intersection of multiple polygons.

Consider a polygon made up of N vertices (xi,yi) where i ranges from 0 to N-1. The last vertex (xN,yN) is assumed to be the same as the first vertex (x0,y0), that is, the polygon is closed. To determine the status of a point (xp,yp) consider a horizontal ray emanating from (xp,yp) and to the right. If the number of times this ray intersects the line segments making up the polygon is even then the point is outside the polygon. Whereas if the number of intersections is odd then the point (xp,yp) lies inside the polygon. The following shows the ray for some sample points and should make the technique clear.

Note: for the purposes of this discussion 0 will be considered even, the test for even or odd will be based on modulus 2, that is, if the number of intersections modulus 2 is 0 then the number is even, if it is 1 then it is odd.

The only trick is what happens in the special cases when an edge or vertex of the polygon lies on the ray from (xp,yp). The possible situations are illustrated below.

The thick lines above are not considered as valid intersections, the thin lines do count as intersections. Ignoring the case of an edge lying along the ray or an edge ending on the ray ensures that the endpoints are only counted once.

Note that this algorithm also works for polygons with holes as illustrated below

The following C function returns INSIDE or OUTSIDE indicating the status of a point P with respect to a polygon with N points.

#define MIN(x,y) (x < y ? x : y)
#define MAX(x,y) (x > y ? x : y)
#define INSIDE 0
#define OUTSIDE 1

typedef struct {
   double x,y;
} Point;

int InsidePolygon(Point *polygon,int N,Point p)
  int counter = 0;
  int i;
  double xinters;
  Point p1,p2;

  p1 = polygon[0];
  for (i=1;i<=N;i++) {
    p2 = polygon[i % N];
    if (p.y > MIN(p1.y,p2.y)) {
      if (p.y <= MAX(p1.y,p2.y)) {
        if (p.x <= MAX(p1.x,p2.x)) {
          if (p1.y != p2.y) {
            xinters = (p.y-p1.y)*(p2.x-p1.x)/(p2.y-p1.y)+p1.x;
            if (p1.x == p2.x || p.x <= xinters)
    p1 = p2;

  if (counter % 2 == 0)

The following code is by Randolph Franklin, it returns 1 for interior points and 0 for exterior points.

    int pnpoly(int npol, float *xp, float *yp, float x, float y)
      int i, j, c = 0;
      for (i = 0, j = npol-1; i < npol; j = i++) {
        if ((((yp[i] <= y) && (y < yp[j])) ||
             ((yp[j] <= y) && (y < yp[i]))) &&
            (x < (xp[j] - xp[i]) * (y - yp[i]) / (yp[j] - yp[i]) + xp[i]))
          c = !c;
      return c;

Contribution by Alexander Motrichuk: InsidePolygonWithBounds.cpp.
Quote: "For most of the algorithms above there is a pathological case if the point being queried lies exactly on a vertex. The easiest way to cope with this is to test that as a separate process and make your own decision as to whether you want to consider them inside or outside."

Contribution written in c# by Jerry Knauss: contains.c#.

Solution 2 (2D)

Another solution forwarded by Philippe Reverdy is to compute the sum of the angles made between the test point and each pair of points making up the polygon. If this sum is 2pi then the point is an interior point, if 0 then the point is an exterior point. This also works for polygons with holes given the polygon is defined with a path made up of coincident edges into and out of the hole as is common practice in many CAD packages.

The inside/outside test might then be defined in C as
typedef struct {
   int h,v;
} Point;

int InsidePolygon(Point *polygon,int n,Point p)
   int i;
   double angle=0;
   Point p1,p2;

   for (i=0;i<n;i++) {
      p1.h = polygon[i].h - p.h;
      p1.v = polygon[i].v - p.v;
      p2.h = polygon[(i+1)%n].h - p.h;
      p2.v = polygon[(i+1)%n].v - p.v;
      angle += Angle2D(p1.h,p1.v,p2.h,p2.v);

   if (ABS(angle) < PI)

   Return the angle between two vectors on a plane
   The angle is from vector 1 to vector 2, positive anticlockwise
   The result is between -pi -> pi
double Angle2D(double x1, double y1, double x2, double y2)
   double dtheta,theta1,theta2;

   theta1 = atan2(y1,x1);
   theta2 = atan2(y2,x2);
   dtheta = theta2 - theta1;
   while (dtheta > PI)
      dtheta -= TWOPI;
   while (dtheta < -PI)
      dtheta += TWOPI;


Solution 3 (2D)

There are other solutions to this problem for polygons with special attributes. If the polygon is convex then one can consider the polygon as a "path" from the first vertex. A point is on the interior of this polygons if it is always on the same side of all the line segments making up the path.

Given a line segment between P0 (x0,y0) and P1 (x1,y1), another point P (x,y) has the following relationship to the line segment.

(y - y0) (x1 - x0) - (x - x0) (y1 - y0)

if it is less than 0 then P is to the right of the line segment, if greater than 0 it is to the left, if equal to 0 then it lies on the line segment.

Solution 4 (3D)

This solution was motivated by solution 2 and correspondence with Reinier van Vliet and Remco Lam. To determine whether a point is on the interior of a convex polygon in 3D one might be tempted to first determine whether the point is on the plane, then determine it's interior status. Both of these can be accomplished at once by computing the sum of the angles between the test point (q below) and every pair of edge points p[i]->p[i+1]. This sum will only be 2pi if both the point is on the plane of the polygon AND on the interior. The angle sum will tend to 0 the further away from the polygon point q becomes.

The following code snippet returns the angle sum between the test point q and all the vertex pairs. Note that the angle sum is returned in radians.

typedef struct {
   double x,y,z;
} XYZ;
#define EPSILON  0.0000001
#define MODULUS(p) (sqrt(p.x*p.x + p.y*p.y + p.z*p.z))
#define TWOPI 6.283185307179586476925287
#define RTOD 57.2957795

double CalcAngleSum(XYZ q,XYZ *p,int n)
   int i;
   double m1,m2;
   double anglesum=0,costheta;
   XYZ p1,p2;

   for (i=0;i<n;i++) {

      p1.x = p[i].x - q.x;
      p1.y = p[i].y - q.y;
      p1.z = p[i].z - q.z;
      p2.x = p[(i+1)%n].x - q.x;
      p2.y = p[(i+1)%n].y - q.y;
      p2.z = p[(i+1)%n].z - q.z;

      m1 = MODULUS(p1);
      m2 = MODULUS(p2);
      if (m1*m2 <= EPSILON)
         return(TWOPI); /* We are on a node, consider this inside */
         costheta = (p1.x*p2.x + p1.y*p2.y + p1.z*p2.z) / (m1*m2);

      anglesum += acos(costheta);

For most of the algorithms above there is a pathological case if the point being queries lies exactly on a vertex. The easiest way to cope with this is to test that as a separate process and make your own decision as to whether you want to consider them inside or outside.

Calculating the area and centroid of a polygon

Written by Paul Bourke
July 1988

See also: centroid.pdf by Robert N├╝rnberg.


The problem of determining the area of a polygon seems at best messy but the final formula is particularly simple. The result and sample source code (C) will be presented here. Consider a polygon made up of line segments between N vertices (xi,yi), i=0 to N-1. The last vertex (xN,yN) is assumed to be the same as the first, ie: the polygon is closed.

The area is given by

Note for polygons with holes. The holes are usually defined by ordering the vertices of the enclosing polygon in the opposite direction to those of the holes. This algorithm still works except that the absolute value should be taken after adding the polygon area to the area of all the holes. That is, the holes areas will be of opposite sign to the bounding polygon area.

The sign of the area expression above (without the absolute value) can be used to determine the ordering of the vertices of the polygon. If the sign is positive then the polygon vertices are ordered counter clockwise about the normal, otherwise clockwise.

To derive this solution, project lines from each vertex to some horizontal line below the lowest part of the polygon. The enclosed region from each line segment is made up of a triangle and rectangle. Sum these areas together noting that the areas outside the polygon eventually cancel as the polygon loops around to the beginning.

The only restriction that will be placed on the polygon for this technique to work is that the polygon must not be self intersecting, for example the solution will fail in the following cases.


The centroid is also known as the "centre of gravity" or the "center of mass". The position of the centroid assuming the polygon to be made of a material of uniform density is given below. As in the calculation of the area above, xN is assumed to be x0, in other words the polygon is closed.

Centroid of a 3D shell described by 3 vertex facets

The centroid C of a 3D object made up of a collection of N triangular faces with vertices (ai,bi,ci) is given below. Ri is the average of the vertices of the i'th face and Ai is twice the area of the i'th face. Note the faces are assumed to be thin sheets of uniform mass, they need not be connected or form a solid object. This reduces to the equations above for a 2D 3 vertex polygon.

Second moment of a polygon

The following assume anticlockwise orientated polygon vertices, use the negative value for clockwise polygons.

Ix = [ yi2 + yi yi+1 + yi+12 ] [ xi yi+1 - xi+1 yi ]

Iy = [ xi2 + xi xi+1 + xi+12 ] [ xi yi+1 - xi+1 yi ]

2 Ixy = [ xi yi+1 + 2 xi yi + 2 xi+1 yi+1 + xi+1 yi ] [ xi yi+1 - xi+1 yi ]

Sample source code

This C function returns the area of a polygon.
VBA version conributed by Rui Vaz Rodrigues
JAVA code submitted by Ramón Talavera.
PolygonUtilities.java contributed by Christopher Fuhrman
Pascal/Dephi example by Rodrigo Alves Pons.
Basic version also by Rodrigo Alves Pons.
JavaScript by Raymond Hill.
Python and example by Jorg Rødscø.

Determining whether a line segment intersects a 3 vertex facet

Written by Paul Bourke
February 1997

The following will find the intersection point (if it exists) between a line segment and a planar 3 vertex facet. The mathematics and solution can also be used to find the intersection between a plane and line, a simpler problem. The intersection between more complex polygons can be found by first triangulating them into multiple 3 vertex facets.


Source code will be provided at the end, it illustrates the solution more than being written for efficiency. The labeling and naming conventions for the line segment and the facet are shown in the following diagram

The procedure will be implemented given the line segment defined by its two end points and the facet bounded by its three vertices.

The solution involves the following steps
  • Check the line and plane are not parallel
  • Find the intersection of the line, on which the given line segment lies, with the plane containing the facet
  • Check that the intersection point lies along the line segment
  • Check that the intersection point lies within the facet

The intersection point P is found by substituting the equation for the line P = P1 + mu (P2 - P1) into the equation for the plane Ax + By + Cz + D = 0.

Note that the values of A,B,C are the components of the normal to the plane which can be found by taking the cross product of any two normalised edge vectors, for example

(A,B,C) = (Pb - Pa) cross (Pc - Pa)

Then D is found by substituting one vertex into the equation for the plane for example

A Pax + B Pay + C Paz = -D

This gives an expression for mu from which the point of intersection P can be found using the equation of the line.

mu = ( D + A P1x + B P1y + C P1z ) / ( A (P1x - P2x) + B (P1y - P2y) + C (P1z - P2z) )

If the denominator above is 0 then the line is parallel to the plane and they don't intersect. For the intersection point to lie on the line segment, mu must be between 0 and 1.

Lastly, we need to check whether or not the intersection point lies within the planar facet bounded by Pa, Pb, Pc

The method used here relies on the fact that the sum of the internal angles of a point on the interior of a triangle is 2pi, points outside the triangular facet will have lower angle sums.

If we form the unit vectors Pa1, Pa2, Pa3 as follows (P is the point being tested to see if it is in the interior)

    Pa1 = (Pa - P) / |(Pa - P)|
    Pa2 = (Pb - P) / |(Pb - P)|
    Pa3 = (Pc - P) / |(Pc - P)|

the angles are

    a1 = acos(Pa1 dot Pa2)
    a2 = acos(Pa2 dot Pa3)
    a3 = acos(Pa3 dot Pa1)

Source code

   Determine whether or not the line segment p1,p2
   Intersects the 3 vertex facet bounded by pa,pb,pc
   Return true/false and the intersection point p

   The equation of the line is p = p1 + mu (p2 - p1)
   The equation of the plane is a x + b y + c z + d = 0
                                n.x x + n.y y + n.z z + d = 0
int LineFacet(p1,p2,pa,pb,pc,p)
XYZ p1,p2,pa,pb,pc,*p;
   double d;
   double a1,a2,a3;
   double total,denom,mu;
   XYZ n,pa1,pa2,pa3;

   /* Calculate the parameters for the plane */
   n.x = (pb.y - pa.y)*(pc.z - pa.z) - (pb.z - pa.z)*(pc.y - pa.y);
   n.y = (pb.z - pa.z)*(pc.x - pa.x) - (pb.x - pa.x)*(pc.z - pa.z);
   n.z = (pb.x - pa.x)*(pc.y - pa.y) - (pb.y - pa.y)*(pc.x - pa.x);
   d = - n.x * pa.x - n.y * pa.y - n.z * pa.z;

   /* Calculate the position on the line that intersects the plane */
   denom = n.x * (p2.x - p1.x) + n.y * (p2.y - p1.y) + n.z * (p2.z - p1.z);
   if (ABS(denom) < EPS)         /* Line and plane don't intersect */
   mu = - (d + n.x * p1.x + n.y * p1.y + n.z * p1.z) / denom;
   p->x = p1.x + mu * (p2.x - p1.x);
   p->y = p1.y + mu * (p2.y - p1.y);
   p->z = p1.z + mu * (p2.z - p1.z);
   if (mu < 0 || mu > 1)   /* Intersection not along line segment */

   /* Determine whether or not the intersection point is bounded by pa,pb,pc */
   pa1.x = pa.x - p->x;
   pa1.y = pa.y - p->y;
   pa1.z = pa.z - p->z;
   pa2.x = pb.x - p->x;
   pa2.y = pb.y - p->y;
   pa2.z = pb.z - p->z;
   pa3.x = pc.x - p->x;
   pa3.y = pc.y - p->y;
   pa3.z = pc.z - p->z;
   a1 = pa1.x*pa2.x + pa1.y*pa2.y + pa1.z*pa2.z;
   a2 = pa2.x*pa3.x + pa2.y*pa3.y + pa2.z*pa3.z;
   a3 = pa3.x*pa1.x + pa3.y*pa1.y + pa3.z*pa1.z;
   total = (acos(a1) + acos(a2) + acos(a3)) * RTOD;
   if (ABS(total - 360) > EPS)


Eulers number and closed surfaces

Written by Paul Bourke
January 1997

Eulers number for geometry made up of planar polygons which are defined in terms of edges and vertices is

V - E + F

Where V, E, and F are the number of vertices, edges, and faces respectively.

For a closed surface the Eulers number is always equal to 2.

This forms a simple test for the closure of computer generated facet geometry.