I have a line and a triangle somewhere in 3D space. In other words, I have 3 points ([x,y,z] each) for the triangle, and two points (also [x,y,z]) for the line.
I need to figure out a way, hopefully using C++, to figure out if the line ever crosses the triangle. A line parallel to the triangle, and with more than one point in common, should be counted as "does not intersect".
I already made some code, but it doesn't work, and I always get false even when a visual representation clearly shows an intersection.
ofVec3f P1, P2;
P1 = ray.s;
P2 = ray.s + ray.t;
ofVec3f p1, p2, p3;
p1 = face.getVertex(0);
p2 = face.getVertex(1);
p3 = face.getVertex(2);
ofVec3f v1 = p1 - p2;
ofVec3f v2 = p3 - p2;
float a, b, c, d;
a = v1.y * v2.z - v1.z * v2.y;
b = -(v1.x * v2.z - v1.z * v2.x);
c = v1.x * v2.y - v1.y * v2.x;
d = -(a * p1.x + b * p1.y + c * p1.z);
ofVec3f O = P1;
ofVec3f V = P2 - P1;
float t;
t = -(a * O.x + b * O.y + c * O.z + d) / (a * V.x + b * V.y + c * V.z);
ofVec3f p = O + V * t;
float xmin = std::min(P1.x, P2.x);
float ymin = std::min(P1.y, P2.y);
float zmin = std::min(P1.z, P2.z);
float xmax = std::max(P1.x, P2.x);
float ymax = std::max(P1.y, P2.y);
float zmax = std::max(P1.z, P2.z);
if (inside(p, xmin, xmax, ymin, ymax, zmin, zmax)) {
*result = p.length();
return true;
}
return false;
And here is the definition of inside()
bool primitive3d::inside(ofVec3f p, float xmin, float xmax, float ymin, float ymax, float zmin, float zmax) const {
if (p.x >= xmin && p.x <= xmax && p.y >= ymin && p.y <= ymax && p.z >= zmin && p.z <= zmax)
return true;
return false;
}
1) If you just want to know whether the line intersects the triangle (without needing the actual intersection point):
Let p1,p2,p3 denote your triangle
Pick two points q1,q2 on the line very far away in both directions.
Let SignedVolume(a,b,c,d) denote the signed volume of the tetrahedron a,b,c,d.
If SignedVolume(q1,p1,p2,p3) and SignedVolume(q2,p1,p2,p3) have different signs AND
SignedVolume(q1,q2,p1,p2), SignedVolume(q1,q2,p2,p3) and SignedVolume(q1,q2,p3,p1) have the same sign, then there is an intersection.
SignedVolume(a,b,c,d) = (1.0/6.0)*dot(cross(b-a,c-a),d-a)
2) Now if you want the intersection, when the test in 1) passes
write the equation of the line in parametric form: p(t) = q1 + t*(q2-q1)
Write the equation of the plane: dot(p-p1,N) = 0 where
N = cross(p2-p1, p3-p1)
Inject p(t) into the equation of the plane: dot(q1 + t*(q2-q1) - p1, N) = 0
Expand: dot(q1-p1,N) + t dot(q2-q1,N) = 0
Deduce t = -dot(q1-p1,N)/dot(q2-q1,N)
The intersection point is q1 + t*(q2-q1)
3) A more efficient algorithm
We now study the algorithm in:
Möller and Trumbore, "Fast, Minimum Storage Ray-Triangle Intersection", Journal of Graphics Tools, vol. 2, 1997, p. 21–28
(see also: https://en.wikipedia.org/wiki/M%C3%B6ller%E2%80%93Trumbore_intersection_algorithm)
The algorithm is in the end simpler (less instructions than what we did in 1) and 2)), but sightly more complicated to understand. Let us derive it step by step.
Notation:
O = origin of the ray,
D = direction vector of the ray,
A,B,C = vertices of the triangle
An arbitrary point P on the ray can be written as P = O + tD
An arbitrary point P on the triangle can be written as P = A + uE1 + vE2 where E1 = B-A and E2 = C-A, u>=0, v>=0 and (u+v)<=1
Writing both expressions of P gives:
O + tD = A + uE1 + vE2
or:
uE1 + vE2 -tD = O-A
in matrix form:
[u]
[E1|E2|-D] [v] = O-A
[t]
(where [E1|E2|-D] is the 3x3 matrix with E1,E2,-D as its columns)
Using Cramer's formula for the solution of:
[a11 a12 a13][x1] [b1]
[a12 a22 a23][x2] = [b2]
[a31 a32 a33][x3] [b3]
gives:
|b1 a12 a13| |a11 a12 a13|
x1 = |b2 a22 a23| / |a21 a22 a23|
|b3 a32 a33| |a31 a32 a33|
|a11 b1 a13| |a11 a12 a13|
x2 = |a21 b2 a23| / |a21 a22 a23|
|a31 b3 a33| |a31 a32 a33|
|a11 a12 b1| |a11 a12 a13|
x3 = |a21 a22 b2| / |a21 a22 a23|
|a31 a32 b3| |a31 a32 a33|
Now we get:
u = (O-A,E2,-D) / (E1,E2,-D)
v = (E1,O-A,-D) / (E1,E2,-D)
t = (E1,E2,O-A) / (E1,E2,-D)
where (A,B,C) denotes the determinant of the 3x3 matrix with A,B,C as its column vectors.
Now we use the following identities:
(A,B,C) = dot(A,cross(B,C)) (develop the determinant w.r.t. first column)
(B,A,C) = -(A,B,C) (swapping two vectors changes the sign)
(B,C,A) = (A,B,C) (circular permutation does not change the sign)
Now we get:
u = -(E2,O-A,D) / (D,E1,E2)
v = (E1,O-A,D) / (D,E1,E2)
t = -(O-A,E1,E2) / (D,E1,E2)
Using:
N=cross(E1,E2);
AO = O-A;
DAO = cross(D,AO)
We obtain finally the following code (here in GLSL, easy to translate to other languages):
bool intersect_triangle(
in Ray R, in vec3 A, in vec3 B, in vec3 C, out float t,
out float u, out float v, out vec3 N
) {
vec3 E1 = B-A;
vec3 E2 = C-A;
N = cross(E1,E2);
float det = -dot(R.Dir, N);
float invdet = 1.0/det;
vec3 AO = R.Origin - A;
vec3 DAO = cross(AO, R.Dir);
u = dot(E2,DAO) * invdet;
v = -dot(E1,DAO) * invdet;
t = dot(AO,N) * invdet;
return (det >= 1e-6 && t >= 0.0 && u >= 0.0 && v >= 0.0 && (u+v) <= 1.0);
}
When the function returns true, the intersection point is given by R.Origin + t * R.Dir. The barycentric coordinates of the intersection in the triangle are u, v, 1-u-v (useful for Gouraud shading or texture mapping). The nice thing is that you get them for free !
Note that the code is branchless.
It is used by some of my shaders on ShaderToy
https://www.shadertoy.com/view/tl3XRN
https://www.shadertoy.com/view/3ltSzM
#BrunoLevi: your algorithm does not seem to work, see the following python implementation:
def intersect_line_triangle(q1,q2,p1,p2,p3):
def signed_tetra_volume(a,b,c,d):
return np.sign(np.dot(np.cross(b-a,c-a),d-a)/6.0)
s1 = signed_tetra_volume(q1,p1,p2,p3)
s2 = signed_tetra_volume(q2,p1,p2,p3)
if s1 != s2:
s3 = signed_tetra_volume(q1,q2,p1,p2)
s4 = signed_tetra_volume(q1,q2,p2,p3)
s5 = signed_tetra_volume(q1,q2,p3,p1)
if s3 == s4 and s4 == s5:
n = np.cross(p2-p1,p3-p1)
t = -np.dot(q1,n-p1) / np.dot(q1,q2-q1)
return q1 + t * (q2-q1)
return None
My test code is:
q0 = np.array([0.0,0.0,1.0])
q1 = np.array([0.0,0.0,-1.0])
p0 = np.array([-1.0,-1.0,0.0])
p1 = np.array([1.0,-1.0,0.0])
p2 = np.array([0.0,1.0,0.0])
print(intersect_line_triangle(q0,q1,p0,p1,p2))
gives:
[ 0. 0. -3.]
instead of the expected
[ 0. 0. 0.]
looking at the line
t = np.dot(q1,n-p1) / np.dot(q1,q2-q1)
Subtracting p1 from the normal doesn't make sense to me, you want to project from q1 onto the plane of the triangle, so you need to project along the normal, with a distance that is proportional to the ratio of the distance from q1 to the plane and q1-q2 along the normal, right?
The following code fixes this:
n = np.cross(p2-p1,p3-p1)
t = np.dot(p1-q1,n) / np.dot(q2-q1,n)
return q1 + t * (q2-q1)
To find the intersection between a line and a triangle in 3D, follow this approach:
Compute the plane supporting the triangle,
Intersect the line with the plane supporting the triangle:
If there is no intersection, then there is no intersection with the triangle.
If there is an intersection, verify that the intersection point indeed lies in the triangle:
Each edge of the triangle together with the normal of the plane supporting the triangle determines a half-space bounding the inside of the triangle (the corresponding bounding plane can be derived from the normal and the edge vertices),
Verify that the intersection point lies on the inside of all the edge half-spaces.
Here is some sample code with detailed computations that should work:
// Compute the plane supporting the triangle (p1, p2, p3)
// normal: n
// offset: d
//
// A point P lies on the supporting plane iff n.dot(P) + d = 0
//
ofVec3f v21 = p2 - p1;
ofVec3f v31 = p3 - p1;
ofVec3f n = v21.getCrossed(v31);
float d = -n.dot(p1);
// A point P belongs to the line from P1 to P2 iff
// P = P1 + t * (P2 - P1)
//
// Find the intersection point P(t) between the line and
// the plane supporting the triangle:
// n.dot(P) + d = 0
// = n.dot(P1 + t (P2 - P1)) + d
// = n.dot(P1) + t n.dot(P2 - P1) + d
//
// t = -(n.dot(P1) + d) / n.dot(P2 - P1)
//
ofVec3f P21 = P2 - P1;
float nDotP21 = n.dot(P21);
// Ignore line parallel to (or lying in) the plane
if (fabs(nDotP21) < Epsilon)
return false;
float t = -(n.dot(P1) + d) / nDotP21;
ofVec3f P = P1 + t * P21;
// Plane bounding the inside half-space of edge (p1, p2):
// normal: n21 = n x (p2 - p1)
// offset: d21 = -n21.dot(p1)
//
// A point P is in the inside half-space iff n21.dot(P) + d21 > 0
//
// Edge (p1, p2)
ofVec3f n21 = n.cross(v21);
float d21 = -n21.dot(p1);
if (n21.dot(P) + d21 <= 0)
return false;
// Edge (p2, p3)
ofVec3f v32 = p3 - p2;
ofVec3f n32 = n.cross(v32);
float d32 = -n32.dot(p2);
if (n32.dot(P) + d32 <= 0)
return false;
// Edge (p3, p1)
ofVec3f n13 = n.cross(-v31);
float d13 = -n13.dot(p3);
if (n13.dot(P) + d13 <= 0)
return false;
return true;
Some comments on the code posted with the question:
Predefined operations of ofVec3f (.dot() and .cross() for geometric products, etc...) should be preferred when available (more readable, avoids implementation mistakes, etc...),
The code initially follows the approach above but then only checks that the intersection point is in the 3D axis-aligned bounding box of the line segment [P1, P2]. This combined with possible other errorscould explain why the results are incorrect.
One can verify that the intersection point is in the 3D axis-aligned bounding box of the (whole) triangle. While this is not enough to guarantee intersection, it can however be used to cull points clearly not intersecting and avoid further complex computations.
I have a different way to do it which I found in my renderer to be far faster than the first way given by BrunoLevy. (I haven't implemented the second way)
Points A, B, C are vertexes of the triangle
O is the origin of the ray
D is the direction of the ray (doesn't need to be normalised, just closer to the origin than the triangle)
Check if the direction (D+O) is inside the tetrahedron A, B, C, O
bool SameSide(vec3 A, vec3 B, vec3 C, vec3 D, vec3 p)
{
vec3 normal = cross(B - A, C - A);
float dotD = dot(normal, D - A);
float dotP = dot(normal, p - A);
return signbit(dotD) == signbit(dotP);
}
bool LineIntersectTri(vec3 A, vec3 B, vec3 C, vec3 O, vec3 D)
{
return SameSide(A, B, C, O, O+D) &&
SameSide(B, C, O, A, O+D) &&
SameSide(C, O, A, B, O+D) &&
SameSide(O, A, B, C, O+D);
}
If D varies, and everything else stays the same (for example in a raycasting renderer) then normal and dotP don't need to be recalculated; This is why I found it so much faster
The code came from this answer https://stackoverflow.com/a/25180294/18244401
I'm referencing a mathematics paper, but the terminology is strange, and I'm unsure of how to code the following:
Return if the orthogonal projection of Point P exists on S(P2, P3).
I found std::inner_product but not sure if thats the correct method to use.
The concept is that you project P onto S and then check whether that projection P' is between P2 and P3.
To make it a little easier you say that P2 is the support-vector of S and P3-P2 is the direction-vector. You then project P-P2 onto the normalized P3-P2 (you compute the scalar-product between them) which gives you the distance D of P' to P2.
Now in your case you only want to know if P' is between P2 and P3. That is true if D is between 0 and 1.
You want the orthogonal projection of P (on the line given by P2 and P3) to be inside the segment [P2,P3]. Mathematically, it writes simply (I'm noting vect(A, B) the vector AB, because I do not know how to use the arrow notation):
0 <= vect(P2, P) . vect (P2, P3) <= vect(P2, P3) . vect(P2, P3)
You can indeed use std::inner_product but if your points are something as simple as:
struct Point {
double x;
double y;
};
You could just use
double operator - (const Point& a, const Point& b) {
return a.x - b.x + a.y - b.y;
}
double operator * (const Point& a, const Point& b) {
return a.x * b.x + a.y * b.y;
}
And the mathematical formula just gives:
bool is_proj_inside(const Point& P, const Point& P2, const Point& P3) {
double p_proj = (P - P2) * (P3 - P2);
double p3_proj = (P3 - P2) * (P3 - P2);
return (p_proj >= 0) && (p_proj <= p3_proj);
}
Yes, you may use inner_product (dot product) to take the result in very simple way.
Make vectors
V2 = P - P2
V3 = P - P3
V = P3 - P2
Find signs of dot products
D2 = Dot(V2,V) and D3 = Dot(V3,V)
Projection of Point P lies at S(P2, P3), if
D2 >=0 and
D3 <=0
Note - there is no need in normalizations, square roots etc. Just some subtractions, multiplications and additions.
(Explanation - angles P-P2-P3 and P-P3-P2 should be acute or right)
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;
}
This function checks if a point is within a polygon. How do I deal with a polygon coordinate that is negative? For example,
float x[3] = { 0.16, 1.2, -10 };
float y[3] = { 1.8, 10, -5.5 };
I tried checking a valid point within the polygon and it returns 0.
There are pretty good implementations from the iSurfer
The two methods used in most cases (and the two I know of) are crossing number and winding number. Both of them are not affected by the signs of the polygon/point coordinates. So it must be a bug in your code.
For completeness I'm placing the code for a crossing number test which seems to be what you're trying to do in your code
// a Point is defined by its coordinates {int x, y;}
// isLeft(): tests if a point is Left|On|Right of an infinite line.
// Input: three points P0, P1, and P2
// Return: >0 for P2 left of the line through P0 and P1
// =0 for P2 on the line
// <0 for P2 right of the line
// See: Algorithm 1 "Area of Triangles and Polygons"
inline int isLeft( Point P0, Point P1, Point P2 )
{
return ( (P1.x - P0.x) * (P2.y - P0.y) - (P2.x - P0.x) * (P1.y - P0.y) );
}
//===================================================================
// cn_PnPoly(): crossing number test for a point in a polygon
// Input: P = a point,
// V[] = vertex points of a polygon V[n+1] with V[n]=V[0]
// Return: 0 = outside, 1 = inside
// This code is patterned after [Franklin, 2000]
int cn_PnPoly( Point P, Point* V, int n )
{
int cn = 0; // the crossing number counter
// loop through all edges of the polygon
for (int i=0; i<n; i++) { // edge from V[i] to V[i+1]
if (((V[i].y <= P.y) && (V[i+1].y > P.y)) // an upward crossing
|| ((V[i].y > P.y) && (V[i+1].y <= P.y))) { // a downward crossing
// compute the actual edge-ray intersect x-coordinate
float vt = (float)(P.y - V[i].y) / (V[i+1].y - V[i].y);
if (P.x < V[i].x + vt * (V[i+1].x - V[i].x)) // P.x < intersect
++cn; // a valid crossing of y=P.y right of P.x
}
}
return (cn&1); // 0 if even (out), and 1 if odd (in)
}
//===================================================================
A special case that can arise with the crossing number number test, is when the ray overlaps an edge of the polygon. In that case it becomes somewhat fuzzy how to count intersections. That's why it's not the actual number of intersections we count, but the number we crossed over semiplanes defined by the ray.
The winding number test is more robust to this respect
Note that if your polygon N and N+1 have the same y value then this test can fail computing vt. For example: determining if a point is inside an axis aligned square.
You need to deal with lines parallel on y
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Closed 10 years ago.
I am attempting to sort three points for a triangle in order to add it successfully into a physics engine. All of the triangles whose position is smaller than 0.0 on the x axis show up correctly. However, after that point the first element (v[0]) is always set to {0.0, 0.0} and the rest seem okay. Here is my main function that does the sorting and adding.
void Core::addPoly(float x1, float y1,
float x2, float y2,
float x3, float y3) {
std::vector<PVector> v(3);
v.push_back(PVector(x1, y1));
v.push_back(PVector(x2, y2));
v.push_back(PVector(x3, y3));
PVector center((x1+x2+x3)/3, (y1+y2+y3)/3);
std::sort(v.begin(), v.end(), [center](PVector b, PVector a) {
if (a.x >= 0 && b.x < 0)
return true;
if (a.x == 0 && b.x == 0)
return a.y > b.y;
// compute the cross product of vectors (center -> a) x (center -> b)
float det = (a.x-center.x) * (b.y-center.y) - (b.x - center.x) * (a.y - center.y);
if (det < 0)
return true;
if (det > 0)
return false;
// points a and b are on the same line from the center
// check which point is closer to the center
float d1 = (a.x-center.x) * (a.x-center.x) + (a.y-center.y) * (a.y-center.y);
float d2 = (b.x-center.x) * (b.x-center.x) + (b.y-center.y) * (b.y-center.y);
return d1 > d2;
});
emap->polys.push_back(Polygon(v[0], v[1], v[2]));
}
I am using the sorting function provided here. It originally had the first element of all triangles pointed to center (however I don't believe this is correct behaviour)- I switched a and b in the lambda declaration and now it only shows up after 0.0 on the x axis.
If we have 3 triangles passed in (Imagine | to be the 0.0 and ___ to be 0.0 on a plane)
^ |
^ | ^
______|_____________
The 0'th vertice of the third triangle will actually make it like this:
^ |
^ | ___/|
______|_/______________
(This is supposed to be a triangle)
However, 0.0 was never passed to addPoly.
This line:
std::vector<PVector> v(3);
Causes your vector to get initialized with three <0.0,0.0> values. Changing it to
std::vector<PVector> v;
should fix your issue.
The code seems to work like this:
struct PVector
{
float x;
float y;
PVector(float x, float y) : x(x),y(y){}
PVector() : x(0.0f), y(0.0f) {}
};
static void addPoly(float x1, float y1,
float x2, float y2,
float x3, float y3) {
std::vector<PVector> v;
v.push_back(PVector(x1, y1));
v.push_back(PVector(x2, y2));
v.push_back(PVector(x3, y3));
PVector center((x1+x2+x3)/3, (y1+y2+y3)/3);
std::sort(v.begin(), v.end(), [center](PVector b, PVector a) -> bool {
if (a.x >= 0 && b.x < 0)
return true;
if (a.x == 0 && b.x == 0)
return a.y > b.y;
// compute the cross product of vectors (center -> a) x (center -> b)
float det = (a.x-center.x) * (b.y-center.y) - (b.x - center.x) * (a.y - center.y);
if (det < 0)
return true;
if (det > 0)
return false;
// points a and b are on the same line from the center
// check which point is closer to the center
float d1 = (a.x-center.x) * (a.x-center.x) + (a.y-center.y) * (a.y-center.y);
float d2 = (b.x-center.x) * (b.x-center.x) + (b.y-center.y) * (b.y-center.y);
return d1 > d2;
});
//emap->polys.push_back(Polygon(v[0], v[1], v[2]));
}