I've implemented line segment intersection formula from GeeksForGeeks.It works great, but I also need to know the point of intersection of these 2 line segments. How can I modify the code to do that? (C++)
First comes the Point class (just a 2d Vector really)
#include <iostream>
using namespace std;
struct Point
{
int x;
int y;
};
This function checks if 2 points are on the same segment.
// Given three colinear points p, q, r, the function checks if
// point q lies on line segment 'pr'
bool onSegment(Point p, Point q, Point r)
{
if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) &&
q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y))
return true;
return false;
}
This one find the rotation of ordered triplet.
// To find orientation of ordered triplet (p, q, r).
// The function returns following values
// 0 --> p, q and r are colinear
// 1 --> Clockwise
// 2 --> Counterclockwise
int orientation(Point p, Point q, Point r)
{
// See https://www.geeksforgeeks.org/orientation-3-ordered-points/
// for details of below formula.
int val = (q.y - p.y) * (r.x - q.x) -
(q.x - p.x) * (r.y - q.y);
if (val == 0) return 0; // colinear
return (val > 0)? 1: 2; // clock or counterclock wise
}
// The main function that returns true if line segment 'p1q1'
// and 'p2q2' intersect.
And finally functions that checks for intersection. But where did it occur?
bool doIntersect(Point p1, Point q1, Point p2, Point q2)
{
// Find the four orientations needed for general and
// special cases
int o1 = orientation(p1, q1, p2);
int o2 = orientation(p1, q1, q2);
int o3 = orientation(p2, q2, p1);
int o4 = orientation(p2, q2, q1);
// General case
if (o1 != o2 && o3 != o4)
return true;
// Special Cases
// p1, q1 and p2 are colinear and p2 lies on segment p1q1
if (o1 == 0 && onSegment(p1, p2, q1)) return true;
// p1, q1 and q2 are colinear and q2 lies on segment p1q1
if (o2 == 0 && onSegment(p1, q2, q1)) return true;
// p2, q2 and p1 are colinear and p1 lies on segment p2q2
if (o3 == 0 && onSegment(p2, p1, q2)) return true;
// p2, q2 and q1 are colinear and q1 lies on segment p2q2
if (o4 == 0 && onSegment(p2, q1, q2)) return true;
return false; // Doesn't fall in any of the above cases
}
You just need to implement one of the equations found in this wiki.
In the following code i chose to implement the first equation
// your previous code
Point CalcIntersection(Point p1, Point q1, Point p2, Point q2){
struct Point pInt;
if(doIntersect(p1, q1, p2, q2)){
pInt.x= ((p1.x * q1.y - p1.y * q1.x)*(p2.x-q2.x) - (p2.x * q2.y - p2.y * q2.x)*(p1.x-q1.x)) / ((p1.x-q1.x) * (p2.y-q2.y) -(p1.y-q1.y) * (p2.x-q2.x));
pInt.y = ((p1.x * q1.y - p1.y * q1.x)*(p2.y-q2.y) - (p2.x * q2.y - p2.y * q2.x)*(p1.y-q1.y)) / ((p1.x-q1.x) * (p2.y-q2.y) -(p1.y-q1.y) * (p2.x-q2.x));
}
return pInt;
}
Rq: i think your x and y should be a float not an int
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
How do I calculate the intersection between a ray and a plane?
Code
This produces the wrong results.
float denom = normal.dot(ray.direction);
if (denom > 0)
{
float t = -((center - ray.origin).dot(normal)) / denom;
if (t >= 0)
{
rec.tHit = t;
rec.anyHit = true;
computeSurfaceHitFields(ray, rec);
return true;
}
}
Parameters
ray represents the ray object.
ray.direction is the direction vector.
ray.origin is the origin vector.
rec represents the result object.
rec.tHit is the value of the hit.
rec.anyHit is a boolean.
My function has access to the plane:
center and normal defines the plane
As wonce commented, you want to also allow the denominator to be negative, otherwise you will miss intersections with the front face of your plane. However, you still want a test to avoid a division by zero, which would indicate the ray being parallel to the plane. You also have a superfluous negation in your computation of t. Overall, it should look like this:
float denom = normal.dot(ray.direction);
if (abs(denom) > 0.0001f) // your favorite epsilon
{
float t = (center - ray.origin).dot(normal) / denom;
if (t >= 0) return true; // you might want to allow an epsilon here too
}
return false;
First consider the math of the ray-plane intersection:
In general one intersects the parametric form of the ray, with the implicit form of the geometry.
So given a ray of the form x = a * t + a0, y = b * t + b0, z = c * t + c0;
and a plane of the form: A x * B y * C z + D = 0;
now substitute the x, y and z ray equations into the plane equation and you will get a polynomial in t. you then solve that polynomial for the real values of t. With those values of t you can back substitute into the ray equation to get the real values of x, y and z.
Here it is in Maxima:
Note that the answer looks like the quotient of two dot products!
The normal to a plane is the first three coefficients of the plane equation A, B, and C.
You still need D to uniquely determine the plane.
Then you code that up in the language of your choice like so:
Point3D intersectRayPlane(Ray ray, Plane plane)
{
Point3D point3D;
// Do the dot products and find t > epsilon that provides intersection.
return (point3D);
}
Math
Define:
Let the ray be given parametrically by q = p + t*v for initial point p and direction vector v for t >= 0.
Let the plane be the set of points r satisfying the equation dot(n, r) + d = 0 for normal vector n = (a, b, c) and constant d. Fully expanded, the plane equation may also be written in the familiar form ax + by + cz + d = 0.
The ray-plane intersection occurs when q satisfies the plane equation. Substituting, we have:
d = -dot(n, q)
= -dot(n, p + t * v)
= -dot(n, p) + t * dot(n, v)
Rearranging:
t = -(dot(n, p) + d) / dot(n, v)
This value of t can be used to determine the intersection by plugging it back into p + t*v.
Example implementation
std::optional<vec3> intersectRayWithPlane(
vec3 p, vec3 v, // ray
vec3 n, float d // plane
) {
float denom = dot(n, v);
// Prevent divide by zero:
if (abs(denom) <= 1e-4f)
return std::nullopt;
// If you want to ensure the ray reflects off only
// the "top" half of the plane, use this instead:
//
// if (-denom <= 1e-4f)
// return std::nullopt;
float t = -(dot(n, p) + d) / dot(n, v);
// Use pointy end of the ray.
// It is technically correct to compare t < 0,
// but that may be undesirable in a raytracer.
if (t <= 1e-4)
return std::nullopt;
return p + t * v;
}
implementation of vwvan's answer
Vector3 Intersect(Vector3 planeP, Vector3 planeN, Vector3 rayP, Vector3 rayD)
{
var d = Vector3.Dot(planeP, -planeN);
var t = -(d + Vector3.Dot(rayP, planeN)) / Vector3.Dot(rayD, planeN);
return rayP + t * rayD;
}