如何在3D中检测两张脸的交集

Question

lets say

struct myFace
{
    3DPoint p0;
    3DPoint p1;
    3DPoint p2;
    3DPoint p3;
    3DPoint pNormal;

};

face1 and face2 is faces which type of myFace.

double ac = face1.pNormal * face2.pNormal;

if (!(ac<1.00000001 && ac>0.99999999) && !(ac>-1.00000001 && ac<-0.99999999))

then faces are not parallel.

but how can detect if they do intersect or do not ?

Answer 1

Oops ignore my comment: thought of another way to do this.

---

• Step 1:

For faces F1 and F2 , take F2 's points as two triangles , eg (p0, p1, p2) and (p1, p2, p3) respectively. Then take F1 's edges, ie (p0, p1) , (p1, p2) , (p2, p3) , and (p3, p0) , then intersect each of them with both of the triangles.

I found some code to do this: (adapted from http://geomalgorithms.com/a06-_intersect-2.html )

#define SMALL_NUM   0.00000001

/* 
   returns: 0 if no intersection 
            1 if parallel but disjoint
            2 if coplanar
*/
int intersect3D_RayTriangle(Vector P0, Vector P1, Vector V0, Vector V1, Vector V2)
{
    Vector    u, v, n;              // triangle vectors
    Vector    dir, w0, w;           // ray vectors
    float     r, a, b;              // params to calc ray-plane intersect

    // get triangle edge vectors and plane normal
    u = V1 - V0;
    v = V2 - V0;
    n = cross(u, v);

    dir = P1 - P0;             // ray direction vector
    w0 = P0 - V0;
    a = -dot(n, w0);
    b = dot(n, dir);
    if (fabs(b) < SMALL_NUM)   // ray is parallel to triangle plane
        return (fabs(a) < SMALL_NUM ? 2 : 0);

    // get intersect point of ray with triangle plane
    r = a / b;
    if (r < 0.0 || r > 1.0)
        return 0;                   // => no intersect
    Vector I = R.P0 + r * dir;      // intersect point of ray and plane

    // is I inside T?
    float uu, uv, vv, wu, wv, D;
    uu = dot(u, u);
    uv = dot(u, v);
    vv = dot(v, v);
    w = I - V0;
    wu = dot(w, u);
    wv = dot(w, v);
    D = uv * uv - uu * vv;

    // get and test parametric coords
    float s, t;
    s = (uv * wv - vv * wu) / D;
    if (s < 0.0 || s > 1.0)         // I is outside T
        return 0;
    t = (uv * wu - uu * wv) / D;
    if (t < 0.0 || (s + t) > 1.0)  // I is outside T
        return 0;

    return 1;                       // I is in T
}

P0 and P1 form one of the edges from F1 , while V0 , V1 and V2 form one the F2 's triangles.

• If one of the checks (there should be 8) returns 1, then they definitely intersect (return true immediately ).
• If all of them return 0, then they don't intersect.
• If one of the checks returns 2 (the first check will probably do that) then we need a different method. Stop these checks and immediately move onto step 2.

---

• Step 2:

This part is for if the polygons are coplanar (ie parallel and in the same plane). This time, take all of F1 's edges and all of F2 's edges; for each of F1 's edges, check if it intersects with any of F2 's edges, and if one pair intersects then return true immediately .

To do such an edge intersection: (adapted from https://gist.github.com/hanigamal/6556506 )

A0 , A1 form the edge from F1 , and B0 , B1 from F2 .

int intersection(Vector A0, Vector A1, Vector B0, Vector B1)
{
   Vector dA = A1 - A0;
   Vector dB = B1 - B0;
   Vector dC = B0 - A0;

   double s = dot(cross(dC, dB), cross(dA, dB)) / norm2(cross(dA, dB));
    return (s >= 0.0 && s <= 1.0);
}