Question

这样做的最容易的方法是什么? 我在数学上失败,我发现互联网上的公式十分复杂。如果有某种更简单的东西,则不希望?

我只需要知道,某一领域是否重叠,即对哪一点进行谨慎,等等。

I m also hoping it would take advantage of the fact that both shapes are symmetric.

Edit: c在X,y,z axises上调。

Answer 1

审视半空间是不够的,你还必须考虑最接近的方法:

~~浏览 Adam s notation:~~

假定轴向对地的异构体,使C1和C2成为对立角,成为该领域的中心,成为该领域的rad,而且这两个物体都是固体:

inline float squared(float v) { return v * v; }
bool doesCubeIntersectSphere(vec3 C1, vec3 C2, vec3 S, float R)
{
    float dist_squared = R * R;
    /* assume C1 and C2 are element-wise sorted, if not, do that now */
    if (S.X < C1.X) dist_squared -= squared(S.X - C1.X);
    else if (S.X > C2.X) dist_squared -= squared(S.X - C2.X);
    if (S.Y < C1.Y) dist_squared -= squared(S.Y - C1.Y);
    else if (S.Y > C2.Y) dist_squared -= squared(S.Y - C2.Y);
    if (S.Z < C1.Z) dist_squared -= squared(S.Z - C1.Z);
    else if (S.Z > C2.Z) dist_squared -= squared(S.Z - C2.Z);
    return dist_squared > 0;
}

Answer 2

Jim Arvo has an algorithm for this in Graphics Gems 2 which works in N-Dimensions. I believe you want "case 3" at the bottom of this page: http://www.ics.uci.edu/~arvo/code/BoxSphereIntersect.c which cleaned up for your case is:

bool BoxIntersectsSphere(Vec3 Bmin, Vec3 Bmax, Vec3 C, float r) {
  float r2 = r * r;
  dmin = 0;
  for( i = 0; i < 3; i++ ) {
    if( C[i] < Bmin[i] ) dmin += SQR( C[i] - Bmin[i] );
    else if( C[i] > Bmax[i] ) dmin += SQR( C[i] - Bmax[i] );     
  }
  return dmin <= r2;
}

Answer 3

// Assume clampTo is a new value. Obviously, don t move the sphere
closestPointBox = sphere.center.clampTo(box)

isIntersecting = sphere.center.distanceTo(closestPointBox) < sphere.radius

其它一切只是优化。

Wow, -2. Tough 人群。 Ok, here s the three.js implementation that basic notes the same letter for term.

intersectsSphere: ( function () {

    var closestPoint;

    return function intersectsSphere( sphere ) {

        if ( closestPoint === undefined ) closestPoint = new Vector3();

        // Find the point on the AABB closest to the sphere center.
        this.clampPoint( sphere.center, closestPoint );

        // If that point is inside the sphere, the AABB and sphere intersect.
        return closestPoint.distanceToSquared( sphere.center ) <= ( sphere.radius * sphere.radius );

    };

} )(),

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