立方体球面相交测试？ [英] Cube sphere intersection test?

问题描述

这是最简单的方法是什么？我在数学上失败了，我发现在互联网上的复杂的公式...我希望如果这些更简单一些？

What's the easiest way of doing this? I fail at math, and i found pretty complicate formulaes over the internet... im hoping if theres some simpler one?

我只需要知道一个球是否重叠一个立方体，我不关心它的哪个点等等。

I just need to know if a sphere is overlapping a cube, i dont care about which point it does that etc.

我也希望它会利用这两个形状是对称的事实。

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

编辑：立方体在x，y，z轴上直线对齐

the cube is aligned straight in the x,y,z axises

推荐答案

看半空间是不够的，你还必须考虑最接近的方法：

Looking at half-spaces is not enough, you have to consider also the point of closest approach:

借用亚当的符号：

假设轴对齐的立方体，让C1和C2是对角，S是球体的中心，R是球体的半径，并且两个对象都是固定的：

Assuming an axis-aligned cube and letting C1 and C2 be opposing corners, S the center of the sphere, and R the radius of the sphere, and that both objects are solid:

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;
}

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