碰撞检测之OBB-OBB检测

来源：互联网发布：农村淘宝五个一是什么编辑：程序博客网时间：2024/05/17 23:19

2D情况

首先回顾一下SAP

两个凸包多边形，当且仅当存在一条线，这两个多边形在这条线上的投影不相交，则这两个多边形也不相交.

这条线称为Separating Axis.垂直Separating Axis存在一条Separating Line将两个多边形分开。

这里还有一个要确定的，就是如果两个矩形之间存在Separating Line，则一定存在一条和两个矩形中的一条边平行。每个矩形对边平行，则我们只需要检查四个方向是否存在Separating Line，如下图

找Separating Line 就是找Separating Axis, Separating Axis也只有四个方向，所以将矩形在四个轴上投影就可以了，如下图

下面来看具体的计算

定义下面几个变量

PA = coordinate position of the center of rectangle A
Ax = unit vector representing the local x-axis of A
Ay = unit vector representing the local y-axis of A
WA = half width of A (corresponds with the local x-axis of A)
HA = half height of A (corresponds with the local y-axis of A)

PB = coordinate position of the center of rectangle B

Bx = unit vector representing the local x-axis of B
By = unit vector representing the local y-axis of B
WB = half width of B (corresponds with the local x-axis of B)
HB = half height of B (corresponds with the local y-axis of B)

T= PB - PA

轴L是Seprating Axis的条件是

Proj是投影计算，展开

| T • L | > | ( WA*Ax ) • L | + | ( HA*Ay ) • L | + | ( WB*Bx ) • L | + |( HB*By ) • L |

L只有四种情况，AX, AY, BX, BY

CASE 1:
// L = Ax
| T • Ax | > | ( WA*Ax ) • Ax | + | ( HA*Ay ) • Ax | + | ( WB*Bx ) • Ax | + |( HB*By ) • Ax |
| T • Ax | > WA + 0 + | ( WB*Bx ) • Ax | + |( HB*By ) • Ax |
| T • Ax | > WA + | ( WB*Bx ) • Ax | + |( HB*By ) • Ax |

如果成立，存在Separating Axis平行Ax。

CASE 2:
// L = Ay
| T • Ay | > HA + | ( WB*Bx ) • Ay | + |( HB*By ) • Ay |

如果成立，存在Separating Axis平行Ay。

CASE 3:
// L = Bx
| T • Bx | > | ( WA* Ax ) • Bx | + | ( HA*Ay ) • Bx | + WB

如果成立，存在Separating Axis平行Bx。

CASE 4:
// L = By
| T • By | > | ( WA* Ax ) • By | + | ( HA*Ay ) • By | + HB

如果成立，存在Separating Axis平行By。

三维情况

在三维情况下的，之前的Separating Line就变成了Separating Plane，如下图

每个Box都有三组面，每组面都是平行的，Separating Plane都是平行其中的一个面。则两个box的SAT中可能的Separating Axis有六个

下图中Separating Plane就平行于右边Box的一个面

然而还有一种情况，如下

这种情况，两个Box并不相交，但是他们并没有发生碰撞，这种情况，Separating Plane的法线是两条红线的叉乘

所以在三维情况下Box和Box的碰撞检测需要判定的情况有6+9种

CASE 1:L = Ax CASE 2:L = Ay CASE 3:L = Az CASE 4:L = Bx CASE 5:L = By CASE 6:L = Bz

CASE 7:L = Ax  Bx CASE 8:L = Ax  By CASE 9:L = Ax  Bz CASE 10:L = Ay  Bx

CASE 11:L = Ay  By CASE 12:L = Ay  Bz CASE 13:L = Az  Bx CASE 14:L = Az  By CASE 15:L = Az  Bz

判断条件还是

| T • L | > | ( WA*Ax ) • L | + | ( HA*Ay ) • L | + |( DA*Az ) • L |+ | ( WB*Bx ) • L | + |( HB*By ) • L | + |( DB*Bz ) • L |

一点优化，关于T • L

T • (Ax * Bx) =(T • Az)(Ay • Bx) - (T• Ay)(Az • Bx)

这里将有叉乘的地方进行了转化，要看证明的请看参考资料。

好，可以上代码了

  public static bool IntersectBoxBox(Box box0, Box box1)        {            Vector3 v = box1.center - box0.center;            //Compute A's basis            Vector3 VAx = box0.rotation * new Vector3(1, 0, 0);            Vector3 VAy = box0.rotation * new Vector3(0, 1, 0);            Vector3 VAz = box0.rotation * new Vector3(0, 0, 1);            Vector3[] VA = new Vector3[3];            VA[0] = VAx;            VA[1] = VAy;            VA[2] = VAz;            //Compute B's basis            Vector3 VBx = box1.rotation * new Vector3(1, 0, 0);            Vector3 VBy = box1.rotation * new Vector3(0, 1, 0);            Vector3 VBz = box1.rotation * new Vector3(0, 0, 1);            Vector3[] VB = new Vector3[3];            VB[0] = VBx;            VB[1] = VBy;            VB[2] = VBz;            Vector3 T = new Vector3(Vector3.Dot(v, VAx), Vector3.Dot(v, VAy), Vector3.Dot(v, VAz));            float[,] R = new float[3, 3];            float[,] FR = new float[3, 3];            float ra, rb, t;            for (int i = 0; i < 3; i++)            {                for (int k = 0; k < 3; k++)                {                    R[i, k] = Vector3.Dot(VA[i], VB[k]);                    FR[i, k] = 1e-6f + Mathf.Abs(R[i, k]);                }            }            // A's basis vectors            for (int i = 0; i < 3; i++)            {                ra = box0.extents[i];                rb = box1.extents[0] * FR[i, 0] + box1.extents[1] * FR[i, 1] + box1.extents[2] * FR[i, 2];                t = Mathf.Abs(T[i]);                if (t > ra + rb) return false;            }            // B's basis vectors            for (int k = 0; k < 3; k++)            {                ra = box0.extents[0] * FR[0, k] + box0.extents[1] * FR[1, k] + box0.extents[2] * FR[2, k];                rb = box1.extents[k];                t = Mathf.Abs(T[0] * R[0, k] + T[1] * R[1, k] + T[2] * R[2, k]);                if (t > ra + rb) return false;            }            //9 cross products            //L = A0 x B0            ra = box0.extents[1] * FR[2, 0] + box0.extents[2] * FR[1, 0];            rb = box1.extents[1] * FR[0, 2] + box1.extents[2] * FR[0, 1];            t = Mathf.Abs(T[2] * R[1, 0] - T[1] * R[2, 0]);            if (t > ra + rb) return false;            //L = A0 x B1            ra = box0.extents[1] * FR[2, 1] + box0.extents[2] * FR[1, 1];            rb = box1.extents[0] * FR[0, 2] + box1.extents[2] * FR[0, 0];            t = Mathf.Abs(T[2] * R[1, 1] - T[1] * R[2, 1]);            if (t > ra + rb) return false;            //L = A0 x B2            ra = box0.extents[1] * FR[2, 2] + box0.extents[2] * FR[1, 2];            rb = box1.extents[0] * FR[0, 1] + box1.extents[1] * FR[0, 0];            t = Mathf.Abs(T[2] * R[1, 2] - T[1] * R[2, 2]);            if (t > ra + rb) return false;            //L = A1 x B0            ra = box0.extents[0] * FR[2, 0] + box0.extents[2] * FR[0, 0];            rb = box1.extents[1] * FR[1, 2] + box1.extents[2] * FR[1, 1];            t = Mathf.Abs(T[0] * R[2, 0] - T[2] * R[0, 0]);            if (t > ra + rb) return false;            //L = A1 x B1            ra = box0.extents[0] * FR[2, 1] + box0.extents[2] * FR[0, 1];            rb = box1.extents[0] * FR[1, 2] + box1.extents[2] * FR[1, 0];            t = Mathf.Abs(T[0] * R[2, 1] - T[2] * R[0, 1]);            if (t > ra + rb) return false;            //L = A1 x B2            ra = box0.extents[0] * FR[2, 2] + box0.extents[2] * FR[0, 2];            rb = box1.extents[0] * FR[1, 1] + box1.extents[1] * FR[1, 0];            t = Mathf.Abs(T[0] * R[2, 2] - T[2] * R[0, 2]);            if (t > ra + rb) return false;            //L = A2 x B0            ra = box0.extents[0] * FR[1, 0] + box0.extents[1] * FR[0, 0];            rb = box1.extents[1] * FR[2, 2] + box1.extents[2] * FR[2, 1];            t = Mathf.Abs(T[1] * R[0, 0] - T[0] * R[1, 0]);            if (t > ra + rb) return false;            //L = A2 x B1            ra = box0.extents[0] * FR[1, 1] + box0.extents[1] * FR[0, 1];            rb = box1.extents[0] * FR[2, 2] + box1.extents[2] * FR[2, 0];            t = Mathf.Abs(T[1] * R[0, 1] - T[0] * R[1, 1]);            if (t > ra + rb) return false;            //L = A2 x B2            ra = box0.extents[0] * FR[1, 2] + box0.extents[1] * FR[0, 2];            rb = box1.extents[0] * FR[2, 1] + box1.extents[1] * FR[2, 0];            t = Mathf.Abs(T[1] * R[0, 2] - T[0] * R[1, 2]);            if (t > ra + rb) return false;            return true;        }

测试代码

public class BoxBoxTester : MonoBehaviour {    public GameObject box;    public GameObject box1;    Box _box;    Box _box1;    // Use this for initialization    void Start()    {        _box = new Box();        _box1 = new Box();    }    // Update is called once per frame    void Update()    {        _box.center = box.transform.position;        _box.rotation = box.transform.rotation;        _box.extents = 0.5f * box.transform.localScale;        _box1.center = box1.transform.position;        _box1.rotation = box1.transform.rotation;        _box1.extents = 0.5f * box1.transform.localScale;        if (NIntersectTests.IntersectBoxBox(_box, _box1))        {            box.GetComponent<MeshRenderer>().materials[0].SetColor("_Color", new Color(1, 0, 0));        }        else        {            box.GetComponent<MeshRenderer>().materials[0].SetColor("_Color", new Color(1, 1, 1));        }    }}

运行结果

参考

Separating Axis Theorem for Oriented Bounding Boxes

1 0