Question

我试图制作一个使用世界坐标和边界框结构的算法检测两个边界框之间的碰撞。我真的不知道我在做什么，但我认为下面的代码会起作用。我的问题是它只检测碰撞，如果边界框位于完全相同的x，y，z位置。

BOOL AABB::isCollidedWith(AABB* bb)
{
if(bb == NULL) return FALSE;

float radX1,radX2;
float radY1,radY2;
float radZ1,radZ2;

float arr[12];

      //please note that all the mins are set to 0
      //and all the maxes are set to 1

radX1 = (bb->maxX - bb->minX) / 2;
radX2 = (this->maxX - this->minX) / 2;
radY1 = (bb->maxY - bb->minY) / 2;
radY2 = (this->maxY - this->minY) / 2;
radZ1 = (bb->maxZ - bb->minZ) / 2;
radZ2 = (this->maxZ - this->minZ) / 2;

//bb coords

arr[1] = bb->bbX - radX1;
arr[2] = bb->bbX + radX1;
arr[3] = bb->bbY - radY1;
arr[4] = bb->bbY + radY1;
arr[5] = bb->bbZ - radZ1;
arr[6] = bb->bbZ + radZ1;

//this coords
arr[7]  = this->bbX - radX2;
arr[8]  = this->bbX + radX2;
arr[9]  = this->bbY - radY2;
arr[10] = this->bbY + radY2;
arr[11] = this->bbZ - radZ2;
arr[12] = this->bbZ + radZ2;

if(arr[2] >= arr[7] && arr[1] <= arr[8])
{
    if(arr[4] >= arr[9] && arr[3] <= arr[10])
    {
        if(arr[6] >= arr[11] && arr[5] <= arr[12])
        {
            this->collided = TRUE;
            OutputDebugStringA("Collided!\n");
            return TRUE;
        }
    }
}
}

我正在比较的结构：

 AABB* aabb1 = new AABB(0.0f,0.0f,0.0f,1.0f,1.0f,1.0f,0.0f,0.0f,0.0f);
 AABB* aabb2 = new AABB(0.0f,0.0f,0.0f,1.0f,1.0f,1.0f,0.0f,0.0f,0.0f);
 aabb2->isCollidedWith(aabb1);

构造函数代码段：另请注意，最后三个参数指示边界框的x，y，z线

AABB::AABB(float minx,float maxx,float miny,float maxy,float minz,float maxz,float x,float y,float z)
{
this->minX = minx;
this->maxX = maxx;
this->minY = miny;
this->maxY = maxy;
this->minZ = minz;
this->maxZ = maxz;

任何帮助，批评或建议都会有所帮助。

Answer 1

在创建minX = 0.0和maxX = 0.0的框时，bbX坐标必须与要碰撞的框相同（因为radX = 0）。 minZ = maxZ = 1.0也是如此。

请注意构造函数中参数的顺序：它是minX, maxX, minY, maxY, minZ, maxZ而不是 minX, minY, minZ, maxX, maxY, maxZ（我想你应该是第二个订单，并希望定义一个1.0 x 1.0的框x 1.0尺寸）。

Answer 2

简单错误！我忽略了列出参数的方式，导致了问题。另外，我从数组“arr”的每个成员中减去0.5，找到AABB的中心。

轴对齐边界框碰撞检测问题

2 个答案: