OpenGL 2D Circle - 旋转的 AABB 碰撞

Question

我找不到检测圆和旋转矩形之间碰撞的方法。我的方法是首先将圆和矩形旋转 -angle，其中 angle 是矩形旋转的弧度。因此，矩形和圆与轴对齐，所以我可以执行基本圆-AABB 碰撞检测。

bool CheckCollision(float circleX, float circleY, float radius, float left, float bottom, float width, float height, float angle){

        // Rotating the circle and the rectangle with -angle
            circleX = circleX * cos(-angle) - circleY * sin(-angle);
            circleY = circleX * sin(-angle) + circleY * cos(-angle);

            left = left * cos(-angle) - bottom* sin(-angle);
            bottom = left * sin(-angle) + bottom * cos(-angle);

       }

glm::vec2 center(circleX, circleY);
        // calculate AABB info (center, half-extents)
        glm::vec2 aabb_half_extents(width / 2.0f, height / 2.0f);
        glm::vec2 aabb_center(
            left + aabb_half_extents.x,
            bottom + aabb_half_extents.y
        );
        // get difference vector between both centers
        glm::vec2 difference = center - aabb_center;
        glm::vec2 clamped = glm::clamp(difference, -aabb_half_extents, aabb_half_extents);
        // add clamped value to AABB_center and we get the value of box closest to circle
        glm::vec2 closest = aabb_center + clamped;
        // retrieve vector between center circle and closest point AABB and check if length <= radius
        difference = closest - center;

        return glm::length(difference) < radius;

Answer 1

设矩形中心为rcx, rcy。以该点为坐标原点，以该点为中心旋转圆心（cx, cy为相对于矩形中心的坐标）：

cx = (circleX - rcx) * cos(-angle) - (circleY - rcy) * sin(-angle);
cy = (circleX - rcx) * sin(-angle) + (circleY - rcy) * cos(-angle);

现在求圆心到矩形最近点的平方距离（零表示圆心在矩形内）：

dx = max(Abs(cx) - rect_width / 2, 0)
dy = max(Abs(cy) - rect_height / 2, 0)
SquaredDistance = dx * dx + dy * dy

然后与平方半径比较

OpenGL 2D Circle - 旋转的 AABB 碰撞

OpenGL 2D Circle - Rotated AABB Collision

c++

geometry

2d

collision-detection

glm-math