圆 - 矩形碰撞检测(交叉)

如果您正在编写任何几何图形,则可能已在库中具有上述功能.否则,pointInRectangle()可以通过多种方式实施; 多边形方法中的任何一般点都可以使用,但是对于一个矩形,您可以检查它是否有效:

0 ? AP·AB ? AB·AB and 0 ? AP·AD ? AD·AD

并且intersectCircle()易于实现:一种方法是检查从P线到线的垂直脚是否足够接近并且在端点之间,否则检查端点.

很酷的是,同样的想法不仅适用于矩形,而且适用于圆与任何简单多边形的交叉- 甚至不必是凸起的!

这是另一个实现起来非常简单的解决方案(也非常快).它将捕获所有交叉点,包括球体完全进入矩形时.

// clamp(value, min, max) - limits value to the range min..max

// Find the closest point to the circle within the rectangle
float closestX = clamp(circle.X, rectangle.Left, rectangle.Right);
float closestY = clamp(circle.Y, rectangle.Top, rectangle.Bottom);

// Calculate the distance between the circle's center and this closest point
float distanceX = circle.X - closestX;
float distanceY = circle.Y - closestY;

// If the distance is less than the circle's radius, an intersection occurs
float distanceSquared = (distanceX * distanceX) + (distanceY * distanceY);
return distanceSquared < (circle.Radius * circle.Radius);

任何体面的数学库,可以缩短为3或4行.

你的球体和矩形与IIF相交
,圆心与矩形的一个顶点之间的距离小于球体的半径,
或者
圆心与矩形的一个边缘之间的距离小于球体的半径( [ 点线距离 ])
或
圆心位于矩形

点点距离内:

P1 = [x1,y1]
P2 = [x2,y2]
Distance = sqrt(abs(x1 - x2)+abs(y1-y2))

点线距离:

L1 = [x1,y1],L2 = [x2,y2] (two points of your line, ie the vertex points)
P1 = [px,py] some point

Distance d =  abs( (x2-x1)(y1-py)-(x1-px)(y2-y1) ) / Distance(L1,L2)

在rect内部的圆心:
采取一个分离轴的方法:如果在一条线上有一个投影,将该矩形与该点分开,它们就不会相交

将点投影在平行于矩形边的线上,然后可以轻松确定它们是否相交.如果它们不与所有4个投影相交,则它们(点和矩形)不能相交.

你只需要内积(x = [x1,x2],y = [y1,y2],x*y = x1*y1 + x2*y2)

你的测试看起来像那样:

//rectangle edges: TL (top left), TR (top right), BL (bottom left), BR (bottom right)
//point to test: POI

seperated = false
for egde in { {TL,TR}, {BL,BR}, {TL,BL},{TR-BR} }:  // the edges
    D = edge[0] - edge[1]
    innerProd =  D * POI
    Interval_min = min(D*edge[0],D*edge[1])
    Interval_max = max(D*edge[0],D*edge[1])
    if not (  Interval_min ≤ innerProd ≤  Interval_max ) 
           seperated = true
           break  // end for loop 
    end if
end for
if (seperated is true)    
      return "no intersection"
else 
      return "intersection"
end if

这并不假设轴对齐的矩形,并且可以很容易地扩展以测试凸集之间的交叉点.

这是最快的解决方案:

public static boolean intersect(Rectangle r, Circle c)
{
    float cx = Math.abs(c.x - r.x - r.halfWidth);
    float xDist = r.halfWidth + c.radius;
    if (cx > xDist)
        return false;
    float cy = Math.abs(c.y - r.y - r.halfHeight);
    float yDist = r.halfHeight + c.radius;
    if (cy > yDist)
        return false;
    if (cx <= r.halfWidth || cy <= r.halfHeight)
        return true;
    float xCornerDist = cx - r.halfWidth;
    float yCornerDist = cy - r.halfHeight;
    float xCornerDistSq = xCornerDist * xCornerDist;
    float yCornerDistSq = yCornerDist * yCornerDist;
    float maxCornerDistSq = c.radius * c.radius;
    return xCornerDistSq + yCornerDistSq <= maxCornerDistSq;
}

请注意执行顺序,并预先计算宽度/高度的一半.平方也是"手动"完成的,以节省一些时钟周期.