如何分辨圆和矩形在二维欧几里得空间中是否相交?(即经典二维几何)
当前回答
为了可视化,拿你的键盘的numpad。如果键“5”代表你的矩形,那么所有的键1-9代表空间的9个象限除以构成矩形的线(5是里面的线)。
1)如果圆的中心在象限5(即在矩形内),则两个形状相交。
这里有两种可能的情况: a)圆与矩形的两条或多条相邻边相交。 b)圆与矩形的一条边相交。
第一种情况很简单。如果圆与矩形的两条相邻边相交,则它必须包含连接这两条边的角。(或者说它的中心在象限5,我们已经讲过了。还要注意,圆只与矩形的两条相对边相交的情况也被覆盖了。)
2)如果矩形的任意角A、B、C、D在圆内,则这两个形状相交。
第二种情况比较棘手。我们应该注意到,只有当圆的中心位于2、4、6或8象限中的一个象限时,才会发生这种情况。(事实上,如果中心在1、3、7、8象限中的任何一个象限上,则相应的角将是离它最近的点。)
现在我们有了圆的中心在一个“边”象限内的情况,它只与相应的边相交。那么,边缘上最接近圆中心的点必须在圆内。
3)对于每条直线AB, BC, CD, DA,构造经过圆中心p的垂直线p(AB, p), p(BC, p), p(CD, p), p(DA, p),对于每条垂直线,如果与原边的交点在圆内,则两个图形相交。
最后一步有一个捷径。如果圆的圆心在象限8,边AB是上边,交点的y坐标是A和B, x坐标是P。
你可以构造四条线的交点并检查它们是否在相应的边上,或者找出P在哪个象限并检查相应的交点。两者都应该化简为相同的布尔方程。要注意的是,上面的步骤2并没有排除P位于“角落”象限之一;它只是在寻找一个十字路口。
编辑:事实证明,我忽略了一个简单的事实,即#2是#3的子情况。毕竟,角也是边缘上的点。请看下面@ShreevatsaR的回答,你会得到很好的解释。与此同时,忘记上面的第二条,除非你想要一个快速但冗余的检查。
其他回答
有效,一周前才发现,现在才开始测试。
double theta = Math.atan2(cir.getX()-sqr.getX()*1.0,
cir.getY()-sqr.getY()*1.0); //radians of the angle
double dBox; //distance from box to edge of box in direction of the circle
if((theta > Math.PI/4 && theta < 3*Math.PI / 4) ||
(theta < -Math.PI/4 && theta > -3*Math.PI / 4)) {
dBox = sqr.getS() / (2*Math.sin(theta));
} else {
dBox = sqr.getS() / (2*Math.cos(theta));
}
boolean touching = (Math.abs(dBox) >=
Math.sqrt(Math.pow(sqr.getX()-cir.getX(), 2) +
Math.pow(sqr.getY()-cir.getY(), 2)));
球面和矩形相交于IIF 圆心和矩形的一个顶点之间的距离小于球体的半径 或 圆心与矩形的一条边之间的距离小于球面的半径([点线距离]) 或 圆的中心在矩形的内部 一点上距离:
P1 = [x1,y1] P2 = [x2,y2] Distance = sqrt(abs(x1 - x2)+abs(y1-y2))
点线路距离:
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)
矩形内圆中心: 采用分离轴的方法:如果存在一个投影到一条直线上,将矩形与点分开,它们就不相交
您将点投影在平行于矩形边的直线上,然后可以很容易地确定它们是否相交。如果它们不在所有4个投影上相交,它们(点和矩形)就不能相交。
你只需要内积(x= [x1,x2],y = [y1,y2],x *y = x1*y1 + x2*y2)
你的测试应该是这样的:
//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
它没有假设一个轴对齐的矩形,并且很容易扩展用于测试凸集之间的交集。
我为处理形状创建了类 希望你喜欢
public class Geomethry {
public static boolean intersectionCircleAndRectangle(int circleX, int circleY, int circleR, int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight){
boolean result = false;
float rectHalfWidth = rectangleWidth/2.0f;
float rectHalfHeight = rectangleHeight/2.0f;
float rectCenterX = rectangleX + rectHalfWidth;
float rectCenterY = rectangleY + rectHalfHeight;
float deltax = Math.abs(rectCenterX - circleX);
float deltay = Math.abs(rectCenterY - circleY);
float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;
do{
// check that distance between the centerse is more than the distance between the circumcircle of rectangle and circle
if(lengthHypotenuseSqure > ((rectHalfWidth+circleR)*(rectHalfWidth+circleR) + (rectHalfHeight+circleR)*(rectHalfHeight+circleR))){
//System.out.println("distance between the centerse is more than the distance between the circumcircle of rectangle and circle");
break;
}
// check that distance between the centerse is less than the distance between the inscribed circle
float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
if(lengthHypotenuseSqure < ((rectMinHalfSide+circleR)*(rectMinHalfSide+circleR))){
//System.out.println("distance between the centerse is less than the distance between the inscribed circle");
result=true;
break;
}
// check that the squares relate to angles
if((deltax > (rectHalfWidth+circleR)*0.9) && (deltay > (rectHalfHeight+circleR)*0.9)){
//System.out.println("squares relate to angles");
result=true;
}
}while(false);
return result;
}
public static boolean intersectionRectangleAndRectangle(int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight, int rectangleX2, int rectangleY2, int rectangleWidth2, int rectangleHeight2){
boolean result = false;
float rectHalfWidth = rectangleWidth/2.0f;
float rectHalfHeight = rectangleHeight/2.0f;
float rectHalfWidth2 = rectangleWidth2/2.0f;
float rectHalfHeight2 = rectangleHeight2/2.0f;
float deltax = Math.abs((rectangleX + rectHalfWidth) - (rectangleX2 + rectHalfWidth2));
float deltay = Math.abs((rectangleY + rectHalfHeight) - (rectangleY2 + rectHalfHeight2));
float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;
do{
// check that distance between the centerse is more than the distance between the circumcircle
if(lengthHypotenuseSqure > ((rectHalfWidth+rectHalfWidth2)*(rectHalfWidth+rectHalfWidth2) + (rectHalfHeight+rectHalfHeight2)*(rectHalfHeight+rectHalfHeight2))){
//System.out.println("distance between the centerse is more than the distance between the circumcircle");
break;
}
// check that distance between the centerse is less than the distance between the inscribed circle
float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
float rectMinHalfSide2 = Math.min(rectHalfWidth2, rectHalfHeight2);
if(lengthHypotenuseSqure < ((rectMinHalfSide+rectMinHalfSide2)*(rectMinHalfSide+rectMinHalfSide2))){
//System.out.println("distance between the centerse is less than the distance between the inscribed circle");
result=true;
break;
}
// check that the squares relate to angles
if((deltax > (rectHalfWidth+rectHalfWidth2)*0.9) && (deltay > (rectHalfHeight+rectHalfHeight2)*0.9)){
//System.out.println("squares relate to angles");
result=true;
}
}while(false);
return result;
}
}
为我工作(只工作时,矩形的角度是180)
function intersects(circle, rect) {
let left = rect.x + rect.width > circle.x - circle.radius;
let right = rect.x < circle.x + circle.radius;
let top = rect.y < circle.y + circle.radius;
let bottom = rect.y + rect.height > circle.y - circle.radius;
return left && right && bottom && top;
}
下面是我的C代码,用于解决球体和非轴对齐的盒子之间的碰撞。它依赖于我自己的几个库例程,但它可能对某些人有用。我在游戏中使用了它,效果非常好。
float physicsProcessCollisionBetweenSelfAndActorRect(SPhysics *self, SPhysics *actor)
{
float diff = 99999;
SVector relative_position_of_circle = getDifference2DBetweenVectors(&self->worldPosition, &actor->worldPosition);
rotateVector2DBy(&relative_position_of_circle, -actor->axis.angleZ); // This aligns the coord system so the rect becomes an AABB
float x_clamped_within_rectangle = relative_position_of_circle.x;
float y_clamped_within_rectangle = relative_position_of_circle.y;
LIMIT(x_clamped_within_rectangle, actor->physicsRect.l, actor->physicsRect.r);
LIMIT(y_clamped_within_rectangle, actor->physicsRect.b, actor->physicsRect.t);
// Calculate the distance between the circle's center and this closest point
float distance_to_nearest_edge_x = relative_position_of_circle.x - x_clamped_within_rectangle;
float distance_to_nearest_edge_y = relative_position_of_circle.y - y_clamped_within_rectangle;
// If the distance is less than the circle's radius, an intersection occurs
float distance_sq_x = SQUARE(distance_to_nearest_edge_x);
float distance_sq_y = SQUARE(distance_to_nearest_edge_y);
float radius_sq = SQUARE(self->physicsRadius);
if(distance_sq_x + distance_sq_y < radius_sq)
{
float half_rect_w = (actor->physicsRect.r - actor->physicsRect.l) * 0.5f;
float half_rect_h = (actor->physicsRect.t - actor->physicsRect.b) * 0.5f;
CREATE_VECTOR(push_vector);
// If we're at one of the corners of this object, treat this as a circular/circular collision
if(fabs(relative_position_of_circle.x) > half_rect_w && fabs(relative_position_of_circle.y) > half_rect_h)
{
SVector edges;
if(relative_position_of_circle.x > 0) edges.x = half_rect_w; else edges.x = -half_rect_w;
if(relative_position_of_circle.y > 0) edges.y = half_rect_h; else edges.y = -half_rect_h;
push_vector = relative_position_of_circle;
moveVectorByInverseVector2D(&push_vector, &edges);
// We now have the vector from the corner of the rect to the point.
float delta_length = getVector2DMagnitude(&push_vector);
float diff = self->physicsRadius - delta_length; // Find out how far away we are from our ideal distance
// Normalise the vector
push_vector.x /= delta_length;
push_vector.y /= delta_length;
scaleVector2DBy(&push_vector, diff); // Now multiply it by the difference
push_vector.z = 0;
}
else // Nope - just bouncing against one of the edges
{
if(relative_position_of_circle.x > 0) // Ball is to the right
push_vector.x = (half_rect_w + self->physicsRadius) - relative_position_of_circle.x;
else
push_vector.x = -((half_rect_w + self->physicsRadius) + relative_position_of_circle.x);
if(relative_position_of_circle.y > 0) // Ball is above
push_vector.y = (half_rect_h + self->physicsRadius) - relative_position_of_circle.y;
else
push_vector.y = -((half_rect_h + self->physicsRadius) + relative_position_of_circle.y);
if(fabs(push_vector.x) < fabs(push_vector.y))
push_vector.y = 0;
else
push_vector.x = 0;
}
diff = 0; // Cheat, since we don't do anything with the value anyway
rotateVector2DBy(&push_vector, actor->axis.angleZ);
SVector *from = &self->worldPosition;
moveVectorBy2D(from, push_vector.x, push_vector.y);
}
return diff;
}