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;
}
其他回答
对于那些需要用SQL在地理坐标中计算圆/矩形碰撞的人, 这是我在oracle 11中实现的e.James建议算法。
在输入中,它需要圆坐标,圆半径km和矩形的两个顶点坐标:
CREATE OR REPLACE FUNCTION "DETECT_CIRC_RECT_COLLISION"
(
circleCenterLat IN NUMBER, -- circle Center Latitude
circleCenterLon IN NUMBER, -- circle Center Longitude
circleRadius IN NUMBER, -- circle Radius in KM
rectSWLat IN NUMBER, -- rectangle South West Latitude
rectSWLon IN NUMBER, -- rectangle South West Longitude
rectNELat IN NUMBER, -- rectangle North Est Latitude
rectNELon IN NUMBER -- rectangle North Est Longitude
)
RETURN NUMBER
AS
-- converts km to degrees (use 69 if miles)
kmToDegreeConst NUMBER := 111.045;
-- Remaining rectangle vertices
rectNWLat NUMBER;
rectNWLon NUMBER;
rectSELat NUMBER;
rectSELon NUMBER;
rectHeight NUMBER;
rectWIdth NUMBER;
circleDistanceLat NUMBER;
circleDistanceLon NUMBER;
cornerDistanceSQ NUMBER;
BEGIN
-- Initialization of remaining rectangle vertices
rectNWLat := rectNELat;
rectNWLon := rectSWLon;
rectSELat := rectSWLat;
rectSELon := rectNELon;
-- Rectangle sides length calculation
rectHeight := calc_distance(rectSWLat, rectSWLon, rectNWLat, rectNWLon);
rectWidth := calc_distance(rectSWLat, rectSWLon, rectSELat, rectSELon);
circleDistanceLat := abs( (circleCenterLat * kmToDegreeConst) - ((rectSWLat * kmToDegreeConst) + (rectHeight/2)) );
circleDistanceLon := abs( (circleCenterLon * kmToDegreeConst) - ((rectSWLon * kmToDegreeConst) + (rectWidth/2)) );
IF circleDistanceLon > ((rectWidth/2) + circleRadius) THEN
RETURN -1; -- -1 => NO Collision ; 0 => Collision Detected
END IF;
IF circleDistanceLat > ((rectHeight/2) + circleRadius) THEN
RETURN -1; -- -1 => NO Collision ; 0 => Collision Detected
END IF;
IF circleDistanceLon <= (rectWidth/2) THEN
RETURN 0; -- -1 => NO Collision ; 0 => Collision Detected
END IF;
IF circleDistanceLat <= (rectHeight/2) THEN
RETURN 0; -- -1 => NO Collision ; 0 => Collision Detected
END IF;
cornerDistanceSQ := POWER(circleDistanceLon - (rectWidth/2), 2) + POWER(circleDistanceLat - (rectHeight/2), 2);
IF cornerDistanceSQ <= POWER(circleRadius, 2) THEN
RETURN 0; -- -1 => NO Collision ; 0 => Collision Detected
ELSE
RETURN -1; -- -1 => NO Collision ; 0 => Collision Detected
END IF;
RETURN -1; -- -1 => NO Collision ; 0 => Collision Detected
END;
稍微改进一下e。james的回答:
double dx = abs(circle.x - rect.x) - rect.w / 2,
dy = abs(circle.y - rect.y) - rect.h / 2;
if (dx > circle.r || dy > circle.r) { return false; }
if (dx <= 0 || dy <= 0) { return true; }
return (dx * dx + dy * dy <= circle.r * circle.r);
这就减去了一次,而不是最多减去三次。
预检查一个完全封装矩形的圆是否与矩形发生碰撞。 检查圆内的矩形角。 对于每条边,看看是否有一条线与圆相交。将中心点C投影到直线AB上,得到点d。如果CD的长度小于半径,则发生了碰撞。
projectionScalar=dot(AC,AB)/(mag(AC)*mag(AB));
if(projectionScalar>=0 && projectionScalar<=1) {
D=A+AB*projectionScalar;
CD=D-C;
if(mag(CD)<circle.radius){
// there was a collision
}
}
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;
}
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)));