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
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;
其他回答
实际上，这要简单得多。你只需要两样东西。
首先，你需要找出从圆中心到矩形每条直线的四个正交距离。如果任意三个圆的半径大于矩形的半径，那么圆就不会与矩形相交。
其次，你需要找到圆中心和矩形中心之间的距离，那么你的圆不会在矩形内部如果距离大于矩形对角线长度的一半。
好运！
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;
有一种非常简单的方法来做到这一点，你必须在x和y上夹住一个点，但在正方形内部，当圆心在一个垂直轴上的两个正方形边界点之间时，你需要将这些坐标夹到平行轴上，只是要确保夹住的坐标不超过正方形的限制。 然后只需得到圆心与夹紧坐标之间的距离，并检查距离是否小于圆的半径。
以下是我是如何做到的(前4个点是方坐标，其余是圆点):
bool DoesCircleImpactBox(float x, float y, float x1, float y1, float xc, float yc, float radius){
float ClampedX=0;
float ClampedY=0;
if(xc>=x and xc<=x1){
ClampedX=xc;
}
if(yc>=y and yc<=y1){
ClampedY=yc;
}
radius = radius+1;
if(xc<x) ClampedX=x;
if(xc>x1) ClampedX=x1-1;
if(yc<y) ClampedY=y;
if(yc>y1) ClampedY=y1-1;
float XDif=ClampedX-xc;
XDif=XDif*XDif;
float YDif=ClampedY-yc;
YDif=YDif*YDif;
if(XDif+YDif<=radius*radius) return true;
return false;
}
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;
}
注意执行顺序，一半的宽度/高度是预先计算好的。此外，平方是“手动”完成的，以节省一些时钟周期。
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;
}
