我有一条从a到B的直线和一个半径为R的圆。
用什么算法来检查直线是否与圆相交?它在圆边的哪个坐标上?
我有一条从a到B的直线和一个半径为R的圆。
用什么算法来检查直线是否与圆相交?它在圆边的哪个坐标上?
当前回答
如果你找到了圆心(因为它是3D的,我想你是指球体而不是圆)和直线之间的距离,然后检查这个距离是否小于可以做到这一点的半径。
碰撞点显然是直线和球面之间最近的点(当你计算球面和直线之间的距离时,会计算出这个点)
点与线之间的距离: http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html
其他回答
好吧,我不会给你代码,但既然你已经标记了这个算法,我认为这对你来说无关紧要。 首先,你要得到一个垂直于这条直线的向量。
y = ax + c是一个未知变量c是未知变量 为了解决这个问题,计算直线经过圆心时的值。
也就是说, 将圆心的位置代入直线方程,解出c。 然后计算原直线与其法线的交点。
这样就能得到直线上离圆最近的点。 计算该点到圆中心之间的距离(使用矢量的大小)。 如果这个小于圆的半径,看,我们有一个交点!
只是这个线程的一个补充… 下面是pahlevan发布的代码版本,但针对c# /XNA,并做了一些整理:
/// <summary>
/// Intersects a line and a circle.
/// </summary>
/// <param name="location">the location of the circle</param>
/// <param name="radius">the radius of the circle</param>
/// <param name="lineFrom">the starting point of the line</param>
/// <param name="lineTo">the ending point of the line</param>
/// <returns>true if the line and circle intersect each other</returns>
public static bool IntersectLineCircle(Vector2 location, float radius, Vector2 lineFrom, Vector2 lineTo)
{
float ab2, acab, h2;
Vector2 ac = location - lineFrom;
Vector2 ab = lineTo - lineFrom;
Vector2.Dot(ref ab, ref ab, out ab2);
Vector2.Dot(ref ac, ref ab, out acab);
float t = acab / ab2;
if (t < 0)
t = 0;
else if (t > 1)
t = 1;
Vector2 h = ((ab * t) + lineFrom) - location;
Vector2.Dot(ref h, ref h, out h2);
return (h2 <= (radius * radius));
}
如果你找到了圆心(因为它是3D的,我想你是指球体而不是圆)和直线之间的距离,然后检查这个距离是否小于可以做到这一点的半径。
碰撞点显然是直线和球面之间最近的点(当你计算球面和直线之间的距离时,会计算出这个点)
点与线之间的距离: http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html
我写了一个小脚本,通过将圆的中心点投影到直线上来测试相交。
vector distVector = centerPoint - projectedPoint;
if(distVector.length() < circle.radius)
{
double distance = circle.radius - distVector.length();
vector moveVector = distVector.normalize() * distance;
circle.move(moveVector);
}
http://jsfiddle.net/ercang/ornh3594/1/
如果需要检查与线段的碰撞,还需要考虑圆心到起点和终点的距离。
vector distVector = centerPoint - startPoint;
if(distVector.length() < circle.radius)
{
double distance = circle.radius - distVector.length();
vector moveVector = distVector.normalize() * distance;
circle.move(moveVector);
}
https://jsfiddle.net/ercang/menp0991/
基于@Joe Skeen的python解决方案
def check_line_segment_circle_intersection(line, point, radious):
""" Checks whether a point intersects with a line defined by two points.
A `point` is list with two values: [2, 3]
A `line` is list with two points: [point1, point2]
"""
line_distance = distance(line[0], line[1])
distance_start_to_point = distance(line[0], point)
distance_end_to_point = distance(line[1], point)
if (distance_start_to_point <= radious or distance_end_to_point <= radious):
return True
# angle between line and point with law of cosines
numerator = (math.pow(distance_start_to_point, 2)
+ math.pow(line_distance, 2)
- math.pow(distance_end_to_point, 2))
denominator = 2 * distance_start_to_point * line_distance
ratio = numerator / denominator
ratio = ratio if ratio <= 1 else 1 # To account for float errors
ratio = ratio if ratio >= -1 else -1 # To account for float errors
angle = math.acos(ratio)
# distance from the point to the line with sin projection
distance_line_to_point = math.sin(angle) * distance_start_to_point
if distance_line_to_point <= radious:
point_projection_in_line = math.cos(angle) * distance_start_to_point
# Intersection occurs whent the point projection in the line is less
# than the line distance and positive
return point_projection_in_line <= line_distance and point_projection_in_line >= 0
return False
def distance(point1, point2):
return math.sqrt(
math.pow(point1[1] - point2[1], 2) +
math.pow(point1[0] - point2[0], 2)
)