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
另一个在c#(部分圆类)。 经过测试，工作就像一个魅力。
public class Circle : IEquatable<Circle>
{
// ******************************************************************
// The center of a circle
private Point _center;
// The radius of a circle
private double _radius;
// ******************************************************************
/// <summary>
/// Find all intersections (0, 1, 2) of the circle with a line defined by its 2 points.
/// Using: http://math.stackexchange.com/questions/228841/how-do-i-calculate-the-intersections-of-a-straight-line-and-a-circle
/// Note: p is the Center.X and q is Center.Y
/// </summary>
/// <param name="linePoint1"></param>
/// <param name="linePoint2"></param>
/// <returns></returns>
public List<Point> GetIntersections(Point linePoint1, Point linePoint2)
{
List<Point> intersections = new List<Point>();
double dx = linePoint2.X - linePoint1.X;
if (dx.AboutEquals(0)) // Straight vertical line
{
if (linePoint1.X.AboutEquals(Center.X - Radius) || linePoint1.X.AboutEquals(Center.X + Radius))
{
Point pt = new Point(linePoint1.X, Center.Y);
intersections.Add(pt);
}
else if (linePoint1.X > Center.X - Radius && linePoint1.X < Center.X + Radius)
{
double x = linePoint1.X - Center.X;
Point pt = new Point(linePoint1.X, Center.Y + Math.Sqrt(Radius * Radius - (x * x)));
intersections.Add(pt);
pt = new Point(linePoint1.X, Center.Y - Math.Sqrt(Radius * Radius - (x * x)));
intersections.Add(pt);
}
return intersections;
}
// Line function (y = mx + b)
double dy = linePoint2.Y - linePoint1.Y;
double m = dy / dx;
double b = linePoint1.Y - m * linePoint1.X;
double A = m * m + 1;
double B = 2 * (m * b - m * _center.Y - Center.X);
double C = Center.X * Center.X + Center.Y * Center.Y - Radius * Radius - 2 * b * Center.Y + b * b;
double discriminant = B * B - 4 * A * C;
if (discriminant < 0)
{
return intersections; // there is no intersections
}
if (discriminant.AboutEquals(0)) // Tangeante (touch on 1 point only)
{
double x = -B / (2 * A);
double y = m * x + b;
intersections.Add(new Point(x, y));
}
else // Secant (touch on 2 points)
{
double x = (-B + Math.Sqrt(discriminant)) / (2 * A);
double y = m * x + b;
intersections.Add(new Point(x, y));
x = (-B - Math.Sqrt(discriminant)) / (2 * A);
y = m * x + b;
intersections.Add(new Point(x, y));
}
return intersections;
}
// ******************************************************************
// Get the center
[XmlElement("Center")]
public Point Center
{
get { return _center; }
set
{
_center = value;
}
}
// ******************************************************************
// Get the radius
[XmlElement]
public double Radius
{
get { return _radius; }
set { _radius = value; }
}
//// ******************************************************************
//[XmlArrayItemAttribute("DoublePoint")]
//public List<Point> Coordinates
//{
// get { return _coordinates; }
//}
// ******************************************************************
// Construct a circle without any specification
public Circle()
{
_center.X = 0;
_center.Y = 0;
_radius = 0;
}
// ******************************************************************
// Construct a circle without any specification
public Circle(double radius)
{
_center.X = 0;
_center.Y = 0;
_radius = radius;
}
// ******************************************************************
// Construct a circle with the specified circle
public Circle(Circle circle)
{
_center = circle._center;
_radius = circle._radius;
}
// ******************************************************************
// Construct a circle with the specified center and radius
public Circle(Point center, double radius)
{
_center = center;
_radius = radius;
}
// ******************************************************************
// Construct a circle based on one point
public Circle(Point center)
{
_center = center;
_radius = 0;
}
// ******************************************************************
// Construct a circle based on two points
public Circle(Point p1, Point p2)
{
Circle2Points(p1, p2);
}
要求:
using System;
namespace Mathematic
{
public static class DoubleExtension
{
// ******************************************************************
// Base on Hans Passant Answer on:
// http://stackoverflow.com/questions/2411392/double-epsilon-for-equality-greater-than-less-than-less-than-or-equal-to-gre
/// <summary>
/// Compare two double taking in account the double precision potential error.
/// Take care: truncation errors accumulate on calculation. More you do, more you should increase the epsilon.
public static bool AboutEquals(this double value1, double value2)
{
if (double.IsPositiveInfinity(value1))
return double.IsPositiveInfinity(value2);
if (double.IsNegativeInfinity(value1))
return double.IsNegativeInfinity(value2);
if (double.IsNaN(value1))
return double.IsNaN(value2);
double epsilon = Math.Max(Math.Abs(value1), Math.Abs(value2)) * 1E-15;
return Math.Abs(value1 - value2) <= epsilon;
}
// ******************************************************************
// Base on Hans Passant Answer on:
// http://stackoverflow.com/questions/2411392/double-epsilon-for-equality-greater-than-less-than-less-than-or-equal-to-gre
/// <summary>
/// Compare two double taking in account the double precision potential error.
/// Take care: truncation errors accumulate on calculation. More you do, more you should increase the epsilon.
/// You get really better performance when you can determine the contextual epsilon first.
/// </summary>
/// <param name="value1"></param>
/// <param name="value2"></param>
/// <param name="precalculatedContextualEpsilon"></param>
/// <returns></returns>
public static bool AboutEquals(this double value1, double value2, double precalculatedContextualEpsilon)
{
if (double.IsPositiveInfinity(value1))
return double.IsPositiveInfinity(value2);
if (double.IsNegativeInfinity(value1))
return double.IsNegativeInfinity(value2);
if (double.IsNaN(value1))
return double.IsNaN(value2);
return Math.Abs(value1 - value2) <= precalculatedContextualEpsilon;
}
// ******************************************************************
public static double GetContextualEpsilon(this double biggestPossibleContextualValue)
{
return biggestPossibleContextualValue * 1E-15;
}
// ******************************************************************
/// <summary>
/// Mathlab equivalent
/// </summary>
/// <param name="dividend"></param>
/// <param name="divisor"></param>
/// <returns></returns>
public static double Mod(this double dividend, double divisor)
{
return dividend - System.Math.Floor(dividend / divisor) * divisor;
}
// ******************************************************************
}
}
其他回答
另一个在c#(部分圆类)。 经过测试，工作就像一个魅力。
public class Circle : IEquatable<Circle>
{
// ******************************************************************
// The center of a circle
private Point _center;
// The radius of a circle
private double _radius;
// ******************************************************************
/// <summary>
/// Find all intersections (0, 1, 2) of the circle with a line defined by its 2 points.
/// Using: http://math.stackexchange.com/questions/228841/how-do-i-calculate-the-intersections-of-a-straight-line-and-a-circle
/// Note: p is the Center.X and q is Center.Y
/// </summary>
/// <param name="linePoint1"></param>
/// <param name="linePoint2"></param>
/// <returns></returns>
public List<Point> GetIntersections(Point linePoint1, Point linePoint2)
{
List<Point> intersections = new List<Point>();
double dx = linePoint2.X - linePoint1.X;
if (dx.AboutEquals(0)) // Straight vertical line
{
if (linePoint1.X.AboutEquals(Center.X - Radius) || linePoint1.X.AboutEquals(Center.X + Radius))
{
Point pt = new Point(linePoint1.X, Center.Y);
intersections.Add(pt);
}
else if (linePoint1.X > Center.X - Radius && linePoint1.X < Center.X + Radius)
{
double x = linePoint1.X - Center.X;
Point pt = new Point(linePoint1.X, Center.Y + Math.Sqrt(Radius * Radius - (x * x)));
intersections.Add(pt);
pt = new Point(linePoint1.X, Center.Y - Math.Sqrt(Radius * Radius - (x * x)));
intersections.Add(pt);
}
return intersections;
}
// Line function (y = mx + b)
double dy = linePoint2.Y - linePoint1.Y;
double m = dy / dx;
double b = linePoint1.Y - m * linePoint1.X;
double A = m * m + 1;
double B = 2 * (m * b - m * _center.Y - Center.X);
double C = Center.X * Center.X + Center.Y * Center.Y - Radius * Radius - 2 * b * Center.Y + b * b;
double discriminant = B * B - 4 * A * C;
if (discriminant < 0)
{
return intersections; // there is no intersections
}
if (discriminant.AboutEquals(0)) // Tangeante (touch on 1 point only)
{
double x = -B / (2 * A);
double y = m * x + b;
intersections.Add(new Point(x, y));
}
else // Secant (touch on 2 points)
{
double x = (-B + Math.Sqrt(discriminant)) / (2 * A);
double y = m * x + b;
intersections.Add(new Point(x, y));
x = (-B - Math.Sqrt(discriminant)) / (2 * A);
y = m * x + b;
intersections.Add(new Point(x, y));
}
return intersections;
}
// ******************************************************************
// Get the center
[XmlElement("Center")]
public Point Center
{
get { return _center; }
set
{
_center = value;
}
}
// ******************************************************************
// Get the radius
[XmlElement]
public double Radius
{
get { return _radius; }
set { _radius = value; }
}
//// ******************************************************************
//[XmlArrayItemAttribute("DoublePoint")]
//public List<Point> Coordinates
//{
// get { return _coordinates; }
//}
// ******************************************************************
// Construct a circle without any specification
public Circle()
{
_center.X = 0;
_center.Y = 0;
_radius = 0;
}
// ******************************************************************
// Construct a circle without any specification
public Circle(double radius)
{
_center.X = 0;
_center.Y = 0;
_radius = radius;
}
// ******************************************************************
// Construct a circle with the specified circle
public Circle(Circle circle)
{
_center = circle._center;
_radius = circle._radius;
}
// ******************************************************************
// Construct a circle with the specified center and radius
public Circle(Point center, double radius)
{
_center = center;
_radius = radius;
}
// ******************************************************************
// Construct a circle based on one point
public Circle(Point center)
{
_center = center;
_radius = 0;
}
// ******************************************************************
// Construct a circle based on two points
public Circle(Point p1, Point p2)
{
Circle2Points(p1, p2);
}
要求:
using System;
namespace Mathematic
{
public static class DoubleExtension
{
// ******************************************************************
// Base on Hans Passant Answer on:
// http://stackoverflow.com/questions/2411392/double-epsilon-for-equality-greater-than-less-than-less-than-or-equal-to-gre
/// <summary>
/// Compare two double taking in account the double precision potential error.
/// Take care: truncation errors accumulate on calculation. More you do, more you should increase the epsilon.
public static bool AboutEquals(this double value1, double value2)
{
if (double.IsPositiveInfinity(value1))
return double.IsPositiveInfinity(value2);
if (double.IsNegativeInfinity(value1))
return double.IsNegativeInfinity(value2);
if (double.IsNaN(value1))
return double.IsNaN(value2);
double epsilon = Math.Max(Math.Abs(value1), Math.Abs(value2)) * 1E-15;
return Math.Abs(value1 - value2) <= epsilon;
}
// ******************************************************************
// Base on Hans Passant Answer on:
// http://stackoverflow.com/questions/2411392/double-epsilon-for-equality-greater-than-less-than-less-than-or-equal-to-gre
/// <summary>
/// Compare two double taking in account the double precision potential error.
/// Take care: truncation errors accumulate on calculation. More you do, more you should increase the epsilon.
/// You get really better performance when you can determine the contextual epsilon first.
/// </summary>
/// <param name="value1"></param>
/// <param name="value2"></param>
/// <param name="precalculatedContextualEpsilon"></param>
/// <returns></returns>
public static bool AboutEquals(this double value1, double value2, double precalculatedContextualEpsilon)
{
if (double.IsPositiveInfinity(value1))
return double.IsPositiveInfinity(value2);
if (double.IsNegativeInfinity(value1))
return double.IsNegativeInfinity(value2);
if (double.IsNaN(value1))
return double.IsNaN(value2);
return Math.Abs(value1 - value2) <= precalculatedContextualEpsilon;
}
// ******************************************************************
public static double GetContextualEpsilon(this double biggestPossibleContextualValue)
{
return biggestPossibleContextualValue * 1E-15;
}
// ******************************************************************
/// <summary>
/// Mathlab equivalent
/// </summary>
/// <param name="dividend"></param>
/// <param name="divisor"></param>
/// <returns></returns>
public static double Mod(this double dividend, double divisor)
{
return dividend - System.Math.Floor(dividend / divisor) * divisor;
}
// ******************************************************************
}
}
也许有另一种方法来解决这个问题，使用坐标系的旋转。
通常，如果一个线段是水平的或垂直的，这意味着平行于x轴或y轴，交点的求解很容易，因为我们已经知道交点的一个坐标，如果有的话。剩下的显然是用圆的方程找到另一个坐标。
受此启发，我们可以利用坐标系旋转，使一个轴的方向与线段的方向重合。
让我们以圆x^2+y^2=1和线段P1-P2为例，P1(-1.5,0.5)和P2(-0.5，-0.5)在x-y系统中。下面的方程提醒你旋转的原理，其中是逆时针方向的角度，x'-y'是旋转后的方程组:
x'=x*cos () + y*sin () y' = - x*sin () + y*cos ()
和反向
X = X ' * cos - y' * sin Y = x' * sin + Y ' * cos
考虑P1-P2方向(用-x表示为45°)，我们可以取=45°。将第二个旋转方程转化为x-y系统中的圆方程:x^2+y^2=1，经过简单的运算，我们得到x'-y'系统中的“相同”方程:x'^2+y'^2=1。
利用第一个旋转方程=> P1(-根号(2)/2，根号(2))，P2(-根号(2)/ 2,0)，线段端点变成x'-y'系统。
假设交点为p，在x'-y'中，Px = -根号2 /2。使用新的圆方程，我们得到Py = +根号(2)/2。将P转换成原始的x-y系统，最终得到P(-1,0)
为了实现这个数值，我们可以先看看线段的方向:水平，垂直或不垂直。如果它属于前两种情况，很简单。如果是最后一种情况，应用上述算法。
为了判断是否有交集，我们可以将解与端点坐标进行比较，看看它们之间是否有一个根。
我相信只要我们有了它的方程，这个方法也可以应用于其他曲线。唯一的缺点是，我们应该在x'-y'坐标系下解方程，这可能很难。
' VB.NET - Code
Function CheckLineSegmentCircleIntersection(x1 As Double, y1 As Double, x2 As Double, y2 As Double, xc As Double, yc As Double, r As Double) As Boolean
Static xd As Double = 0.0F
Static yd As Double = 0.0F
Static t As Double = 0.0F
Static d As Double = 0.0F
Static dx_2_1 As Double = 0.0F
Static dy_2_1 As Double = 0.0F
dx_2_1 = x2 - x1
dy_2_1 = y2 - y1
t = ((yc - y1) * dy_2_1 + (xc - x1) * dx_2_1) / (dy_2_1 * dy_2_1 + dx_2_1 * dx_2_1)
If 0 <= t And t <= 1 Then
xd = x1 + t * dx_2_1
yd = y1 + t * dy_2_1
d = Math.Sqrt((xd - xc) * (xd - xc) + (yd - yc) * (yd - yc))
Return d <= r
Else
d = Math.Sqrt((xc - x1) * (xc - x1) + (yc - y1) * (yc - y1))
If d <= r Then
Return True
Else
d = Math.Sqrt((xc - x2) * (xc - x2) + (yc - y2) * (yc - y2))
If d <= r Then
Return True
Else
Return False
End If
End If
End If
End Function
我会用这个算法来计算点(圆心)和线(线AB)之间的距离。这可以用来确定直线与圆的交点。
假设有点A B c, Ax和Ay是A点的x和y分量。B和c也是一样，标量R是圆半径。
该算法要求A B C是不同的点，且R不为0。
这是算法
// compute the euclidean distance between A and B
LAB = sqrt( (Bx-Ax)²+(By-Ay)² )
// compute the direction vector D from A to B
Dx = (Bx-Ax)/LAB
Dy = (By-Ay)/LAB
// the equation of the line AB is x = Dx*t + Ax, y = Dy*t + Ay with 0 <= t <= LAB.
// compute the distance between the points A and E, where
// E is the point of AB closest the circle center (Cx, Cy)
t = Dx*(Cx-Ax) + Dy*(Cy-Ay)
// compute the coordinates of the point E
Ex = t*Dx+Ax
Ey = t*Dy+Ay
// compute the euclidean distance between E and C
LEC = sqrt((Ex-Cx)²+(Ey-Cy)²)
// test if the line intersects the circle
if( LEC < R )
{
// compute distance from t to circle intersection point
dt = sqrt( R² - LEC²)
// compute first intersection point
Fx = (t-dt)*Dx + Ax
Fy = (t-dt)*Dy + Ay
// compute second intersection point
Gx = (t+dt)*Dx + Ax
Gy = (t+dt)*Dy + Ay
}
// else test if the line is tangent to circle
else if( LEC == R )
// tangent point to circle is E
else
// line doesn't touch circle
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
