圆线段碰撞检测算法？
我有一条从A到B的线和一条位于C的圆，半径为R.
用于检查线是否与圆相交的好算法是什么？它沿圆圈边缘的坐标发生了什么？
28 个答案:
答案 0 :(得分：188)
采取
- E 是光线的起点，
- L 是光线的终点，
- C 是您正在测试的球体的中心
- r 是该球体的半径
- <强>展开
x 2 - 2xh + h 2 + y 2 - 2yk + k 2 - r 2 < / sup> = 0 - <强>插头
x = e x + td x
y = e y + td y
（e x + td x ） 2 - 2（e x + td x ）h + h 2 + （e y + td y ） 2 - 2（e y + td y ）k + k 2 - r 2 = 0 - 爆炸
e x 2 + 2e x td x + t 2 d x 2 - 2e x h - 2td x h + h 2 + e y 2 + 2e y td y + t 2 d y 2 - 2e y k - 2td y k + k 2 - r 2 = 0 - 的组
t 2 （d x 2 + d y 2 ）+ 2t（e x d x + e y d y - d x h - d y k）+ e x 2 + e y 2 - 2e x h - 2e y k + h 2 + k 2 - r 2 = 0 - 的最后，
t 2 （_ d * _d）+ 2t（_e * _d - _d * _c）+ _e * _e - 2（_e * _c）+ _c * _c - r 2 = 0
*其中_d是向量d，*是点积。* - 然后，
t 2 （_ d * _d）+ 2t（_d *（_ e - _c））+（_ e - _c）*（_ e - _c） - r 2 = 0 < / LI> - 让_f = _e - _c
t 2 （_ d * _d）+ 2t（_d * _f）+ _f * _f - r 2 = 0
计算：
d = L - E（光线的方向矢量，从开始到结束）
f = E - C（从中心球到射线开始的矢量）
然后通过...发现交叉点
堵：
P = E + t * d
这是一个参数方程：
P x = E x + td x
P y = E y + td y
到
（x - h） 2 +（y - k） 2 = r 2
（h，k）=圆心。
注意：我们在这里将问题简化为2D，我们得到的解决方案也适用于3D
得到：
所以我们得到：
t 2 *（d DOT d）+ 2t *（f DOT d）+（f DOT f - r 2 ）= 0
所以求解二次方程：
float a = d.Dot( d ) ;
float b = 2*f.Dot( d ) ;
float c = f.Dot( f ) - r*r ;
float discriminant = b*b-4*a*c;
if( discriminant < 0 )
{
// no intersection
}
else
{
// ray didn't totally miss sphere,
// so there is a solution to
// the equation.
discriminant = sqrt( discriminant );
// either solution may be on or off the ray so need to test both
// t1 is always the smaller value, because BOTH discriminant and
// a are nonnegative.
float t1 = (-b - discriminant)/(2*a);
float t2 = (-b + discriminant)/(2*a);
// 3x HIT cases:
// -o-> --|--> | | --|->
// Impale(t1 hit,t2 hit), Poke(t1 hit,t2>1), ExitWound(t1<0, t2 hit),
// 3x MISS cases:
// -> o o -> | -> |
// FallShort (t1>1,t2>1), Past (t1<0,t2<0), CompletelyInside(t1<0, t2>1)
if( t1 >= 0 && t1 <= 1 )
{
// t1 is the intersection, and it's closer than t2
// (since t1 uses -b - discriminant)
// Impale, Poke
return true ;
}
// here t1 didn't intersect so we are either started
// inside the sphere or completely past it
if( t2 >= 0 && t2 <= 1 )
{
// ExitWound
return true ;
}
// no intn: FallShort, Past, CompletelyInside
return false ;
}
答案 1 :(得分：128)
似乎没有人考虑投影，我完全偏离这里吗？
将向量AC投影到AB。投影向量AD给出新点D
如果D和C之间的距离小于（或等于）R，我们就会有一个交集点。
像这样：
答案 2 :(得分：47)
我会用算法计算点（圆心）和直线（AB线）之间的距离。然后可以使用它来确定直线与圆的交点。
假设我们有点A，B，C。Ax和Ay是A点的x和y分量。 B和C相同。标量R是圆半径。
该算法要求A，B和C是不同的点，R不是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
答案 3 :(得分：18)
好的，我不会给你代码，但是因为你已经标记了这个algorithm，我认为这对你不重要。 首先，你必须得到一个垂直于线的矢量。
y = ax + c 中会有一个未知变量（ c未知）
要解决这个问题，请在线穿过圆心时计算它的值。
即，
将圆心的位置插入线方程并求解c
然后计算原始线与其法线的交点。
这将为您提供圆圈线上的最近点 计算此点与圆心之间的距离（使用矢量的大小） 如果这小于圆的半径 - 瞧，我们有一个交叉点！
答案 4 :(得分：9)
另一种方法使用三角形ABC区域公式。相交测试比投影方法更简单，更有效，但找到交点的坐标需要更多的工作。至少它会延迟到需要的程度。
计算三角形区域的公式为：area = bh / 2
其中b是基本长度，h是高度。我们选择AB段作为基础，因此h是从C，圆心到线的最短距离。
由于三角形区域也可以通过矢量点积计算，我们可以确定h。
// compute the triangle area times 2 (area = area2/2)
area2 = abs( (Bx-Ax)*(Cy-Ay) - (Cx-Ax)(By-Ay) )
// compute the AB segment length
LAB = sqrt( (Bx-Ax)² + (By-Ay)² )
// compute the triangle height
h = area2/LAB
// if the line intersects the circle
if( h < R )
{
...
}
更新1：
您可以使用here描述的快速平方根计算来优化代码，以获得1 / LAB的良好近似值。
计算交叉点并不困难。在这里
// compute the line AB direction vector components
Dx = (Bx-Ax)/LAB
Dy = (By-Ay)/LAB
// compute the distance from A toward B of closest point to C
t = Dx*(Cx-Ax) + Dy*(Cy-Ay)
// t should be equal to sqrt( (Cx-Ax)² + (Cy-Ay)² - h² )
// compute the intersection point distance from t
dt = sqrt( R² - h² )
// compute first intersection point coordinate
Ex = Ax + (t-dt)*Dx
Ey = Ay + (t-dt)*Dy
// compute second intersection point coordinate
Fx = Ax + (t+dt)*Dx
Fy = Ay + (t+dt)*Dy
如果h = R，那么AB线与圆相切，值dt = 0，E = F.点坐标是E和F的坐标。
如果您的应用程序中可能发生这种情况，则应检查A是否与B不同，并且段长度不为空。
答案 5 :(得分：7)
我写了一个小脚本，通过将圆圈的中心点投射到线上来测试交叉点。
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/
如果您需要检查与细分的碰撞，还需要考虑圆心开始和结束点的距离。
vector distVector = centerPoint - startPoint;
if(distVector.length() < circle.radius)
{
double distance = circle.radius - distVector.length();
vector moveVector = distVector.normalize() * distance;
circle.move(moveVector);
}
答案 6 :(得分：5)
我发现这个解决方案似乎比其他一些更容易理解。
服用：
p1 and p2 as the points for the line, and
c as the center point for the circle and r for the radius
我会用斜率截距形式求解线的方程。但是，我不想处理以c为点的困难方程式，所以我只是将坐标系移动到圆圈位于0,0
p3 = p1 - c
p4 = p2 - c
顺便说一下，每当我从彼此中减去点数时，我减去x然后减去y，然后将它们放到一个新点，以防万一有人没有不知道。
无论如何，我现在用p3和p4来解决这一行的等式：
m = (p4_y - p3_y) / (p4_x - p3) (the underscore is an attempt at subscript)
y = mx + b
y - mx = b (just put in a point for x and y, and insert the m we found)
确定。现在我需要将这些方程设置为相等。首先，我需要解决x
x^2 + y^2 = r^2
y^2 = r^2 - x^2
y = sqrt(r^2 - x^2)
然后我将它们设置为相等：
mx + b = sqrt(r^2 - x^2)
求解二次方程（0 = ax^2 + bx + c）：
(mx + b)^2 = r^2 - x^2
(mx)^2 + 2mbx + b^2 = r^2 - x^2
0 = m^2 * x^2 + x^2 + 2mbx + b^2 - r^2
0 = (m^2 + 1) * x^2 + 2mbx + b^2 - r^2
现在我有a，b和c。
a = m^2 + 1
b = 2mb
c = b^2 - r^2
所以我把它放到二次公式中：
(-b ± sqrt(b^2 - 4ac)) / 2a
然后用值替换然后尽可能地简化：
(-2mb ± sqrt(b^2 - 4ac)) / 2a
(-2mb ± sqrt((-2mb)^2 - 4(m^2 + 1)(b^2 - r^2))) / 2(m^2 + 1)
(-2mb ± sqrt(4m^2 * b^2 - 4(m^2 * b^2 - m^2 * r^2 + b^2 - r^2))) / 2m^2 + 2
(-2mb ± sqrt(4 * (m^2 * b^2 - (m^2 * b^2 - m^2 * r^2 + b^2 - r^2))))/ 2m^2 + 2
(-2mb ± sqrt(4 * (m^2 * b^2 - m^2 * b^2 + m^2 * r^2 - b^2 + r^2)))/ 2m^2 + 2
(-2mb ± sqrt(4 * (m^2 * r^2 - b^2 + r^2)))/ 2m^2 + 2
(-2mb ± sqrt(4) * sqrt(m^2 * r^2 - b^2 + r^2))/ 2m^2 + 2
(-2mb ± 2 * sqrt(m^2 * r^2 - b^2 + r^2))/ 2m^2 + 2
(-2mb ± 2 * sqrt(m^2 * r^2 + r^2 - b^2))/ 2m^2 + 2
(-2mb ± 2 * sqrt(r^2 * (m^2 + 1) - b^2))/ 2m^2 + 2
这几乎可以简化。最后，用±：
分离出方程式(-2mb + 2 * sqrt(r^2 * (m^2 + 1) - b^2))/ 2m^2 + 2 or
(-2mb - 2 * sqrt(r^2 * (m^2 + 1) - b^2))/ 2m^2 + 2
然后只需将这两个方程的结果插入x中的mx + b即可。为清楚起见，我写了一些JavaScript代码来说明如何使用它：
function interceptOnCircle(p1,p2,c,r){
//p1 is the first line point
//p2 is the second line point
//c is the circle's center
//r is the circle's radius
var p3 = {x:p1.x - c.x, y:p1.y - c.y} //shifted line points
var p4 = {x:p2.x - c.x, y:p2.y - c.y}
var m = (p4.y - p3.y) / (p4.x - p3.x); //slope of the line
var b = p3.y - m * p3.x; //y-intercept of line
var underRadical = Math.pow((Math.pow(r,2)*(Math.pow(m,2)+1)),2)-Math.pow(b,2)); //the value under the square root sign
if (underRadical < 0){
//line completely missed
return false;
} else {
var t1 = (-2*m*b+2*Math.sqrt(underRadical))/(2 * Math.pow(m,2) + 2); //one of the intercept x's
var t2 = (-2*m*b-2*Math.sqrt(underRadical))/(2 * Math.pow(m,2) + 2); //other intercept's x
var i1 = {x:t1,y:m*t1+b} //intercept point 1
var i2 = {x:t2,y:m*t2+b} //intercept point 2
return [i1,i2];
}
}
我希望这有帮助！
P.S。如果有人发现任何错误或有任何建议，请发表评论。我很新，欢迎所有的帮助/建议。
答案 7 :(得分：4)
很奇怪我可以回答但不是评论...... 我喜欢Multitaskpro的方法来改变一切，使圆圈的中心落在原点上。不幸的是，他的代码存在两个问题。首先在平方根部分，您需要移除双倍功率。所以不是：
var underRadical = Math.pow((Math.pow(r,2)*(Math.pow(m,2)+1)),2)-Math.pow(b,2));
但：
var underRadical = Math.pow(r,2)*(Math.pow(m,2)+1)) - Math.pow(b,2);
在最终坐标中，他忘记将解决方案移回。所以不是：
var i1 = {x:t1,y:m*t1+b}
但：
var i1 = {x:t1+c.x, y:m*t1+b+c.y};
然后整个功能变为：
function interceptOnCircle(p1, p2, c, r) {
//p1 is the first line point
//p2 is the second line point
//c is the circle's center
//r is the circle's radius
var p3 = {x:p1.x - c.x, y:p1.y - c.y}; //shifted line points
var p4 = {x:p2.x - c.x, y:p2.y - c.y};
var m = (p4.y - p3.y) / (p4.x - p3.x); //slope of the line
var b = p3.y - m * p3.x; //y-intercept of line
var underRadical = Math.pow(r,2)*Math.pow(m,2) + Math.pow(r,2) - Math.pow(b,2); //the value under the square root sign
if (underRadical < 0) {
//line completely missed
return false;
} else {
var t1 = (-m*b + Math.sqrt(underRadical))/(Math.pow(m,2) + 1); //one of the intercept x's
var t2 = (-m*b - Math.sqrt(underRadical))/(Math.pow(m,2) + 1); //other intercept's x
var i1 = {x:t1+c.x, y:m*t1+b+c.y}; //intercept point 1
var i2 = {x:t2+c.x, y:m*t2+b+c.y}; //intercept point 2
return [i1, i2];
}
}
答案 8 :(得分：4)
通过将矢量AC投影到矢量AB上，可以在无限直线上找到最接近圆心的点。计算该点与圆心之间的距离。如果R大于R，则没有交叉点。如果距离等于R，则线是圆的切线，最接近圆心的点实际上是交点。如果距离小于R，则有2个交叉点。它们与距离圆心最近的点位于相同的距离。使用毕达哥拉斯定理可以很容易地计算出这个距离。这是伪代码的算法：
{
dX = bX - aX;
dY = bY - aY;
if ((dX == 0) && (dY == 0))
{
// A and B are the same points, no way to calculate intersection
return;
}
dl = (dX * dX + dY * dY);
t = ((cX - aX) * dX + (cY - aY) * dY) / dl;
// point on a line nearest to circle center
nearestX = aX + t * dX;
nearestY = aY + t * dY;
dist = point_dist(nearestX, nearestY, cX, cY);
if (dist == R)
{
// line segment touches circle; one intersection point
iX = nearestX;
iY = nearestY;
if (t < 0 || t > 1)
{
// intersection point is not actually within line segment
}
}
else if (dist < R)
{
// two possible intersection points
dt = sqrt(R * R - dist * dist) / sqrt(dl);
// intersection point nearest to A
t1 = t - dt;
i1X = aX + t1 * dX;
i1Y = aY + t1 * dY;
if (t1 < 0 || t1 > 1)
{
// intersection point is not actually within line segment
}
// intersection point farthest from A
t2 = t + dt;
i2X = aX + t2 * dX;
i2Y = aY + t2 * dY;
if (t2 < 0 || t2 > 1)
{
// intersection point is not actually within line segment
}
}
else
{
// no intersection
}
}
编辑：添加代码以检查找到的交叉点是否实际位于线段内。
答案 9 :(得分：3)
你需要一些数学：
假设A =（Xa，Ya），B =（Xb，Yb）和C =（Xc，Yc）。从A到B的线上的任何点都有坐标（alpha * Xa +（1-alpha） Xb，alpha Ya +（1-alpha）* Yb）= P
如果点P的距离为R到C，则它必须在圆上。你想要的是解决问题
distance(P, C) = R
即
(alpha*Xa + (1-alpha)*Xb)^2 + (alpha*Ya + (1-alpha)*Yb)^2 = R^2
alpha^2*Xa^2 + alpha^2*Xb^2 - 2*alpha*Xb^2 + Xb^2 + alpha^2*Ya^2 + alpha^2*Yb^2 - 2*alpha*Yb^2 + Yb^2=R^2
(Xa^2 + Xb^2 + Ya^2 + Yb^2)*alpha^2 - 2*(Xb^2 + Yb^2)*alpha + (Xb^2 + Yb^2 - R^2) = 0
如果你将ABC公式应用于此公式以解决alpha问题，并使用alpha解决方案计算P的坐标，则得到交点（如果存在）。
答案 10 :(得分：3)
如果你发现球体中心之间的距离（因为它是3D我假设你的意思是球体而不是圆形）和线条，那么检查那个距离是否小于将要做到这一点的半径。
碰撞点显然是直线和球体之间的最近点（当你计算球体和直线之间的距离时会计算出来）
点与线之间的距离：
http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html
答案 11 :(得分：3)
这是Javascript中的一个实现。我的方法是首先将线段转换为无限线，然后找到交叉点。从那里我检查找到的点是否在线段上。代码有详细记录，您应该能够遵循。
您可以在此live demo上试用此处的代码。 代码来自我的algorithms repo。
// Small epsilon value
var EPS = 0.0000001;
// point (x, y)
function Point(x, y) {
this.x = x;
this.y = y;
}
// Circle with center at (x,y) and radius r
function Circle(x, y, r) {
this.x = x;
this.y = y;
this.r = r;
}
// A line segment (x1, y1), (x2, y2)
function LineSegment(x1, y1, x2, y2) {
var d = Math.sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) );
if (d < EPS) throw 'A point is not a line segment';
this.x1 = x1; this.y1 = y1;
this.x2 = x2; this.y2 = y2;
}
// An infinite line defined as: ax + by = c
function Line(a, b, c) {
this.a = a; this.b = b; this.c = c;
// Normalize line for good measure
if (Math.abs(b) < EPS) {
c /= a; a = 1; b = 0;
} else {
a = (Math.abs(a) < EPS) ? 0 : a / b;
c /= b; b = 1;
}
}
// Given a line in standard form: ax + by = c and a circle with
// a center at (x,y) with radius r this method finds the intersection
// of the line and the circle (if any).
function circleLineIntersection(circle, line) {
var a = line.a, b = line.b, c = line.c;
var x = circle.x, y = circle.y, r = circle.r;
// Solve for the variable x with the formulas: ax + by = c (equation of line)
// and (x-X)^2 + (y-Y)^2 = r^2 (equation of circle where X,Y are known) and expand to obtain quadratic:
// (a^2 + b^2)x^2 + (2abY - 2ac + - 2b^2X)x + (b^2X^2 + b^2Y^2 - 2bcY + c^2 - b^2r^2) = 0
// Then use quadratic formula X = (-b +- sqrt(a^2 - 4ac))/2a to find the
// roots of the equation (if they exist) and this will tell us the intersection points
// In general a quadratic is written as: Ax^2 + Bx + C = 0
// (a^2 + b^2)x^2 + (2abY - 2ac + - 2b^2X)x + (b^2X^2 + b^2Y^2 - 2bcY + c^2 - b^2r^2) = 0
var A = a*a + b*b;
var B = 2*a*b*y - 2*a*c - 2*b*b*x;
var C = b*b*x*x + b*b*y*y - 2*b*c*y + c*c - b*b*r*r;
// Use quadratic formula x = (-b +- sqrt(a^2 - 4ac))/2a to find the
// roots of the equation (if they exist).
var D = B*B - 4*A*C;
var x1,y1,x2,y2;
// Handle vertical line case with b = 0
if (Math.abs(b) < EPS) {
// Line equation is ax + by = c, but b = 0, so x = c/a
x1 = c/a;
// No intersection
if (Math.abs(x-x1) > r) return [];
// Vertical line is tangent to circle
if (Math.abs((x1-r)-x) < EPS || Math.abs((x1+r)-x) < EPS)
return [new Point(x1, y)];
var dx = Math.abs(x1 - x);
var dy = Math.sqrt(r*r-dx*dx);
// Vertical line cuts through circle
return [
new Point(x1,y+dy),
new Point(x1,y-dy)
];
// Line is tangent to circle
} else if (Math.abs(D) < EPS) {
x1 = -B/(2*A);
y1 = (c - a*x1)/b;
return [new Point(x1,y1)];
// No intersection
} else if (D < 0) {
return [];
} else {
D = Math.sqrt(D);
x1 = (-B+D)/(2*A);
y1 = (c - a*x1)/b;
x2 = (-B-D)/(2*A);
y2 = (c - a*x2)/b;
return [
new Point(x1, y1),
new Point(x2, y2)
];
}
}
// Converts a line segment to a line in general form
function segmentToGeneralForm(x1,y1,x2,y2) {
var a = y1 - y2;
var b = x2 - x1;
var c = x2*y1 - x1*y2;
return new Line(a,b,c);
}
// Checks if a point 'pt' is inside the rect defined by (x1,y1), (x2,y2)
function pointInRectangle(pt,x1,y1,x2,y2) {
var x = Math.min(x1,x2), X = Math.max(x1,x2);
var y = Math.min(y1,y2), Y = Math.max(y1,y2);
return x - EPS <= pt.x && pt.x <= X + EPS &&
y - EPS <= pt.y && pt.y <= Y + EPS;
}
// Finds the intersection(s) of a line segment and a circle
function lineSegmentCircleIntersection(segment, circle) {
var x1 = segment.x1, y1 = segment.y1, x2 = segment.x2, y2 = segment.y2;
var line = segmentToGeneralForm(x1,y1,x2,y2);
var pts = circleLineIntersection(circle, line);
// No intersection
if (pts.length === 0) return [];
var pt1 = pts[0];
var includePt1 = pointInRectangle(pt1,x1,y1,x2,y2);
// Check for unique intersection
if (pts.length === 1) {
if (includePt1) return [pt1];
return [];
}
var pt2 = pts[1];
var includePt2 = pointInRectangle(pt2,x1,y1,x2,y2);
// Check for remaining intersections
if (includePt1 && includePt2) return [pt1, pt2];
if (includePt1) return [pt1];
if (includePt2) return [pt2];
return [];
}
答案 12 :(得分：3)
在此后圆线中，将通过检查圆心和线段（Ipoint）之间的距离来检查线圈，该线段表示从圆心到线段的法线N（图像2）之间的交点。
（https://i.stack.imgur.com/3o6do.png）
在图像1上显示一个圆和一条线，矢量A点到线起点，矢量B指向线端点，矢量C指向圆心。现在我们必须找到矢量E（从行起点到圆心）和矢量D（从行起点到行终点），这个计算显示在图像1上。
（https://i.stack.imgur.com/7098a.png）
在图像2中，我们可以看到矢量E通过矢量E和单位矢量D的“点积”投射在矢量D上，点积的结果是标量Xp，表示线起点和交点之间的距离（ Ipoint）向量N和向量D. 通过乘以单位矢量D和标量Xp找到下一个矢量X.
现在我们需要找到向量Z（向量到Ipoint），它很容易添加向量A（起始点在线）和向量X的简单向量。接下来我们需要处理特殊情况我们必须检查是否在线Ipoint如果它不是我们必须找到它是它的左侧还是右侧，我们将使用最接近的向量来确定哪个点最接近圆。
（https://i.stack.imgur.com/p9WIr.png）
当投影Xp为负时，Ipoint为线段左侧，矢量最接近等于线起点的矢量，当投影Xp大于矢量D的幅度时，则Ipoint为线段右侧，则最近矢量等于矢量在任何其他情况下，最终向量的行结束点等于向量Z.
现在当我们有最接近的矢量时，我们需要找到从圆心到Ipoint（dist矢量）的矢量，其简单我们只需要从中心矢量中减去最接近的矢量。接下来只检查矢量dist幅度是否小于圆半径，如果它是碰撞，如果它没有碰撞那么。
（https://i.stack.imgur.com/QJ63q.png）
为了结束，我们可以返回一些用于解决碰撞的值，最简单的方法是返回碰撞的重叠（从矢量dist幅度减去半径）和返回碰撞轴，它的矢量D.如果需要，交点也是矢量Z.
答案 13 :(得分：2)
只是这个帖子的补充...... 下面是pahlevan发布的代码版本，但对于C＃/ XNA并整理了一点：
/// <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));
}
答案 14 :(得分：2)
' 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
答案 15 :(得分：2)
我已根据chmike
+ (NSArray *)intersectionPointsOfCircleWithCenter:(CGPoint)center withRadius:(float)radius toLinePoint1:(CGPoint)p1 andLinePoint2:(CGPoint)p2
{
NSMutableArray *intersectionPoints = [NSMutableArray array];
float Ax = p1.x;
float Ay = p1.y;
float Bx = p2.x;
float By = p2.y;
float Cx = center.x;
float Cy = center.y;
float R = radius;
// compute the euclidean distance between A and B
float LAB = sqrt( pow(Bx-Ax, 2)+pow(By-Ay, 2) );
// compute the direction vector D from A to B
float Dx = (Bx-Ax)/LAB;
float Dy = (By-Ay)/LAB;
// Now the line equation is x = Dx*t + Ax, y = Dy*t + Ay with 0 <= t <= 1.
// compute the value t of the closest point to the circle center (Cx, Cy)
float t = Dx*(Cx-Ax) + Dy*(Cy-Ay);
// This is the projection of C on the line from A to B.
// compute the coordinates of the point E on line and closest to C
float Ex = t*Dx+Ax;
float Ey = t*Dy+Ay;
// compute the euclidean distance from E to C
float LEC = sqrt( pow(Ex-Cx, 2)+ pow(Ey-Cy, 2) );
// test if the line intersects the circle
if( LEC < R )
{
// compute distance from t to circle intersection point
float dt = sqrt( pow(R, 2) - pow(LEC,2) );
// compute first intersection point
float Fx = (t-dt)*Dx + Ax;
float Fy = (t-dt)*Dy + Ay;
// compute second intersection point
float Gx = (t+dt)*Dx + Ax;
float Gy = (t+dt)*Dy + Ay;
[intersectionPoints addObject:[NSValue valueWithCGPoint:CGPointMake(Fx, Fy)]];
[intersectionPoints addObject:[NSValue valueWithCGPoint:CGPointMake(Gx, Gy)]];
}
// else test if the line is tangent to circle
else if( LEC == R ) {
// tangent point to circle is E
[intersectionPoints addObject:[NSValue valueWithCGPoint:CGPointMake(Ex, Ey)]];
}
else {
// line doesn't touch circle
}
return intersectionPoints;
}
答案 16 :(得分：2)
如果线的坐标是A.x，A.y和B.x，B.y和圆心是C.x，C.y那么线公式是：
x = A.x * t + B.x *（1-t）
y = A.y * t + B.y *（1 - t）
其中0 <= t <= 1
，圆圈是
（C.x-x）^ 2 +（C.y-y）^ 2 = R ^ 2
如果将线的x和y公式替换为圆公式，则得到t的二阶方程，其解是交点（如果有的话）。如果你得到一个小于0或大于1的t，那么它不是一个解决方案，但它表明这条线是“指向”圆的方向。
答案 17 :(得分：1)
此Java函数返回DVec2对象。圆的中心，圆的半径和直线需要DVec2。
public static DVec2 CircLine(DVec2 C, double r, Line line)
{
DVec2 A = line.p1;
DVec2 B = line.p2;
DVec2 P;
DVec2 AC = new DVec2( C );
AC.sub(A);
DVec2 AB = new DVec2( B );
AB.sub(A);
double ab2 = AB.dot(AB);
double acab = AC.dot(AB);
double t = acab / ab2;
if (t < 0.0)
t = 0.0;
else if (t > 1.0)
t = 1.0;
//P = A + t * AB;
P = new DVec2( AB );
P.mul( t );
P.add( A );
DVec2 H = new DVec2( P );
H.sub( C );
double h2 = H.dot(H);
double r2 = r * r;
if(h2 > r2)
return null;
else
return P;
}
答案 18 :(得分：1)
也许还有另一种方法可以使用坐标系的旋转来解决此问题。
通常，如果一个线段是水平或垂直的，这意味着平行于x或y轴，则求解交点非常容易，因为我们已经知道交点的一个坐标（如果有）。其余的显然是使用圆的方程式找到另一个坐标。
受此想法启发，我们可以应用坐标系旋转以使一个轴的方向与线段的方向一致。
让我们以在x-y系统中圆为x^2+y^2=1且段为P1-P2且具有P1（-1.5,0.5）和P2（-0.5，-0.5）的示例为例。以下方程式提醒您旋转原理，其中theta是逆时针的角度，x'-y'是旋转后的系统：
x'= x * cos（theta）+ y * sin（theta）
y'=-x * sin（θ）+ y * cos（θ）
反之
x = x'* cos（theta）-y'* sin（theta）
y = x'* sin（θ）+ y'* cos（theta）
考虑线段P1-P2的方向（以-x表示45°），我们可以采用theta=45°。在x-y系统x^2+y^2=1中将第二个旋转方程转化为圆方程，简单操作后，在x'-y'系统x'^2+y'^2=1中得到“相同”方程。
使用第一个旋转方程，分段端点成为x'-y'系统=> P1（-sqrt（2）/ 2，sqrt（2）），P2（-sqrt（2）/ 2，0）。
假设交点为P。我们在x'-y'中具有Px = -sqrt（2）/ 2。使用新的圆方程，我们得到Py = + sqrt（2）/ 2。将P转换为原始的x-y系统，我们最终得到P（-1,0）。
要以数字方式实现此目的，我们首先可以查看片段的方向：水平，垂直还是不垂直。如果它属于前两种情况，就像我说的那样简单。如果是最后一种情况，请应用上述算法。
要判断是否有交点，可以将解与端点坐标进行比较，以查看它们之间是否有一个根。
我相信只要有方程式，该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'系统中为另一个坐标求解方程，这可能很困难。
答案 19 :(得分：1)
我知道距此线程打开已经有一段时间了。从chmike给出的答案到Aqib Mumtaz的改进。他们给出了很好的答案，但仅适用于无限行，如Aqib所述。因此，我添加了一些比较，以了解线段是否接触到圆，我用Python编写了它。
def LineIntersectCircle(c, r, p1, p2):
#p1 is the first line point
#p2 is the second line point
#c is the circle's center
#r is the circle's radius
p3 = [p1[0]-c[0], p1[1]-c[1]]
p4 = [p2[0]-c[0], p2[1]-c[1]]
m = (p4[1] - p3[1]) / (p4[0] - p3[0])
b = p3[1] - m * p3[0]
underRadical = math.pow(r,2)*math.pow(m,2) + math.pow(r,2) - math.pow(b,2)
if (underRadical < 0):
print("NOT")
else:
t1 = (-2*m*b+2*math.sqrt(underRadical)) / (2 * math.pow(m,2) + 2)
t2 = (-2*m*b-2*math.sqrt(underRadical)) / (2 * math.pow(m,2) + 2)
i1 = [t1+c[0], m * t1 + b + c[1]]
i2 = [t2+c[0], m * t2 + b + c[1]]
if p1[0] > p2[0]: #Si el punto 1 es mayor al 2 en X
if (i1[0] < p1[0]) and (i1[0] > p2[0]): #Si el punto iX esta entre 2 y 1 en X
if p1[1] > p2[1]: #Si el punto 1 es mayor al 2 en Y
if (i1[1] < p1[1]) and (i1[1] > p2[1]): #Si el punto iy esta entre 2 y 1
print("Intersection")
if p1[1] < p2[1]: #Si el punto 2 es mayo al 2 en Y
if (i1[1] > p1[1]) and (i1[1] < p2[1]): #Si el punto iy esta entre 1 y 2
print("Intersection")
if p1[0] < p2[0]: #Si el punto 2 es mayor al 1 en X
if (i1[0] > p1[0]) and (i1[0] < p2[0]): #Si el punto iX esta entre 1 y 2 en X
if p1[1] > p2[1]: #Si el punto 1 es mayor al 2 en Y
if (i1[1] < p1[1]) and (i1[1] > p2[1]): #Si el punto iy esta entre 2 y 1
print("Intersection")
if p1[1] < p2[1]: #Si el punto 2 es mayo al 2 en Y
if (i1[1] > p1[1]) and (i1[1] < p2[1]): #Si el punto iy esta entre 1 y 2
print("Intersection")
if p1[0] > p2[0]: #Si el punto 1 es mayor al 2 en X
if (i2[0] < p1[0]) and (i2[0] > p2[0]): #Si el punto iX esta entre 2 y 1 en X
if p1[1] > p2[1]: #Si el punto 1 es mayor al 2 en Y
if (i2[1] < p1[1]) and (i2[1] > p2[1]): #Si el punto iy esta entre 2 y 1
print("Intersection")
if p1[1] < p2[1]: #Si el punto 2 es mayo al 2 en Y
if (i2[1] > p1[1]) and (i2[1] < p2[1]): #Si el punto iy esta entre 1 y 2
print("Intersection")
if p1[0] < p2[0]: #Si el punto 2 es mayor al 1 en X
if (i2[0] > p1[0]) and (i2[0] < p2[0]): #Si el punto iX esta entre 1 y 2 en X
if p1[1] > p2[1]: #Si el punto 1 es mayor al 2 en Y
if (i2[1] < p1[1]) and (i2[1] > p2[1]): #Si el punto iy esta entre 2 y 1
print("Intersection")
if p1[1] < p2[1]: #Si el punto 2 es mayo al 2 en Y
if (i2[1] > p1[1]) and (i2[1] < p2[1]): #Si el punto iy esta entre 1 y 2
print("Intersection")
答案 20 :(得分：1)
以下是@Mizipzor建议的想法（使用投影），这是我在TypeScript中的解决方案：
/**
* Determines whether a line segment defined by a start and end point intersects with a sphere defined by a center point and a radius
* @param a the start point of the line segment
* @param b the end point of the line segment
* @param c the center point of the sphere
* @param r the radius of the sphere
*/
export function lineSphereIntersects(
a: IPoint,
b: IPoint,
c: IPoint,
r: number
): boolean {
// find the three sides of the triangle formed by the three points
const ab: number = distance(a, b);
const ac: number = distance(a, c);
const bc: number = distance(b, c);
// check to see if either ends of the line segment are inside of the sphere
if (ac < r || bc < r) {
return true;
}
// find the angle between the line segment and the center of the sphere
const numerator: number = Math.pow(ac, 2) + Math.pow(ab, 2) - Math.pow(bc, 2);
const denominator: number = 2 * ac * ab;
const cab: number = Math.acos(numerator / denominator);
// find the distance from the center of the sphere and the line segment
const cd: number = Math.sin(cab) * ac;
// if the radius is at least as long as the distance between the center and the line
if (r >= cd) {
// find the distance between the line start and the point on the line closest to
// the center of the sphere
const ad: number = Math.cos(cab) * ac;
// intersection occurs when the point on the line closest to the sphere center is
// no further away than the end of the line
return ad <= ab;
}
return false;
}
export function distance(a: IPoint, b: IPoint): number {
return Math.sqrt(
Math.pow(b.z - a.z, 2) + Math.pow(b.y - a.y, 2) + Math.pow(b.x - a.x, 2)
);
}
export interface IPoint {
x: number;
y: number;
z: number;
}
答案 21 :(得分：1)
圈子确实是一个坏人:)所以一个好方法是避免真正的圈子，如果可以的话。如果您正在对游戏进行碰撞检查，您可以进行一些简化，只需要3个点产品，并进行一些比较。
我称之为“胖点”或“薄圈”。它是一种在平行于一个段的方向上具有零半径的椭圆。但在垂直于一段的方向上的全半径
首先，我会考虑重命名和切换坐标系以避免过多的数据：
s0s1 = B-A;
s0qp = C-A;
rSqr = r*r;
其次，hvec2f中的索引h意味着比矢量必须支持水平操作，如dot（）/ det（）。这意味着它的组件将被放置在一个单独的xmm寄存器中，以避免混乱/充电/ hsub'ing。在这里，我们将为2D游戏提供最简单的最简单的碰撞检测版本：
bool fat_point_collides_segment(const hvec2f& s0qp, const hvec2f& s0s1, const float& rSqr) {
auto a = dot(s0s1, s0s1);
//if( a != 0 ) // if you haven't zero-length segments omit this, as it would save you 1 _mm_comineq_ss() instruction and 1 memory fetch
{
auto b = dot(s0s1, s0qp);
auto t = b / a; // length of projection of s0qp onto s0s1
//std::cout << "t = " << t << "\n";
if ((t >= 0) && (t <= 1)) //
{
auto c = dot(s0qp, s0qp);
auto r2 = c - a * t * t;
return (r2 <= rSqr); // true if collides
}
}
return false;
}
我怀疑你是否可以进一步优化它。我正在将它用于神经网络驱动的赛车碰撞检测，以处理数亿次迭代步骤。
答案 22 :(得分：1)
这是JavaScript中的好解决方案（包含所有必需的数学和实时插图） https://bl.ocks.org/milkbread/11000965
虽然该解决方案中的task_hello = BashOperator(task_id='print_hello',
bash_command="mkdir ~/airflow/test_airflow", dag=dag)
函数需要修改：
＆＃13;
＆＃13;
＆＃13;
＆＃13;
is_on
答案 23 :(得分：1)
c＃中的另一个（部分Circle类）。 经过测试，就像魅力一样。
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);
}
必需：
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;
}
// ******************************************************************
}
}
答案 24 :(得分：0)
这是用golang编写的解决方案。该方法类似于此处发布的其他一些答案，但不完全相同。它易于实施，并已经过测试。以下是步骤：
- 翻译坐标，使圆圈位于原点。
- 将线段表示为x和y坐标的t的参数化函数。如果t为0，则函数的值是段的一个端点，如果t为1，则函数的值是另一个端点。
- 如果可能的话，解决由约束t的值产生的二次方程，该值产生x，y坐标，距离原点的距离等于圆的半径。
- 丢弃t为＆lt;的解决方案。 0或＆gt; 1（对于开放段，＆lt; = 0或＆gt; = 1）。这些点不包含在细分中。
- 翻译回原始坐标。
这里导出了二次方的A，B和C的值，其中（n-et）和（m-dt）分别是线的x和y坐标的方程。 r是圆的半径。
(n-et)(n-et) + (m-dt)(m-dt) = rr
nn - 2etn + etet + mm - 2mdt + dtdt = rr
(ee+dd)tt - 2(en + dm)t + nn + mm - rr = 0
因此A = ee + dd，B = - 2（en + dm），C = nn + mm - rr。
这是函数的golang代码：
package geom
import (
"math"
)
// SegmentCircleIntersection return points of intersection between a circle and
// a line segment. The Boolean intersects returns true if one or
// more solutions exist. If only one solution exists,
// x1 == x2 and y1 == y2.
// s1x and s1y are coordinates for one end point of the segment, and
// s2x and s2y are coordinates for the other end of the segment.
// cx and cy are the coordinates of the center of the circle and
// r is the radius of the circle.
func SegmentCircleIntersection(s1x, s1y, s2x, s2y, cx, cy, r float64) (x1, y1, x2, y2 float64, intersects bool) {
// (n-et) and (m-dt) are expressions for the x and y coordinates
// of a parameterized line in coordinates whose origin is the
// center of the circle.
// When t = 0, (n-et) == s1x - cx and (m-dt) == s1y - cy
// When t = 1, (n-et) == s2x - cx and (m-dt) == s2y - cy.
n := s2x - cx
m := s2y - cy
e := s2x - s1x
d := s2y - s1y
// lineFunc checks if the t parameter is in the segment and if so
// calculates the line point in the unshifted coordinates (adds back
// cx and cy.
lineFunc := func(t float64) (x, y float64, inBounds bool) {
inBounds = t >= 0 && t <= 1 // Check bounds on closed segment
// To check bounds for an open segment use t > 0 && t < 1
if inBounds { // Calc coords for point in segment
x = n - e*t + cx
y = m - d*t + cy
}
return
}
// Since we want the points on the line distance r from the origin,
// (n-et)(n-et) + (m-dt)(m-dt) = rr.
// Expanding and collecting terms yeilds the following quadratic equation:
A, B, C := e*e+d*d, -2*(e*n+m*d), n*n+m*m-r*r
D := B*B - 4*A*C // discriminant of quadratic
if D < 0 {
return // No solution
}
D = math.Sqrt(D)
var p1In, p2In bool
x1, y1, p1In = lineFunc((-B + D) / (2 * A)) // First root
if D == 0.0 {
intersects = p1In
x2, y2 = x1, y1
return // Only possible solution, quadratic has one root.
}
x2, y2, p2In = lineFunc((-B - D) / (2 * A)) // Second root
intersects = p1In || p2In
if p1In == false { // Only x2, y2 may be valid solutions
x1, y1 = x2, y2
} else if p2In == false { // Only x1, y1 are valid solutions
x2, y2 = x1, y1
}
return
}
我使用此功能对其进行了测试，确认解决方案点位于线段和圆上。它会创建一个测试段并围绕给定的圆圈进行扫描：
package geom_test
import (
"testing"
. "**put your package path here**"
)
func CheckEpsilon(t *testing.T, v, epsilon float64, message string) {
if v > epsilon || v < -epsilon {
t.Error(message, v, epsilon)
t.FailNow()
}
}
func TestSegmentCircleIntersection(t *testing.T) {
epsilon := 1e-10 // Something smallish
x1, y1 := 5.0, 2.0 // segment end point 1
x2, y2 := 50.0, 30.0 // segment end point 2
cx, cy := 100.0, 90.0 // center of circle
r := 80.0
segx, segy := x2-x1, y2-y1
testCntr, solutionCntr := 0, 0
for i := -100; i < 100; i++ {
for j := -100; j < 100; j++ {
testCntr++
s1x, s2x := x1+float64(i), x2+float64(i)
s1y, s2y := y1+float64(j), y2+float64(j)
sc1x, sc1y := s1x-cx, s1y-cy
seg1Inside := sc1x*sc1x+sc1y*sc1y < r*r
sc2x, sc2y := s2x-cx, s2y-cy
seg2Inside := sc2x*sc2x+sc2y*sc2y < r*r
p1x, p1y, p2x, p2y, intersects := SegmentCircleIntersection(s1x, s1y, s2x, s2y, cx, cy, r)
if intersects {
solutionCntr++
//Check if points are on circle
c1x, c1y := p1x-cx, p1y-cy
deltaLen1 := (c1x*c1x + c1y*c1y) - r*r
CheckEpsilon(t, deltaLen1, epsilon, "p1 not on circle")
c2x, c2y := p2x-cx, p2y-cy
deltaLen2 := (c2x*c2x + c2y*c2y) - r*r
CheckEpsilon(t, deltaLen2, epsilon, "p2 not on circle")
// Check if points are on the line through the line segment
// "cross product" of vector from a segment point to the point
// and the vector for the segment should be near zero
vp1x, vp1y := p1x-s1x, p1y-s1y
crossProd1 := vp1x*segy - vp1y*segx
CheckEpsilon(t, crossProd1, epsilon, "p1 not on line ")
vp2x, vp2y := p2x-s1x, p2y-s1y
crossProd2 := vp2x*segy - vp2y*segx
CheckEpsilon(t, crossProd2, epsilon, "p2 not on line ")
// Check if point is between points s1 and s2 on line
// This means the sign of the dot prod of the segment vector
// and point to segment end point vectors are opposite for
// either end.
wp1x, wp1y := p1x-s2x, p1y-s2y
dp1v := vp1x*segx + vp1y*segy
dp1w := wp1x*segx + wp1y*segy
if (dp1v < 0 && dp1w < 0) || (dp1v > 0 && dp1w > 0) {
t.Error("point not contained in segment ", dp1v, dp1w)
t.FailNow()
}
wp2x, wp2y := p2x-s2x, p2y-s2y
dp2v := vp2x*segx + vp2y*segy
dp2w := wp2x*segx + wp2y*segy
if (dp2v < 0 && dp2w < 0) || (dp2v > 0 && dp2w > 0) {
t.Error("point not contained in segment ", dp2v, dp2w)
t.FailNow()
}
if s1x == s2x && s2y == s1y { //Only one solution
// Test that one end of the segment is withing the radius of the circle
// and one is not
if seg1Inside && seg2Inside {
t.Error("Only one solution but both line segment ends inside")
t.FailNow()
}
if !seg1Inside && !seg2Inside {
t.Error("Only one solution but both line segment ends outside")
t.FailNow()
}
}
} else { // No intersection, check if both points outside or inside
if (seg1Inside && !seg2Inside) || (!seg1Inside && seg2Inside) {
t.Error("No solution but only one point in radius of circle")
t.FailNow()
}
}
}
}
t.Log("Tested ", testCntr, " examples and found ", solutionCntr, " solutions.")
}
以下是测试的输出：
=== RUN TestSegmentCircleIntersection
--- PASS: TestSegmentCircleIntersection (0.00s)
geom_test.go:105: Tested 40000 examples and found 7343 solutions.
最后，该方法可以容易地扩展到光线从一点开始，穿过另一点并延伸到无穷大的情况，仅通过测试是否t> 1。 0或t <0 1但不是两者。
答案 25 :(得分：0)
我只需要那个，所以我想出了这个解决方案。该语言是maxscript，但应轻松翻译成任何其他语言。 sideA，sideB和CircleRadius是标量，其余变量为[x，y，z]点。我假设z = 0可以在XY平面上求解
fn projectPoint p1 p2 p3 = --project p1 perpendicular to the line p2-p3
(
local v= normalize (p3-p2)
local p= (p1-p2)
p2+((dot v p)*v)
)
fn findIntersectionLineCircle CircleCenter CircleRadius LineP1 LineP2=
(
pp=projectPoint CircleCenter LineP1 LineP2
sideA=distance pp CircleCenter
--use pythagoras to solve the third side
sideB=sqrt(CircleRadius^2-sideA^2) -- this will return NaN if they don't intersect
IntersectV=normalize (pp-CircleCenter)
perpV=[IntersectV.y,-IntersectV.x,IntersectV.z]
--project the point to both sides to find the solutions
solution1=pp+(sideB*perpV)
solution2=pp-(sideB*perpV)
return #(solution1,solution2)
)
答案 26 :(得分：0)
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)
)
答案 27 :(得分：0)
另一种解决方案，首先考虑您不关心碰撞位置的情况。请注意，此特定功能是在假设xB和yB为矢量输入的情况下构建的，但如果不是这种情况，则可以轻松进行修改。变量名在函数开始处定义
#Line segment points (A0, Af) defined by xA0, yA0, xAf, yAf; circle center denoted by xB, yB; rB=radius of circle, rA = radius of point (set to zero for your application)
def staticCollision_f(xA0, yA0, xAf, yAf, rA, xB, yB, rB): #note potential speed up here by casting all variables to same type and/or using Cython
#Build equations of a line for linear agents (convert y = mx + b to ax + by + c = 0 means that a = -m, b = 1, c = -b
m_v = (yAf - yA0) / (xAf - xA0)
b_v = yAf - m_v * xAf
rEff = rA + rB #radii are added since we are considering the agent path as a thin line
#Check if points (circles) are within line segment (find center of line segment and check if circle is within radius of this point)
segmentMask = np.sqrt( (yB - (yA0+yAf)/2)**2 + (xB - (xA0+xAf)/2)**2 ) < np.sqrt( (yAf - yA0)**2 + (xAf - xA0)**2 ) / 2 + rEff
#Calculate perpendicular distance between line and a point
dist_v = np.abs(-m_v * xB + yB - b_v) / np.sqrt(m_v**2 + 1)
collisionMask = (dist_v < rEff) & segmentMask
#return True if collision is detected
return collisionMask, collisionMask.any()
如果需要碰撞的位置，则可以使用此站点上详述的方法，并将其中一个特工的速度设置为零。这种方法也适用于向量输入：http://twobitcoder.blogspot.com/2010/04/circle-collision-detection.html
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