另一种解决方案，首先考虑不关心碰撞位置的情况。请注意，这个特定的函数是在假设xB和yB为向量输入的情况下构建的，但如果情况并非如此，则可以轻松修改。变量名在函数的开头定义

#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()

如果您需要碰撞的位置，您可以使用这个站点上详细介绍的方法，并将其中一个代理的速度设置为零。这种方法也适用于矢量输入:http://twobitcoder.blogspot.com/2010/04/circle-collision-detection.html

这里你需要一些数学知识:

假设A = (Xa, Ya)， B = (Xb, Yb)， C = (Xc, Yc)。从A到B的直线上的任意一点都有坐标(*Xa + (1-)Xb， * ya + (1-)*Yb) = P

如果点P的距离是R到C，它一定在圆上。你想要的是解决

distance(P, C) = R

这是

(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公式应用到这个方程来求解，并使用alpha的解来计算P的坐标，你会得到交点，如果存在的话。

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/

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

如果直线的坐标为A.x, A.y和B.x, B.y，圆心为C.x, C.y，则直线公式为:

x = A.x * t + B.x * (1 - t)

y = A.y * t + B.y * (1 - t)

0 < = t < = 1

这个圆是

(C.x - x)²+ (C.y - y)²= R²

如果你把直线的x和y公式代入圆公式，你会得到一个t的二阶方程，它的解是交点(如果有的话)。如果你得到的t小于0或大于1，那么它不是一个解，但它表明这条线“指向”圆的方向。