Circle line-segment collision detection algorithm?

Question

I have a line from A to B and a circle positioned at C with the radius R.

What is a good alg

Answer 1

I wrote a small script to test intersection by projecting circle's center point on to line.

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/

If you need to check the collision with the segment, you also need to consider circle center's distance to start and end points.

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/

Answer 2

If you find the distance between the center of the sphere (since it's 3D I assume you mean sphere and not circle) and the line, then check if that distance is less than the radius that will do the trick.

The collision point is obviously the closest point between the line and the sphere (which will be calculated when you're calculating the distance between the sphere and the line)

Distance between a point and a line:
http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html

Answer 3

Another solution, first considering the case where you don't care about collision location. Note that this particular function is built assuming vector input for xB and yB but can easily be modified if that is not the case. Variable names are defined at the start of the function

#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()

If you need the location of the collision, you can use the approach detailed on this site, and set the velocity of one of the agents to zero. This approach works well with vector inputs as well: http://twobitcoder.blogspot.com/2010/04/circle-collision-detection.html

Answer 4

' 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

Answer 5

This Java Function returns a DVec2 Object. It takes a DVec2 for the center of the circle, the radius of the circle, and a Line.

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;
}

Answer 6

In this post circle line collision will be checked by checking distance between circle center and point on line segment (Ipoint) that represent intersection point between normal N (Image 2) from circle center to line segment.

(https://i.stack.imgur.com/3o6do.png)

On image 1 one circle and one line are shown, vector A point to line start point, vector B point to line end point, vector C point to circle center. Now we must find vector E (from line start point to circle center) and vector D (from line start point to line end point) this calculation is shown on image 1.

(https://i.stack.imgur.com/7098a.png)

At image 2 we can see that vector E is projected on Vector D by "dot product" of vector E and unit vector D, result of dot product is scalar Xp that represent the distance between line start point and point of intersection (Ipoint) of vector N and vector D. Next vector X is found by multiplying unit vector D and scalar Xp.

Now we need to find vector Z (vector to Ipoint), its easy its simple vector addition of vector A (start point on line) and vector X. Next we need to deal with special cases we must check is Ipoint on line segment, if its not we must find out is it left of it or right of it, we will use vector closest to determine which point is closest to circle.

(https://i.stack.imgur.com/p9WIr.png)

When projection Xp is negative Ipoint is left of line segment, vector closest is equal to vector of line start point, when projection Xp is greater then magnitude of vector D then Ipoint is right of line segment then closest vector is equal to vector of line end point in any other case closest vector is equal to vector Z.

Now when we have closest vector , we need to find vector from circle center to Ipoint (dist vector), its simple we just need to subtract closest vector from center vector. Next just check if vector dist magnitude is less then circle radius if it is then they collide, if its not there is no collision.

(https://i.stack.imgur.com/QJ63q.png)

For end, we can return some values for resolving collision , easiest way is to return overlap of collision (subtract radius from vector dist magnitude) and return axis of collision, its vector D. Also intersection point is vector Z if needed.