Intersection between a line and a sphere

Question

I\'m trying to find the point of intersection between a sphere and a line but honestly, I don\'t have any idea of how to do so. Could anyone help me on this one ?

Answer 1

Express the line as an function of t:

{ x(t) = x0*(1-t) + t*x1
{ y(t) = y0*(1-t) + t*y1
{ z(t) = z0*(1-t) + t*z1

When t = 0, it will be at one end-point (x0,y0,z0). When t = 1, it will be at the other end-point (x1,y1,z1).

Write a formula for the distance to the center of the sphere (squared) in t (where (xc,yc,zc) is the center of the sphere):

f(t) = (x(t) - xc)^2 + (y(t) - yc)^2 + (z(t) - zc)^2

Solve for t when f(t) equals R^2 (R being the radius of the sphere):

(x(t) - xc)^2 + (y(t) - yc)^2 + (z(t) - zc)^2 = R^2

A = (x0-xc)^2 + (y0-yc)^2 + (z0-zc)^2 - R^2
B = (x1-xc)^2 + (y1-yc)^2 + (z1-zc)^2 - A - C - R^2
C = (x0-x1)^2 + (y0-y1)^2 + (z0-z1)^2

Solve A + B*t + C*t^2 = 0 for t. This is a normal quadratic equation.

You can get up to two solutions. Any solution where t lies between 0 and 1 are valid.

If you got a valid solution for t, plug it in the first equations to get the point of intersection.

I assumed you meant a line segment (two end-points). If you instead want a full line (infinite length), then you could pick two points along the line (not too close), and use them. Also let t be any real value, not just between 0 and 1.

Edit: I fixed the formula for B. I was mixing up the signs. Thanks M Katz, for mentioning that it didn't work.

Answer 2

I believe there is an inaccuracy in the solution by Markus Jarderot. Not sure what the problem is, but I'm pretty sure I translated it faithfully to code, and when I tried to find the intersection of a line segment known to cross into a sphere, I got a negative discriminant (no solutions).

I found this: http://www.codeproject.com/Articles/19799/Simple-Ray-Tracing-in-C-Part-II-Triangles-Intersec, which gives a similar but slightly different derivation.

I turned that into the following C# code and it works for me:

    public static Point3D[] FindLineSphereIntersections( Point3D linePoint0, Point3D linePoint1, Point3D circleCenter, double circleRadius )
    {
        // http://www.codeproject.com/Articles/19799/Simple-Ray-Tracing-in-C-Part-II-Triangles-Intersec

        double cx = circleCenter.X;
        double cy = circleCenter.Y;
        double cz = circleCenter.Z;

        double px = linePoint0.X;
        double py = linePoint0.Y;
        double pz = linePoint0.Z;

        double vx = linePoint1.X - px;
        double vy = linePoint1.Y - py;
        double vz = linePoint1.Z - pz;

        double A = vx * vx + vy * vy + vz * vz;
        double B = 2.0 * (px * vx + py * vy + pz * vz - vx * cx - vy * cy - vz * cz);
        double C = px * px - 2 * px * cx + cx * cx + py * py - 2 * py * cy + cy * cy +
                   pz * pz - 2 * pz * cz + cz * cz - circleRadius * circleRadius;

        // discriminant
        double D = B * B - 4 * A * C;

        if ( D < 0 )
        {
            return new Point3D[ 0 ];
        }

        double t1 = ( -B - Math.Sqrt ( D ) ) / ( 2.0 * A );

        Point3D solution1 = new Point3D( linePoint0.X * ( 1 - t1 ) + t1 * linePoint1.X,
                                         linePoint0.Y * ( 1 - t1 ) + t1 * linePoint1.Y,
                                         linePoint0.Z * ( 1 - t1 ) + t1 * linePoint1.Z );
        if ( D == 0 )
        {
            return new Point3D[] { solution1 };
        }

        double t2 = ( -B + Math.Sqrt( D ) ) / ( 2.0 * A );
        Point3D solution2 = new Point3D( linePoint0.X * ( 1 - t2 ) + t2 * linePoint1.X,
                                         linePoint0.Y * ( 1 - t2 ) + t2 * linePoint1.Y,
                                         linePoint0.Z * ( 1 - t2 ) + t2 * linePoint1.Z );

        // prefer a solution that's on the line segment itself

        if ( Math.Abs( t1 - 0.5 ) < Math.Abs( t2 - 0.5 ) )
        {
            return new Point3D[] { solution1, solution2 };
        }

        return new Point3D[] { solution2, solution1 };
    }

Answer 3

I don't have the reputation to comment on Ashavsky's solution, but the check at the end needed a bit more tweaking.

if (D < 0)
    return new Point3D[0];
else if ((t1 > 1 || t1 < 0) && (t2 > 1 || t2 < 0))
    return new Point3D[0];
else if (!(t1 > 1 || t1 < 0) && (t2 > 1 || t2 < 0))
    return new [] { solution1 };
else if ((t1 > 1 || t1 < 0) && !(t2 > 1 || t2 < 0))
    return new [] { solution2 };
else if (D == 0)
    return new [] { solution1 };
else
    return new [] { solution1, solution2 };

Answer 4

Find the solution of the two equations in (x,y,z) describing the line and the sphere.

There may be 0, 1 or 2 solutions.

• 0 implies they don't intersect
• 1 implies the line is a tangent to the sphere
• 2 implies the line passes through the sphere.

Answer 5

Here's a more concise formulation using inner products, less than 100 LOCs, and no external links. Also, the question was asked for a line, not a line segment.

Assume that the sphere is centered at C with radius r. The line is described by P+l*D where D*D=1. P and C are points, D is a vector, l is a number.

We set PC = P-C, pd = PC*D and s = pd*pd - PC*PC + r*r. If s < 0 there are no solutions, if s == 0 there is just one, otherwise there are two. For the solutions we set l = -pd +- sqrt(s), then plug into P+l*D.

Answer 6

You may use Wolfram Alpha to solve it in the coordinate system where the sphere is centered.

In this system, the equations are:

Sphere:

     x^2 + y^2 + z^2 = r^2  

Straight line:

    x = x0 + Cos[x1] t
    y = y0 + Cos[y1] t
    z = z0 + Cos[z1] t 

Then we ask Wolfram Alpha to solve for t: (Try it!)

and after that you may change again to your original coordinate system (a simple translation)