Line–sphere intersection

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The three possible line-sphere intersections:
1. No intersection.
2. Point intersection.
3. Two point intersection.

In analytic geometry, a line and a sphere can intersect in three ways: no intersection at all, at exactly one point, or in two points. Methods for distinguishing these cases, and determining equations for the points in the latter cases, are useful in a number of circumstances. For example, this is a common calculation to perform during ray tracing (Eberly 2006:698).

Calculation using vectors in 3D[edit]

In vector notation, the equations are as follows:

Equation for a sphere

\left\Vert \mathbf{x} - \mathbf{c} \right\Vert^2=r^2
  • \mathbf{c} - center point
  • r - radius
  • \mathbf{x} - points on the sphere

Equation for a line starting at \mathbf{o}

\mathbf{x}=\mathbf{o} + d\mathbf{l}
  • d - distance along line from starting point
  • \mathbf{l} - direction of line (a unit vector)
  • \mathbf{o} - origin of the line
  • \mathbf{x} - points on the line

Searching for points that are on the line and on the sphere means combining the equations and solving for d:

Equations combined
\left\Vert \mathbf{o} + d\mathbf{l} - \mathbf{c} \right\Vert^2=r^2 \Leftrightarrow (\mathbf{o} + d\mathbf{l} - \mathbf{c}) \cdot (\mathbf{o} + d\mathbf{l} - \mathbf{c}) = r^2
Expanded
d^2(\mathbf{l}\cdot\mathbf{l})+2d(\mathbf{l}\cdot(\mathbf{o}-\mathbf{c}))+(\mathbf{o}-\mathbf{c})\cdot(\mathbf{o}-\mathbf{c})=r^2
Rearranged
d^2(\mathbf{l}\cdot\mathbf{l})+2d(\mathbf{l}\cdot(\mathbf{o}-\mathbf{c}))+(\mathbf{o}-\mathbf{c})\cdot(\mathbf{o}-\mathbf{c})-r^2=0
The form of a Quadratic formula is now observable. (This quadratic equation is an example of Joachimsthal's Equation [1].)
a d^2 + b d + c = 0
where
  • a=\mathbf{l}\cdot\mathbf{l}
  • b=2(\mathbf{l}\cdot(\mathbf{o}-\mathbf{c}))
  • c=(\mathbf{o}-\mathbf{c})\cdot(\mathbf{o}-\mathbf{c})-r^2
Simplified
d=\frac{-(\mathbf{l}\cdot(\mathbf{o}-\mathbf{c})) \pm \sqrt{(\mathbf{l}\cdot(\mathbf{o}-\mathbf{c}))^2-\mathbf{l}^2((\mathbf{o}-\mathbf{c})^2-r^2)}}{\mathbf{l}^2}
Note that \mathbf{l} is a unit vector, and thus \mathbf{l}^2=1. Thus, we can simplify this further to
d=-(\mathbf{l}\cdot(\mathbf{o}-\mathbf{c})) \pm \sqrt{(\mathbf{l}\cdot(\mathbf{o}-\mathbf{c}))^2-(\mathbf{o}-\mathbf{c})^2+r^2}
  • If the value under the square-root ((\mathbf{l}\cdot(\mathbf{o}-\mathbf{c}))^2-(\mathbf{o}-\mathbf{c})^2+r^2) is less than zero, then it is clear that no solutions exist, i.e. the line does not intersect the sphere (case 1).
  • If it is zero, then exactly one solution exists, i.e. the line just touches the sphere in one point (case 2).
  • If it is greater than zero, two solutions exist, and thus the line touches the sphere in two points (case 3).

See also[edit]

References[edit]

  • David H. Eberly (2006), 3D game engine design: a practical approach to real-time computer graphics, 2nd edition, Morgan Kaufmann. ISBN 0-12-229063-1