Line–sphere intersection
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Geometrical problem
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
- Intersection in exactly one point
- Intersection in two points.
Methods for distinguishing these cases, and determining the coordinates for the points in the latter cases, are useful in a number of circumstances. For example, it is a common calculation to perform during ray tracing.[1]
Calculation using vectors in 3D
[edit ]In vector notation, the equations are as follows:
Equation for a sphere
- {\displaystyle \left\Vert \mathbf {x} -\mathbf {c} \right\Vert ^{2}=r^{2}}
- {\displaystyle \mathbf {x} } : points on the sphere
- {\displaystyle \mathbf {c} } : center point
- {\displaystyle r} : radius of the sphere
Equation for a line starting at {\displaystyle \mathbf {o} }
- {\displaystyle \mathbf {x} =\mathbf {o} +d\mathbf {u} }
- {\displaystyle \mathbf {x} } : points on the line
- {\displaystyle \mathbf {o} } : origin of the line
- {\displaystyle d} : distance from the origin of the line
- {\displaystyle \mathbf {u} } : direction of line (a non-zero vector)
Searching for points that are on the line and on the sphere means combining the equations and solving for {\displaystyle d}, involving the dot product of vectors:
- Equations combined
- {\displaystyle \left\Vert \mathbf {o} +d\mathbf {u} -\mathbf {c} \right\Vert ^{2}=r^{2}\Leftrightarrow (\mathbf {o} +d\mathbf {u} -\mathbf {c} )\cdot (\mathbf {o} +d\mathbf {u} -\mathbf {c} )=r^{2}}
- Expanded and rearranged:
- {\displaystyle d^{2}(\mathbf {u} \cdot \mathbf {u} )+2d[\mathbf {u} \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 instance of Joachimsthal's equation.[2] )
- {\displaystyle ad^{2}+bd+c=0}
- where
- {\displaystyle a=\mathbf {u} \cdot \mathbf {u} =\left\Vert \mathbf {u} \right\Vert ^{2}}
- {\displaystyle b=2[\mathbf {u} \cdot (\mathbf {o} -\mathbf {c} )]}
- {\displaystyle c=(\mathbf {o} -\mathbf {c} )\cdot (\mathbf {o} -\mathbf {c} )-r^{2}=\left\Vert \mathbf {o} -\mathbf {c} \right\Vert ^{2}-r^{2}}
- Simplified
- {\displaystyle d={\frac {-2[\mathbf {u} \cdot (\mathbf {o} -\mathbf {c} )]\pm {\sqrt {(2[\mathbf {u} \cdot (\mathbf {o} -\mathbf {c} )])^{2}-4\left\Vert \mathbf {u} \right\Vert ^{2}(\left\Vert \mathbf {o} -\mathbf {c} \right\Vert ^{2}-r^{2})}}}{2\left\Vert \mathbf {u} \right\Vert ^{2}}}}
- Note that in the specific case where {\displaystyle \mathbf {u} } is a unit vector, and thus {\displaystyle \left\Vert \mathbf {u} \right\Vert ^{2}=1}, we can simplify this further to (writing {\displaystyle {\hat {\mathbf {u} }}} instead of {\displaystyle \mathbf {u} } to indicate a unit vector):
- {\displaystyle \nabla =[{\hat {\mathbf {u} }}\cdot (\mathbf {o} -\mathbf {c} )]^{2}-(\left\Vert \mathbf {o} -\mathbf {c} \right\Vert ^{2}-r^{2})}
- {\displaystyle d=-[{\hat {\mathbf {u} }}\cdot (\mathbf {o} -\mathbf {c} )]\pm {\sqrt {\nabla }}}
- If {\displaystyle \nabla <0}, then it is clear that no solutions exist, i.e. the line does not intersect the sphere (case 1).
- If {\displaystyle \nabla =0}, then exactly one solution exists, i.e. the line just touches the sphere in one point (case 2).
- If {\displaystyle \nabla >0}, two solutions exist, and thus the line touches the sphere in two points (case 3).
See also
[edit ]- Intersection (geometry) § A line and a circle
- Analytic geometry
- Line–plane intersection
- Plane–plane intersection
- Plane–sphere intersection
References
[edit ]- ^ Eberly, David H. (2006). 3D game engine design: a practical approach to real-time computer graphics, 2nd edition. Morgan Kaufmann. p. 698. ISBN 0-12-229063-1.
- ^ "Joachimsthal's Equation".