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#import "/common.typ": post, static, img
#show: post.with(
title: "Ray vs. sphere",
date: datetime(year: 2018, month: 5, day: 9),
)
#figure(
img(static("math/ray_vs_sphere.png")),
caption: [A drawing of a ray intersecting a sphere in two dimensions.],
)
In $RR^N$, a ray of origin $bold(x)_r$ and direction $bold(v)$ and an N-sphere of origin $bold(x)_s$ and radius $r$ intersect at $bold(x)_I$ given by:
$
bold(x)_I = bold(x)_r + t_(plus.minus) bold(v)
$
where:
$
t_(plus.minus) &= (bold(v) dot bold(u) plus.minus sqrt((bold(v) dot bold(u))^2 - v^2 (u^2 - r^2))) / v^2 \
bold(u) &= bold(x)_s - bold(x)_r
$
If the expression under the square root is negative, there is no intersection. If it's zero, $t_+ = t_-$ and there is only one intersection.
Notice that since $t v$ is the distance to intersection, $t_-$ corresponds to the *closer one* and $t_+$ to the *further*. Depending on the application, you might want to discard negative $t$'s (intersections "behind" the ray).
= Surface normal
The surface normal $hat(bold(n))$ at an intersection is given by:
$
bold(r) &= t_(plus.minus) bold(v) - bold(u) \
hat(bold(n)) &= hat(bold(r))
$
= Normalized v
Notice that the expression for $t_(plus.minus)$ becomes appreciably simpler if $v = 1$:
$
t_(plus.minus) &= bold(v) dot bold(u) plus.minus sqrt((bold(v) dot bold(u))^2 - (u^2 - r^2))
$
= Derivation
Notice that the point of intersection lies somewhere on the line defined by the ray:
$
bold(x)_I = bold(x)_r + t bold(v)
$
for some real $t$. It also lies somewhere on the sphere:
$
bold(x)_I = bold(x)_s + bold(r)
$
for some vector $bold(r)$ of length $r$. Combining the two equations and solving for $bold(r)$ yields:
$
bold(x)_r + t bold(v) - bold(x)_s = bold(r)
$
To simplify the algebra, substitute $bold(x)_s - bold(x)_r = bold(u)$:
$
t bold(v) - bold(u) = bold(r)
$
Since only the length of $bold(r)$ is known, the only thing that can be done at this point is equating the norms of both sides. Squared norms are more convenient in this particular case:
$
(t bold(v) - bold(u))^2 &= r^2 \
t^2 v^2 + u^2 - 2 t bold(v) dot bold(u) &= r^2 \
v^2 t^2 - 2 bold(v) dot bold(u) t + u^2 - r^2 &= 0
$
Solving the quadratic equation and simplifying:
$
t_(plus.minus) &= (2 bold(v) dot bold(u) plus.minus sqrt(4 (bold(v) dot bold(u))^2 - 4 v^2 (u^2 - r^2))) / (2 v^2) \
t_(plus.minus) &= (bold(v) dot bold(u) plus.minus sqrt((bold(v) dot bold(u))^2 - v^2 (u^2 - r^2))) / v^2
$
As with any quadratic equation, there is no real solution if the expression under the square root is negative and if it's zero, the solutions coincide.
Notice that $bold(r)$ is perpendicular to the sphere at an intersection. Normalizing this vector yields the surface normal $hat(bold(n))$.
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