Smallest sphere to encapsulate a triangle in 3D? - algorithm

Question

At first I figured you sum the vertices and scale by 1/3 to find the origin then take the largest distance from the vertex to the origin. This results in a sphere that contains the triangle, but it isn't necessarily the smallest.
Is there a known method for determining the smallest sphere to fully encapsulate an arbitrary triangle in 3D?

Answer 1

Used the answers here and wikipedia to come up with something in c++ that works for me, I hope this helps someone!
static Sphere makeMinimumBoundingSphere(const Vec3 &p1, const Vec3 &p2, const Vec3 &p3) {
Sphere s;
// Calculate relative distances
float A = (p1 - p2).distance();
float B = (p2 - p3).distance();
float C = (p3 - p1).distance();
// Re-orient triangle (make A longest side)
const Vec3 *a = &p3, *b = &p1, *c = &p2;
if (B < C) swap(B, C), swap(b, c);
if (A < B) swap(A, B), swap(a, b);
// If obtuse, just use longest diameter, otherwise circumscribe
if ((B*B) + (C*C) <= (A*A)) {
s.radius = A / 2.f;
s.position = (*b + *c) / 2.f;
} else {
// http://en.wikipedia.org/wiki/Circumscribed_circle
precision cos_a = (B*B + C*C - A*A) / (B*C*2);
s.radius = A / (sqrt(1 - cos_a*cos_a)*2.f);
Vec3 alpha = *a - *c, beta = *b - *c;
s.position = (beta * alpha.dot(alpha) - alpha * beta.dot(beta)).cross(alpha.cross(beta)) /
(alpha.cross(beta).dot(alpha.cross(beta)) * 2.f) + *c;
}
return s;
}

Answer 2

The smallest sphere to encapsulate the triangle is just the circumsribed cirlce extended into the third dimension.
Update: Scratch that, of course it isn't. It's the sphere that you get if you rotate the smallest circle around its diameter. The reason being that for any containing sphere that has its origin out of the plane of the triangle there is a smaller one that has its origin on the plane (by projecting the origin orthogonally onto the plane).

Answer 3

You are trying to find the smallest enclosing ball MB(P) of a point set P, so you could use an algorithm as implemented here https://github.com/hbf/miniball. (Note: "ball" and "sphere" are synonyms in this context.)
However, this is overkill in your case, since the point set P at hand contains exactly 3 points (the vertices of the triangle). In this particular case, you can use the fact that the smallest enclosing ball MB(P) of P={p,q,r} equals either:
B(p,q) if r is contained in B(p,q), or
B(p,r) if q is contained in B(p,r), or
B(q,r) if p is contained in B(q,r), or
B(p,q,r) otherwise.
Here, B(x,y) is the smallest ball containing the points x,y and B(x,y,z) is the smallest ball containing the points x,y,z on the boundary. B(x,y) and B(x,y,z) can be computed by solving a linear system of equations.
Note: I am the author of https://github.com/hbf/miniball.

Answer 4

Assuming that the sphere is simply a trivial extension of a circle (2-D) into 3-D (using both the same center point and the same radius), I believe what you are looking for is called circumscribed circle of a triangle.
Apparently I didn't consider the case of an obtuse triangle which if you have the vertices (points) of the triangle on the circle, then the circle is not the smallest bounding circle (and thus smallest bounding sphere).
Now I believe that you are looking for the minimum bounding sphere, which is a known and studied problem in mathematics, and computer graphics. "Smallest Enclosing Circle Problem" is a description of an O( n^{2} ) and a linear O(n) algorithms.
And as far as I know the minimal bounding circle does produce the minimal bounding sphere, using the same parameters (center point and radius) projected into three dimensions.