kbi*_*irk 3 algorithm math linear-algebra
首先,我想你将顶点和比例相加1/3以找到原点然后从顶点到原点的最大距离.这会产生一个包含三角形的球体,但它不一定是最小的.
有没有一种已知的方法来确定最小的球体以完全封装3D中的任意三角形?
使用这里的答案和维基百科来提出适合我的c ++的东西,我希望这有助于某人!
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;
}
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