我们有 3（三个）xyz 点在 3D 空间中定义了一个圆，这个圆需要转换为折线（用于进一步渲染）。我正在寻找可以完成这项工作的现成 C 或 C++ 函数或免费库。
不明白为什么关闭。我什至不能在那里回答我自己的问题。为你们感到羞耻。但你不会阻止知识的传播！
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3 回答
11
有一个更简单的解决方案可以在真实 3D 中找到圆参数，只需查看http://en.wikipedia.org/wiki/Circumscribed_circle中的“重心坐标”部分。您可以从中提取以下优化代码：
// triangle "edges"
const Vector3d t = p2-p1;
const Vector3d u = p3-p1;
const Vector3d v = p3-p2;
// triangle normal
const Vector3d w = t.crossProduct(u);
const double wsl = w.getSqrLength();
if (wsl<10e-14) return false; // area of the triangle is too small (you may additionally check the points for colinearity if you are paranoid)
// helpers
const double iwsl2 = 1.0 / (2.0*wsl);
const double tt = t*t;
const double uu = u*u;
// result circle
Vector3d circCenter = p1 + (u*tt*(u*v) - t*uu*(t*v)) * iwsl2;
double circRadius = sqrt(tt * uu * (v*v) * iwsl2*0.5);
Vector3d circAxis = w / sqrt(wsl);
然后，您也可以在真实 3D 中计算圆上的点，例如使用 OpenGL 中的 GL_LINE_STRIP 绘制它们。这应该比使用 2D sin/cos 方法快得多。
// find orthogonal vector to the circle axis
const Vector3d an = circAxis.getNormalized();
const Vector3d ao = Vector3d(4.0+an[0], 4.0+an[0]+an[1], 4.0+an[0]+an[1]+an[2]).crossProduct(an).getNormalized();
// 4x4 rotation matrix around the circle axis
const int steps = 360; // maybe adjust according to circle size on screen
Matrix4d R = makeRotMatrix4d(circCenter, circAxis, 2.0*M_PI/double(steps));
// one point on the circle
Vector3d cp = circCenter + ao*circRadius;
// rotate point on the circle
for (int i=0; i<steps; ++i)
{
circlePoints.push_back(cp);
cp = transformPoint(cp, R); // apply the matrix
}
对于变换矩阵的创建（即 makeRotMatrix4d()），请参见http://paulbourke.net/geometry/rotate/例如。
请注意，我没有测试上面的代码是否真的编译，但它应该给你足够的提示。
于 2012-12-21T15:27:03.990 回答
3
有一篇不错的文章和代码示例，介绍了如何在 2D、XY 平面中通过 3 个点构建一个圆。
http://paulbourke.net/geometry/circlesphere/
http://paulbourke.net/geometry/circlesphere/Circle.cpp
要构建一个 3D 圆，我们必须：
- 将我们的 3 个点旋转到 XY 平面
- 计算圆心
- 使用文章中的代码在 XY 平面上构建一个圆
- 将其旋转回原来的平面
对于旋转，最好使用四元数。
为了找到正确的四元数，我查看了 Ogre3d 源代码： void Quaternion::FromAngleAxis (const Radian& rfAngle, const Vector3& rkAxis)
还有一个更有用的函数： Quaternion getRotationTo(const Vector3& dest, const Vector3& fallbackAxis = Vector3::ZERO) const 但我没有使用它。
对于四元数和向量，我使用了我们自己的类。这是完成这项工作的函数的完整源代码：
bool IsPerpendicular(Point3d *pt1, Point3d *pt2, Point3d *pt3);
double CalcCircleCenter(Point3d *pt1, Point3d *pt2, Point3d *pt3, Point3d *center);
void FindCircleCenter(const Point3d *V1, const Point3d *V2, const Point3d *V3, Point3d *center)
{
Point3d *pt1=new Point3d(*V1);
Point3d *pt2=new Point3d(*V2);
Point3d *pt3=new Point3d(*V3);
if (!IsPerpendicular(pt1, pt2, pt3) ) CalcCircleCenter(pt1, pt2, pt3, center);
else if (!IsPerpendicular(pt1, pt3, pt2) ) CalcCircleCenter(pt1, pt3, pt2, center);
else if (!IsPerpendicular(pt2, pt1, pt3) ) CalcCircleCenter(pt2, pt1, pt3, center);
else if (!IsPerpendicular(pt2, pt3, pt1) ) CalcCircleCenter(pt2, pt3, pt1, center);
else if (!IsPerpendicular(pt3, pt2, pt1) ) CalcCircleCenter(pt3, pt2, pt1, center);
else if (!IsPerpendicular(pt3, pt1, pt2) ) CalcCircleCenter(pt3, pt1, pt2, center);
else {
delete pt1;
delete pt2;
delete pt3;
return;
}
delete pt1;
delete pt2;
delete pt3;
}
bool IsPerpendicular(Point3d *pt1, Point3d *pt2, Point3d *pt3)
// Check the given point are perpendicular to x or y axis
{
double yDelta_a= pt2->y - pt1->y;
double xDelta_a= pt2->x - pt1->x;
double yDelta_b= pt3->y - pt2->y;
double xDelta_b= pt3->x - pt2->x;
// checking whether the line of the two pts are vertical
if (fabs(xDelta_a) <= 0.000000001 && fabs(yDelta_b) <= 0.000000001){
return false;
}
if (fabs(yDelta_a) <= 0.0000001){
return true;
}
else if (fabs(yDelta_b) <= 0.0000001){
return true;
}
else if (fabs(xDelta_a)<= 0.000000001){
return true;
}
else if (fabs(xDelta_b)<= 0.000000001){
return true;
}
else
return false ;
}
double CalcCircleCenter(Point3d *pt1, Point3d *pt2, Point3d *pt3, Point3d *center)
{
double yDelta_a = pt2->y - pt1->y;
double xDelta_a = pt2->x - pt1->x;
double yDelta_b = pt3->y - pt2->y;
double xDelta_b = pt3->x - pt2->x;
if (fabs(xDelta_a) <= 0.000000001 && fabs(yDelta_b) <= 0.000000001){
center->x= 0.5*(pt2->x + pt3->x);
center->y= 0.5*(pt1->y + pt2->y);
center->z= pt1->z;
return 1;
}
// IsPerpendicular() assure that xDelta(s) are not zero
double aSlope=yDelta_a/xDelta_a; //
double bSlope=yDelta_b/xDelta_b;
if (fabs(aSlope-bSlope) <= 0.000000001){ // checking whether the given points are colinear.
return -1;
}
// calc center
center->x= (aSlope*bSlope*(pt1->y - pt3->y) + bSlope*(pt1->x + pt2 ->x)
- aSlope*(pt2->x+pt3->x) )/(2* (bSlope-aSlope) );
center->y = -1*(center->x - (pt1->x+pt2->x)/2)/aSlope + (pt1->y+pt2->y)/2;
return 1;
}
//! Builds a circle in 3D space by 3 points on it and an optional center
void buildCircleBy3Pt(const float *pt1,
const float *pt2,
const float *pt3,
const float *c, // center, can be NULL
std::vector<float> *circle)
{
/* Get the normal vector to the triangle formed by 3 points
Calc a rotation quaternion from that normal to the 0,0,1 axis
Rotate 3 points using quaternion. Points will be in XY plane
Build a circle by 3 points on XY plane
Rotate a circle back into original plane using quaternion
*/
Point3d p1(pt1[0], pt1[1], pt1[2]);
Point3d p2(pt2[0], pt2[1], pt2[2]);
Point3d p3(pt3[0], pt3[1], pt3[2]);
Point3d center;
if (c)
{
center.set(c[0], c[1], c[2]);
}
const Vector3d p2top1 = p1 - p2;
const Vector3d p2top3 = p3 - p2;
const Vector3d circle_normal = p2top1.crossProduct(p2top3).normalize();
const Vector3d xy_normal(0, 0, 1);
Quaternion rot_quat;
// building rotation quaternion
{
// Rotation axis around which we will rotate our circle into XY plane
Vector3d rot_axis = xy_normal.crossProduct(circle_normal).normalize();
const double rot_angle = xy_normal.angleTo(circle_normal); // radians
const double w = cos(rot_angle * 0.5);
rot_axis *= sin(rot_angle * 0.5);
rot_quat.set(w, rot_axis.x, rot_axis.y, rot_axis.z);
}
Quaternion rot_back_quat;
// building backward rotation quaternion, same as prev. but -angle
{
const double rot_angle = -(xy_normal.angleTo(circle_normal)); // radians
const double w_back = cos(rot_angle * 0.5);
Vector3d rot_back_axis = xy_normal.crossProduct(circle_normal).normalize();
rot_back_axis *= sin(rot_angle * 0.5);
rot_back_quat.set(w_back, rot_back_axis.x, rot_back_axis.y, rot_back_axis.z);
}
rot_quat.rotate(p1);
rot_quat.rotate(p2);
rot_quat.rotate(p3);
rot_quat.rotate(center);
if (!c)
{
// calculate 2D center
FindCircleCenter(&p1, &p2, &p3, ¢er);
}
// calc radius
const double radius = center.distanceTo(p1);
const float DEG2RAD = 3.14159f / 180.0f;
// build circle
for (int i = 0; i < 360; ++i)
{
float degInRad = i * DEG2RAD;
Point3d pt(cos(degInRad) * radius + center.x, sin(degInRad) * radius + center.y, 0);
// rotate the point back into original plane
rot_back_quat.rotate(pt);
circle->push_back(pt.x);
circle->push_back(pt.y);
circle->push_back(pt.z);
}
}
于 2012-12-20T17:22:45.687 回答
1
以下是Mark的答案的 C#/Unity 端口。它使用Unity 脚本 API 中的类型和实用功能。
// triangle "edges"
var t = p2 - p1;
var u = p3 - p1;
var v = p3 - p2;
// triangle normal
var w = Vector3.Cross(t, u);
var wsl = Vector3.Dot(w, w);
// TODO: if (wsl<10e-14) return false; // area of the triangle is too small (you may additionally check the points for colinearity if you are paranoid)
// helpers
var iwsl2 = 1f / (2f * wsl);
var tt = Vector3.Dot(t, t);
var uu = Vector3.Dot(u, u);
// result circle
Vector3 circCenter = p1 + (u * tt * (Vector3.Dot(u, v)) - t * uu * (Vector3.Dot(t, v))) * iwsl2;
var circRadius = Mathf.Sqrt(tt * uu * (Vector3.Dot(v, v)) * iwsl2 * 0.5f);
Vector3 circAxis = w / Mathf.Sqrt(wsl);
使用Unity 的 Gizmos，可以如下绘制圆（在这种情况下使用 30 条线段来近似它）：
// Draw the circle:
Gizmos.color = Color.white;
for (int i = 0; i < 30; ++i)
{
Gizmos.DrawLine(
circCenter + Quaternion.AngleAxis(360f / 30f * i , circAxis) * (p1 - circCenter),
circCenter + Quaternion.AngleAxis(360f / 30f * (i + 1), circAxis) * (p1 - circCenter)
);
}
顶点位置的结果如下所示var p1 = new Vector3(0f, 1.44f, 0f); var p2 = new Vector3(0f, 0.73f, 0.65f); var p3 = new Vector3(0f, -1.04f, 0f);：
于 2021-07-23T16:53:05.033 回答