如何在飞机上获得三个非共线点？ - C ++ [英] How do I get three non-colinear points on a plane? - C++
问题描述
我想在线平面交集算法中实现。根据维基百科，我需要三个非共线点在飞机上这样做。
因此，我尝试实施但是，使用C ++的算法。事情绝对是错误的，因为它没有任何意义，我可以选择任何x和y坐标，他们将适合在飞机上。如果平面是垂直的并且沿着x轴怎么办？没有点y = 1会在平面中。
我意识到这个问题已经在StackOverflow上发布了很多，我看到很多解决方案，其中飞机由3个点定义。但我只有一个正常和一个位置。在我整理出非共线点搜索器之前，我不能测试我的线平面交点算法。
现在的问题是，我除以normal.z，当normal.z为0时，显然不会工作。我用这个飞机测试：Plane * p = new Plane ），Vec3d（0.0,0.0,0.0），Vec3d（0.0,1.0,0.0））; //第二个参数：position，第三个参数：normal
当前代码给出了不正确的答案：
{0，0，0} //好吧，这是原来的
{12.8377，17.2728，-inf} //显然这不是给定平面上的非共线点
这是我的代码：
std :: vector< Vec3d> * Plane :: getThreeNonColinearPoints（）{
std :: vector< Vec3d> * v = new std :: vector< Vec3d>
v-> push_back（Vec3d（position.x，position.y，position.z））; //原始位置可以作为三个非共线点之一。
srandom（time（NULL））;
double rx，ry，rz，start;
rx = Plane :: fRand（10.0，20.0）;
ry = Plane :: fRand（10.0，20.0）;
//这里的公式：http://en.wikipedia.org/wiki/Plane_(geometry)#Definition_with_a_point_and_a_normal_vector
// nx（x-x0）+ ny（y-y0）+ nz z-z0）= 0
// | ----------------- | < - 这是开始
//我将尝试插入位置为x0，y0，z0，正常为nx，ny，nz，并求解等式
start = normal.x * （rx-position.x）+ normal.y *（ry-position.y）;
// nz（z-z0）= -start
start = -start;
//（z-z0）= start / nz
start / = normal.z; //除以零
// z = start + z0
start + = position.z;
rz = start;
v-> push_back（Vec3d（rx，ry，rz））;
// TODO一点
return v;
}
我意识到我可能会试图解决这个问题。如果是这样，请链接具体的实现。我确信它必须存在，当我看到这么多的线平面交集的实现。
提前感谢。
飞机可以用几种方法定义。通常使用平面上的点和法线向量。要从三个点获得法向量（ P1 ， P2 ， P3 ）取三角形边的叉积
P1 = {x1，y1，z1}
P2 = {x2，y2，z2};
P3 = {x3，y3，z3};
N = UNIT（CROSS（P2-P1，P3-P1））;
Plane P = {P1，N}
code> P1 和正常 N 到三个点，从任何方向开始 G DOT（G，N）！= 0 沿着正常 N >。沿着平面的两个正交方向是
// try G = {0,0,1}或{0,1 ，0}或{1,0,0}
G = {0,0,1};
if（MAG（CROSS（G，N））
V = CROSS（U，N）;
P2 = P1 + U;
P3 = P1 + V;
线由点和方向定义。通常两点（ Q1 ， Q2 ）定义行
Q1 = {x1，y1，z1};
Q2 = {x2，y2，z2};
E = UNIT（Q2-Q1）;
Line L = {Q1，E}
定义线和平面的交点通过与所述平面相交的线 r = Q1 + t * E 上的点使得 DOT（r-P1，N） code>。这是为标量距离 t 沿线解决的
t = DOT（P1-Q1，N）/ DOT（E，N）;
且位置为
r = Q1 +（t * E）;
注意： DOT（）两个向量的小点积， CROSS（）交叉乘积和 UNIT（）单位向量（P，Q）= P [0] * Q [0] + P [1] * Q [1] + P [2] * Q [2];
CROSS（P，Q）= {P [1] * Q [2] -P [2] * Q [1]，P [2] * Q [0] -P [0] * Q [2 ]，P [0] * Q [1] -P [1] * Q [0]};
UNIT（P）= {P [0] / sqrt（DOT（P，P）），P [1] / sqrt（DOT（P，P）），P [2] / sqrt ，P））};
t * P = {t * P [0]，t * P [1]，t * P [2]
MAG（P）= sqrt（P [0] * P [0] + P [1] * P [1] + P [2] * P [2]）;
I'm trying to implement at line-plane intersection algorithm. According to Wikipedia I need three non-colinear points on the plane to do that.
I therefore tried implementing this algorithm in C++, however. Something is definitely wrong, cause it makes no sense that I can choose whatever x and y coordinates and they'll fit in the plane. What if the plane is vertical and along the x-axis? No point with y=1 would then be in the plane.
I realize that this problem has been posted a lot on StackOverflow, and I see lots of solutions where the plane is defined by 3 points. But I only have a normal and a position. And I can't test my line-plane intersection algorithm before I sort out my non-colinear point finder.
The problem right now, is that I'm dividing by normal.z, and that obviously won't work when normal.z is 0.
I'm testing with this plane: Plane* p = new Plane(Color(), Vec3d(0.0,0.0,0.0), Vec3d(0.0,1.0,0.0)); // second parameter : position, third parameter : normal
The current code gives this incorrect answer:
{0 , 0 , 0} // alright, this is the original
{12.8377 , 17.2728 , -inf} // obviously this is not a non-colinear point on the given plane
Here's my code:
std::vector<Vec3d>* Plane::getThreeNonColinearPoints() {
std::vector<Vec3d>* v = new std::vector<Vec3d>();
v->push_back(Vec3d(position.x, position.y, position.z)); // original position can serve as one of the three non-colinear points.
srandom(time(NULL));
double rx, ry, rz, start;
rx = Plane::fRand(10.0, 20.0);
ry = Plane::fRand(10.0, 20.0);
// Formula from here: http://en.wikipedia.org/wiki/Plane_(geometry)#Definition_with_a_point_and_a_normal_vector
// nx(x-x0) + ny(y-y0) + nz(z-z0) = 0
// |-----------------| <- this is "start"
//I'll try to insert position as x0,y0,z0 and normal as nx,ny,nz, and solve the equation
start = normal.x * (rx - position.x) + normal.y * (ry - position.y);
// nz(z-z0) = -start
start = -start;
// (z-z0) = start/nz
start /= normal.z; // division by zero
// z = start+z0
start += position.z;
rz = start;
v->push_back(Vec3d(rx, ry, rz));
// TODO one more point
return v;
}
I realize that I might be trying to solve this totally wrong. If so, please link a concrete implementation of this. I'm sure it must exist, when I see so many line-plane intersection implementations.
Thanks in advance.
A plane can be defined with several ways. Typically a point on the plane and a normal vector is used. To get the normal vector from three points (P1, P2, P3 ) take the cross product of the side of the triangle
P1 = {x1, y1, z1};
P2 = {x2, y2, z2};
P3 = {x3, y3, z3};
N = UNIT( CROSS( P2-P1, P3-P1 ) );
Plane P = { P1, N }
The reverse, to go from a point P1 and normal N to three points, you start from any direction G not along the normal N such that DOT(G,N)!=0. The two orthogonal directions along the plane are then
//try G={0,0,1} or {0,1,0} or {1,0,0}
G = {0,0,1};
if( MAG(CROSS(G,N))<TINY ) { G = {0,1,0}; }
if( MAG(CROSS(G,N))<TINY ) { G = {1,0,0}; }
U = UNIT( CROSS(N, G) );
V = CROSS(U,N);
P2 = P1 + U;
P3 = P1 + V;
A line is defined by a point and a direction. Typically two points (Q1, Q2) define the line
Q1 = {x1, y1, z1};
Q2 = {x2, y2, z2};
E = UNIT( Q2-Q1 );
Line L = { Q1, E }
The intersection of the line and plane are defined by the point on the line r=Q1+t*E that intersects the plane such that DOT(r-P1,N)=0. This is solved for the scalar distance t along the line as
t = DOT(P1-Q1,N)/DOT(E,N);
and the location as
r = Q1+(t*E);
NOTE: The DOT() returns the dot-product of two vector, CROSS() the cross-product, and UNIT() the unit vector (with magnitude=1).
DOT(P,Q) = P[0]*Q[0]+P[1]*Q[1]+P[2]*Q[2];
CROSS(P,Q) = { P[1]*Q[2]-P[2]*Q[1], P[2]*Q[0]-P[0]*Q[2], P[0]*Q[1]-P[1]*Q[0] };
UNIT(P) = {P[0]/sqrt(DOT(P,P)), P[1]/sqrt(DOT(P,P)), P[2]/sqrt(DOT(P,P))};
t*P = { t*P[0], t*P[1], t*P[2] };
MAG(P) = sqrt(P[0]*P[0]+P[1]*P[1]+P[2]*P[2]);
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