旋转球面系统的点

假设我在一个球体上有10个点(随机分布)，我想旋转整个系统以确保有一个点位于北极。我该如何使用c++实现这一点呢?

我通过观察3D旋转矩阵来解决这个问题:

http://en.wikipedia.org/wiki/Rotation_matrix

我绕x轴旋转我的点直到y分量为零，然后绕y轴旋转直到x分量为零。这应该使问题的点在北极或南极，对吧?

我的代码是这样的:

#include <stdio.h> 
#include <string.h>
#include <math.h>
#include <iostream>
#include <iomanip>
#include <fstream>
#include <time.h>
#include <stdlib.h>
#include <sstream>
using namespace std;
#define PI 3.14159265358979323846
int main()
{
  int a,b,c,f,i,j,k,m,n,s;
  double p,Time,Averagetime,Energy,energy,Distance,Length,DotProdForce,Forcemagnitude,
         ForceMagnitude[101],Force[101][4],E[1000001],En[501],x[101][4],y[101][4],
         Dist[101][101],Epsilon,z[101],theta,phi;
    /*  set the number of points */
    n=10;
    /* check that there are no more than 100 points */
    if(n>100){
      cout << n << " is too many points for me :-( n";
      exit(0);
    }
    /* reset the random number generator */
    srand((unsigned)time(0));  
    for (i=1;i<=n;i++){
      x[i][1]=((rand()*1.0)/(1.0*RAND_MAX)-0.5)*2.0;
      x[i][2]=((rand()*1.0)/(1.0*RAND_MAX)-0.5)*2.0;
      x[i][3]=((rand()*1.0)/(1.0*RAND_MAX)-0.5)*2.0;
      Length=sqrt(pow(x[i][1],2)+pow(x[i][2],2)+pow(x[i][3],2));
      for (k=1;k<=3;k++){
        x[i][k]=x[i][k]/Length;
      }
    }
    /* calculate the energy */
    Energy=0.0;
    for(i=1;i<=n;i++){
      for(j=i+1;j<=n;j++){
        Distance=sqrt(pow(x[i][1]-x[j][1],2)+pow(x[i][2]-x[j][2],2)
                    +pow(x[i][3]-x[j][3],2));
        Energy=Energy+1.0/Distance;
      }
    }
   cout << fixed << setprecision(10) << "energy=" << Energy << "n";  
  /* Save Values so far */
  for(i=1;i<=n;i++){
    for(j=1;j<=3;j++){
      y[i][j]=x[i][j];
    }
  }
  /* Choose each point in turn and make it the north pole note this is what the while loop os for, but have set it to a<2 to just look at first point */
  a=1;
  b=0;
  c=0;
  while(a<2){
  /* Find theta and phi to rotate points by */
  cout << fixed << setprecision(5) << "x[" << a << "][1]=" << x[a][1] << 
  " x[" << a << "][2]=" << x[a][2] << " x[" << a << "][3]=" << x[a][3] << "n";
  theta=x[a][2]/x[a][3];
  theta=b*PI+atan(theta);
  /* Rotate Points by theta around x axis and then by phi around y axis */
  for(i=1;i<=n;i++){
    x[i][1]=x[i][1];
    x[i][2]=x[i][2]*cos(theta)-x[i][3]*sin(theta);
    x[i][3]=x[i][2]*sin(theta)+x[i][3]*cos(theta);
  }
  phi=x[a][1]/x[a][3];
  phi=c*PI+atan(phi);
  for(i=1;i<=n;i++){
    x[i][1]=x[i][1]*cos(phi)-x[i][3]*sin(phi);
    x[i][2]=x[i][2];
    x[i][3]=x[i][1]*sin(phi)+x[i][3]*cos(phi);
  }
  cout << fixed << setprecision(5) << "x[" << a << "][1]=" << x[a][1] << 
  " x[" << a << "][2]=" << x[a][2] << " x[" << a << "][3]=" << x[a][3] << "n";
   if(x[a][3]==1.0 && x[a][2]==x[a][3]==0)
    a=a+1;
  else if(b==0 && c==0)
    for(i=1;i<=n;i++){
      for(j=1;j<=3;j++){
        x[i][j]=y[i][j];
        c=1;
      }
    }
  else if(b==0 && c==1)
    for(i=1;i<=n;i++){
      for(j=1;j<=3;j++){
        x[i][j]=y[i][j];
        b=1;
        c=0;
      }
    }
  else if(b==1 && c==0)
    for(i=1;i<=n;i++){
      for(j=1;j<=3;j++){
        x[i][j]=y[i][j];
        c=1;
      }
    }
  else if(b==1 && c==1)
    break;
  }
  energy=0.0;
  for(i=1;i<=n;i++){
    for(j=i+1;j<=n;j++){
      Distance=sqrt(pow(x[i][1]-x[j][1],2)+pow(x[i][2]-x[j][2],2)
                    +pow(x[i][3]-x[j][3],2));
      energy=energy+1.0/Distance;
    }
  }  
  cout << fixed << setprecision(10) << "ENERGY=" << energy << "n";  
  cout << fixed << setprecision(5) << "x[" << a << "][1]=" << x[a][1] << 
  " x[" << a << "][2]=" << x[a][2] << " x[" << a << "][3]=" << x[a][3] << "n";
  /* Output to help with gnuin.txt */
  ofstream File4 ("mypoints");
  for(i=1;i<=n;i++){
    File4 << x[i][1] << " " <<   x[i][2] << " " << x[i][3] << "n";
  }
  File4.close(); 
  return 0;
}

好的，这里有很多问题，比如if语句(第103行)不应该有一个等于双精度的条件，因为它永远不会工作，但我可以稍后使用间接比较(一些epsilon的东西)来解决这个问题。我真正的问题是，为什么旋转即使作用于所有的点也会使球面上的点离开?(正如你所看到的，这些值已经被标准化，使它们都在第38行单位球体上)。

注意:b, c的东西是检查点是在北极还是南极

x[i][1]=x[i][1];
x[i][2]=x[i][2]*cos(theta)-x[i][3]*sin(theta);
x[i][3]=x[i][2]*sin(theta)+x[i][3]*cos(theta);

double new_y, new_z;
new_y=x[i][2]*cos(theta)-x[i][3]*sin(theta);
new_z=x[i][2]*sin(theta)+x[i][3]*cos(theta);
x[i][2] = new_y;
x[i][3] = new_z;