我正在尝试编写一个代码来解决n-body问题,当我使用一个包含所有实体的数组而不是单独使用不同的实体时遇到了麻烦。我目前不知道我的代码中出了什么问题,但是当我为任何一个身体绘制y函数时,我得到一条直线,显然是不对的。
这是我目前的代码:
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <fstream>
#define h 10000.0
#define N 3
#define G 6.67384*pow(10.0,-11)
using namespace std;
class particle{
public:
double kx1,kx2,kx3,kx4, kv1, kv2, kv3, kv4;
double ky1, ky2, ky3, ky4, kvy1, kvy2, kvy3, kvy4;
double x,y,vx,vy,m;
double dist(particle aap){
double dx = x - aap.x;
double dy = y - aap.y;
return sqrt(pow(dx,2.0)+pow(dy,2.0));
}
double g(double x1, double y1,particle aap){
return G*aap.m*(aap.x-x1)/pow(dist(aap),3.0);
}
double p(double x1, double y1, particle aap){
return G*aap.m*(aap.y-y1)/pow(dist(aap),3.0);
}
void update(){ //object advances 1 step
x = x + (1/6.0)*(kx1+2*kx2+2*kx3+kx4);
vx = vx + (1/6.0)*(kv1+2*kv2+2*kv3+kv4);
y = y + (1/6.0)*(ky1+2*ky2+2*ky3+ky4);
vy = vy + (1/6.0)*(kvy1+2*kvy2+2*kvy3+kvy4);
}
void create(double x1, double y1, double vx1, double vy1, double m1){
x = x1;
y = y1;
vx = vx1;
vy = vy1;
m =m1;
}
bool operator ==(particle &other){
if(x == other.x && y == other.y && vx == other.vx && vy == other.vy){
return true;
}
}
};
particle bodies[N];
void set(particle (&bodies)[N]){
bodies[0].create(1, 1, -2, 1, 2*pow(10.0,30));
bodies[1].create(2870671*pow(10.0,6), 0, 0, 6800, 8.6810*pow(10.0,25));
bodies[2].create(4498542*pow(10.0,6),0 ,0, 5430, 1.0243*pow(10.0,26));
}
double xforce(double x1, double y1, particle aap, particle bodies[N]){ //force in the x- direction
double fx = 0;
for (int i = 0; i <= N; i++){
if (bodies[i] == aap ){;}
else{
fx += aap.g(x1,y1,bodies[i]);
}
}
return fx;
}
double yforce(double x1, double y1, particle aap, particle bodies[N]){ //force in the y- direction
double fy = 0;
for (int i = 0; i <= N; i++){
if (bodies[i] == aap) {;}
else{
fy += aap.p(x1,y1,bodies[i]);
}
}
return fy;
}
void corr(double t, particle bodies[N]){ //runge kutta 4
for(int i =0; i <= N; i++){
bodies[i].kx1 = t*bodies[i].vx;
bodies[i].kv1 = t*xforce(bodies[i].x, bodies[i].y, bodies[i], bodies);
bodies[i].ky1 = t*bodies[i].vy;
bodies[i].kvy1 = t*yforce(bodies[i].x, bodies[i].y, bodies[i], bodies);
bodies[i].kx2 = t*(bodies[i].vx + 0.5*bodies[i].kv1);
bodies[i].kv2 = t*xforce(bodies[i].x + 0.5*bodies[i].kx1, bodies[i].y + 0.5*bodies[i].ky1, bodies[i], bodies);
bodies[i].ky2 = t*(bodies[i].vy + 0.5*bodies[i].kvy1);
bodies[i].kvy2 = t*yforce(bodies[i].x + 0.5*bodies[i].kx1, bodies[i].y + 0.5*bodies[i].ky1, bodies[i], bodies);
bodies[i].kx3 = t*(bodies[i].vx+ 0.5*bodies[i].kv2);
bodies[i].kv3 = t*xforce(bodies[i].x + 0.5*bodies[i].kx2, bodies[i].y + 0.5*bodies[i].ky2, bodies[i], bodies);
bodies[i].ky3 = t*(bodies[i].vy+ 0.5*bodies[i].kvy2);
bodies[i].kvy3 = t*yforce(bodies[i].x + 0.5*bodies[i].kx2, bodies[i].y + 0.5*bodies[i].ky2,bodies[i], bodies);
bodies[i].kx4 = t*(bodies[i].vx + bodies[i].kv3);
bodies[i].kv4 = t*xforce(bodies[i].x+ bodies[i].kx3, bodies[i].y + bodies[i].ky3, bodies[i], bodies);
bodies[i].ky4 = t*(bodies[i].vy + bodies[i].kvy3);
bodies[i].kvy4 = t*yforce(bodies[i].x + bodies[i].kx3, bodies[i].y + bodies[i].ky3, bodies[i], bodies);
}
}
void calculate(particle (&bodies)[N]){
set(bodies);
ofstream file;
file.open("tester.txt");
for(int i =0; i <=50000; i++){
corr(h, bodies);
for(int j = 0; j <= N; j++){
bodies[j].update();
}
if( i%1000 == 0){
file << i*h;
for(int j = 0; j <=N ; j++){
file <<" "<<bodies[j].x << " "<< bodies[j].y;
}
file <<" "<<"\n";
}
else{;}
}
file.close();
}
int main()
{
calculate(bodies);
system("pause");
return 0;
}
问题可能在于类粒子之外,因为程序在我开始使用数组体之前就已经工作了。任何关于非必要改进的建议都是值得欢迎的。我试图做的另一件事是使用std :: vector而不是数组,但我不知道如何定义我的函数之外的向量,就像我定义数组体一样。
答案 0 :(得分:1)
首先,您的所有i <= N
都是错误的,因为您的循环将执行4次(0,1,2,3)而不是i < N
的3次。
答案 1 :(得分:0)
您可能遇到energy drift,因为RK4不是symplectic,n体问题是Hamiltonian system。尝试将RK4用于太阳系n体时,我也遇到了同样的问题。 this person也是如此。您应该尝试使用另一种辛数值方法,例如Euler,Verlet,Ruth's 3rd或Ruth's 4th阶辛积分器。