C#中的N体仿真
作者：互联网
我正在尝试使用Runge Kutta 4或Velocity Verlet集成算法在C#中实现N体模拟.
在我移动到更大数量的粒子之前,我想通过模拟围绕太阳的地球轨道来测试模拟,然而,由于某种原因,我得到了一个奇怪的螺旋而不是椭圆轨道.
我无法弄清楚这个问题,因为我使用相同的算法对太阳系进行了简单的模拟,其中太阳被固定在适当的位置并且一切都运行良好.积分器工作得很好,因为我使用哪一个并不重要,我得到两者的螺旋.
任何帮助,将不胜感激.
这是代码：
class NBODY
{
public static double G = 4 * Math.PI * Math.PI;
class Particle
{
public double[] r; // position vector
public double[] v; // velocity vector
public double mass;
//constructor
public Particle() {}
public Particle(double x, double y, double z, double vx, double vy, double vz, double m)
{
this.r = new double[3];
this.v = new double[3];
this.r[0] = x;
this.r[1] = y;
this.r[2] = z;
this.v[0] = vx;
this.v[1] = vy;
this.v[2] = vz;
this.mass = m;
}
public void Update(Particle[] particles, double t, double h, int particleNumber)
{
RungeKutta4(particles, t, h, particleNumber);
}
private double acc(double r, Particle[] particles, int particleNumber, double[] r_temp, int l)
{
// dv/dt = f(x) = -G * m_i * (x - x_i) / [(x - x_i)^2 + (y - y_i)^2 + (z - z_i)^2]^(3/2)
double sum = 0;
switch (l)
{
case 0:
for (int i = 0; i < particles.Length; i++)
if (i != particleNumber)
sum += particles[i].mass * (r - particles[i].r[l]) / Math.Pow( Math.Pow(r - particles[i].r[l], 2)
+ Math.Pow(r_temp[1] - particles[i].r[1], 2) + Math.Pow(r_temp[2] - particles[i].r[2], 2), 1.5);
break;
case 1:
for (int i = 0; i < particles.Length; i++)
if (i != particleNumber)
sum += particles[i].mass * (r - particles[i].r[l]) / Math.Pow(Math.Pow(r - particles[i].r[l], 2)
+ Math.Pow(r_temp[0] - particles[i].r[0], 2) + Math.Pow(r_temp[2] - particles[i].r[2], 2), 1.5);
break;
case 2:
for (int i = 0; i < particles.Length; i++)
if (i != particleNumber)
sum += particles[i].mass * (r - particles[i].r[l]) / Math.Pow(Math.Pow(r - particles[i].r[l], 2)
+ Math.Pow(r_temp[0] - particles[i].r[0], 2) + Math.Pow(r_temp[1] - particles[i].r[1], 2), 1.5);
break;
}
return -G * sum;
}
private void RungeKutta4(Particle[] particles, double t, double h, int particleNumber)
{
//current position of the particle is saved in a vector
double[] r_temp = new double[3];
for (int j = 0; j < 3; j++)
r_temp[j] = this.r[j];
//loop going over all the coordinates and updating each using RK4 algorithm
for (int l = 0; l < 3; l++)
{
double[,] k = new double[4, 2];
k[0, 0] = this.v[l]; //k1_r
k[0, 1] = acc(this.r[l], particles, particleNumber, r_temp, l); //k1_v
k[1, 0] = this.v[l] + k[0, 1] * 0.5 * h; //k2_r
k[1, 1] = acc(this.r[l] + k[0, 0] * 0.5 * h, particles, particleNumber, r_temp, l); //k2_v
k[2, 0] = this.v[l] + k[1, 1] * 0.5 * h; //k3_r
k[2, 1] = acc(this.r[l] + k[1, 0] * 0.5 * h, particles, particleNumber, r_temp, l); //k3_v
k[3, 0] = this.v[l] + k[2, 1] * h; //k4_r
k[3, 1] = acc(this.r[l] + k[2, 0] * h, particles, particleNumber, r_temp, l); //k4_v
this.r[l] += (h / 6.0) * (k[0, 0] + 2 * k[1, 0] + 2 * k[2, 0] + k[3, 0]);
this.v[l] += (h / 6.0) * (k[0, 1] + 2 * k[1, 1] + 2 * k[2, 1] + k[3, 1]);
}
}
/*
Velocity Verlet algorithm:
1. Calculate y(t+h) = y(t) + v(t)h + 0.5a(t)h*h
2. Derive a(t+h) from dv/dt = -y using y(t+h)
3. Calculate v(t+h) = v(t) + 0.5*(a(t) + a(t+h))*h
*/
private void VelocityVerlet(Particle[] particles, double t, double h, int particleNumber)
{
double[] r_temp = new double[3];
for (int j = 0; j < 3; j++)
r_temp[j] = this.r[j];
//loop going over all the coordinates and updating each using RK4 algorithm
for (int l = 0; l < 3; l++)
{
//position
this.r[l] += h * this.v[l] + 0.5 * h * h * acc(this.r[l], particles, particleNumber, r_temp, l);
//velocity
this.v[l] += 0.5 * h * (acc(r_temp[l], particles, particleNumber, r_temp,l)
+ acc(this.r[l], particles, particleNumber, r_temp,l));
}
}
}
static void Main(string[] args)
{
//output file
TextWriter output = new StreamWriter("ispis.txt");
// declarations of variables
Particle[] particles = new Particle[2];
particles[0] = new Particle(0, 0, 0, 0, 0, 0, 1); //sun
particles[1] = new Particle(1, 0, 0, 0, 6.28, 0, 3.003467E-06); //earth
int N = 200;
double h, t, tmax;
double[,,] x = new double[particles.Length, N, 3]; //output
// setting initial values, step size and max time tmax
h = 0.01; // the step size in years
tmax = h * N;
// initial time
t = 0;
int i = 0;
while (t <= tmax) {
//updates position of all particles
for (int z = 1; z < particles.Length; z++)
particles[z].Update(particles, t, h, z);
//saves the position for output
for (int j = 1; j < particles.Length ; j++)
for (int z = 0; z < 3; z++ )
x[j,i,z] = particles[j].r[z];
t += h;
i++;
}
//output to file
for (int k = 0; k < particles.Length; k++ )
{
for (int f = 0; f < 3; f++)
{
for (int l = 0; l < N; l++)
output.Write(string.Format("{0,-15:0.########},", x[k,l,f]));
output.Write(string.Format("\n\n"));
}
output.Write(string.Format("\n\n\n\n"));
}
output.Close();
}
}
这是地球轨道输出数据的图表：
解决方法:
您的模型计算两个粒子之间的重力两次：对于第一个粒子,力基于其原始坐标,对于第二个粒子,它基于第一个粒子的更新位置.这显然违反了牛顿第3定律.您必须在任何更新之前预先计算所有力.
标签：c,algorithm,physics,simulation,numerical-methods 来源： https://codeday.me/bug/20190519/1138180.html