什么是 -nan(ind) 以及如何解决该错误？

Question

当迭代次数增加时，我得到的是

-nan(ind)

，而不是数值。我在程序中应用了四阶龙格库塔方法，函数在

dvdt

和

drdt

中给出。

#include <iostream>
#include <fstream>
#include <cmath>
#include <iomanip>
using namespace std;

double dvdt(double t, double v);
double drdt(double t, double r);

double RKv(double t, double v, double h);
double RKr(double t, double r, double h);

double xo = 0;    //initial X position
double yo = 0;    //initial Y position
double x1;      //final X position
double y11;      //final Y position
double ti = 0;  //initial time = 0
double tf = 30;      //time entered
double dt = 0.01;   //time step (iteration time)
double rud = 0;   //rudder angle
double m = 53330750;   //mass of submarine (Kg)
double Iz = 57.5;     //moment of inertia of submarine (kg.m^2)
double xg = -0.012;   //x position of COG
double yg = 0;     // position of COG
double u = 10; // surge velocity
double v, vo = 0; //sway velocity
double r, ro = 0; // yaw rate
double psio = 0; //initial ship heading
double psi1;  //final ship heading
double rudmax; // max rudder angle
double au; //surge acceleration
double av; // sway acceleration
double ar; //angular acceleration
double X; //surge force
double Y; //sway force
double N; //turning moment
double h = 1;


//matrices
double M[2][2];         //mass matrix
double Nu0[2][2];       //inertia matrix ig
double b[2][1];         //force matrix
double A[2][1];         //acceleration matrix
double V[2][1];         //velocity matrix


//maneuvering and hydrodynamic coefficents
double Xu1 = -1.0467 * pow(10, -3);
double Yv1 = -23.889 * pow(10, -3);
double Yr1 = -1.0510 * pow(10, -3);
double Nv1 = 1.1531 * pow(10, -3);
double Nr1 = -0.50717 * pow(10, -3);
double Xu = -2.8763 * pow(10, -3);
double Yv = -38.948 * pow(10, -3);
double Yr = 2.0031 * pow(10, -3);
double Nv = -14.287 * pow(10, -3);
double Nr = -4.2267 * pow(10, -3);
double Yrud = 1.4308 * pow(10, -3);
double Nrud = -0.71540 * pow(10, -3);

double a1 = m - Yv1;
double b1 = m * xg - Yr1;
double a2 = m * xg - Yr1;
double b2 = Iz - Nr1;
double det = a1 * b2 - a2 * b1;

double q1 = (-Yv * (Iz - Nr1) / det);
double q2 = ((m - Xu1) * u - Yr) * (-m * xg - Yr1) / det;
double q3 = ((Xu1 - Yv1) * u - Nv) * (-m * xg + Yr) / det;
double q4 = ((m * xg - Yr1) * u - Nr)* (m - Yr1) / det;
int main()
{
    while (ti <= tf)
    {
        vo = RKv(ti, vo, h);
        ro = RKr(ti, ro, h);

        v += vo;
        r += ro;
        
        cout << "v = " << v << "    " << "r = " << r << "\n";
        ti = ti + 0.5;
        rud += 0.05;
    }
}

double RKv(double t, double v, double h)
{
    double k1, k2, k3, k4, k5;

    


    for (int i = 0.1; i <= h; i++)
    {
        
        k1 = h * dvdt(t, v);
        k2 = h * dvdt(t + 0.5 * h, v + 0.5 * k1);
        k3 = h * dvdt(t + 0.5 * h, v + 0.5 * k2);
        k4 = h * dvdt(t + h, v + k3);

        
        t = t + h;
        v = v + (1.0 / 6.0) * (k1 + 2 * k2 + 2 * k3 + k4);

        
    }
    return v;
}

double RKr(double t, double r, double h)
{
    double k1, k2, k3, k4, k5;

    for (int i = 1; i <= h; i++)
    {
        
        k1 = h * drdt(t, r);
        k2 = h * drdt(t + 0.5 * h, r + 0.5 * k1);
        k3 = h * drdt(t + 0.5 * h, r + 0.5 * k2);
        k4 = h * drdt(t + h, r + k3);

        
        t = t + h;
        r = r + (1.0 / 6.0) * (k1 + 2 * k2 + 2 * k3 + k4);

    }
    return r;
}


double dvdt(double t, double v)
{
//  return(-Yrud * rud - (-Yv * (Iz - Nr1) / det * v + ((m - Xu1) * u - Yr) * (-m * xg - Yr1) / det * r));
    return(-Yrud * rud - q1 * v + q2 * r);
    
}

double drdt(double t, double r)
{
    //return(-Nrud * rud - (((Xu1 - Yv1) * u - Nv) * (-m * xg + Yr) / det * v + ((m * xg - Yr1) * u - Nr) * (m - Yr1) / det * r));
    return(-Nrud * rud - q3 * v + q4 * r);
    
}

以下是该计划的结果。我得到前几次迭代的数值，之后我得到

-nan(ind)

:

Answer 1

例如，您超出了号码范围。如何解决这个问题，你可以使用“long double”类型而不是常规的“double”