我正在尝试编写一个代码，该代码给出1D框中N个粒子的质量，速度，位移和半径将以0.03s的小增量返回它们的位置。 该计划符合。但是不会产生正确的值。
#include <iostream>
#include <cmath>
#include <fstream>
#include <vector>
using namespace std;
struct particle {
double x; //position of particle
double im; //inverse mass
double v; //velocity of particle
double r; //radius of particle
};
void bonkers(vector<particle> &array, int NN, double t, double ddt, int T);
double time(vector<particle> &a, int NN, int &whichn);
void collision(particle &a, particle &b);
void position(particle &a, double ddt);
void position(particle &a, double ddt)//position of particles
{
a.x += ddt * a.v; //position = t * velocity
}
void collision(particle &a, particle &b)//velocities after collision
{
double realativeV = a.v - b.v;
double af = a.im / (a.im + b.im);
double bf = b.im / (a.im + b.im);
a.v -= 2.0 * af * realativeV;
b.v += 2.0 * bf * realativeV;
}
double time(vector<particle> &a, int NN, int &whichn)//Finds time of first collision.
{
double dt = 1e100; //really large number
for (int n = 0; n < NN - 1 ; n++)
{
double RelativeV = a[n].v - a[n + 1].v;
if (RelativeV > 0)//this means dt won't be 0, no negative times
{
double Collisiont = ((a[n + 1].x - a[n + 1].r) - (a[n].x + a[n].r))/ RelativeV;//gives time of collision
//time of collision worked out as time when the distance between the two balls centre is equal to their combined radi
if (Collisiont < dt)//finds smallest possible time of collision
{
dt = Collisiont; //returns time of first collision
whichn = n; //gives which particles collide. Therefore which particles velocities need to change
}
}
}
return dt;
}
void bonkers(vector<particle> &array, int NN, double t, double ddt, int T)
{
ofstream myfile;
myfile.open("example.txt");//sends all information to file called example.txt
int whichn = 0;
double k;
double l = 0;
for (int i = 0; i <= T; i++)//set arbitrary number of collision i.e. T
{
k = t;//after every collision k reset to 0 as time function works from 0
while( k <= time(array, NN, whichn) )
{
myfile << l; //prints the time
k += ddt; //incriments time
l += ddt; //increments time, but doesn't reset time so gives total time
for (int n = 1; n < NN -1 ; n++)
{
position(array[n], ddt);//calculates new position
myfile << "\t" << array[n].x;//prints the positions of all the particles at said time
}
myfile << endl;
}
collision(array[whichn], array[whichn + 1]);//asings new velocities to the two particles which collided
}
myfile.close();
}
int main()
{
int N; //number of balls
double m; //mass of balls
int T = 50; //number of collisions
double t = 0.0; //starts from 0 time
double ddt = 0.03; //invrements by 0.03 seconds each time
cout << "Input number of balls" << endl;
cin >> N;
vector<particle> a(N + 2); //extra two for the two walls, wall1 at 0 and wall2 at 20
for (int k = 1; k <= N; k++)
{
cout << "Input mass of ball " << k << endl;
cin >> m;
a[k].im = 1.0 / m; //finds inverse mass
cout << "Input position of ball " << k << " between 0 and 20" << endl;
cin >> a[k].x;
cout << "Input speed of ball " << k << endl;
cin >> a[k].v;
cout << "Input radius of ball " << k << endl;
cin >> a[k].r;
}
//asign wall variables
a[0].im = 0; //infinte mass, so walls don't move
a[0].r = 0;
a[0].x = 0; //wall1 at 0
a[0].v = 0;
a[N + 1].im = 0;
a[N + 1].r = 0;
a[N + 1].x = 20; //wall2 at 20
a[N + 1].v = 0;
bonkers(a, N + 2, t, ddt, T);
return 0;
}
当你只有一个球时，有一个非常明显的问题。半径，质量和速度都是1。 由于某种原因，球的速度在位置10.03处变化，而不是在20处弹出壁。
我认为问题出在代码的“时间”功能部分。 通过在bonkers函数中输入“k”到example.txt而不是“l”，这个问题变得更加明显。
非常感谢：）
答案 0 :(得分：0)
问题在于这一行。
while( k <= time(array, NN, whichn) )
您在while循环的每次迭代中重新计算碰撞时间，并且还会延长经过的时间k，然后比较它们。事件和经过时间之间的时间比较没有意义。你可以计算碰撞时间和增加时间k，直到你到达那里，或者你可以重新计算碰撞时间并在每次迭代中与0比较。