Simulation of a point mass in a box (3D space) - c++

Question

I would like to simulate a point mass within a closed box. There is no friction and the point mass obeys the impact law. So there are only elastic collisions with the walls of the box. The output of the program is the time, position (rx,ry ,rz) and velocity (vx,vy,vz). I plot the trajectory by using GNUplot.
The problem I have now is, that the point mass gets energy from somewhere. So their jumps get each time more intense.
Is someone able to check my code?
/* Start of the code */
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
struct pointmass
{
double m; // mass
double r[3]; // coordinates
double v[3]; // velocity
};
// Grav.constant
const double G[3] = {0, -9.81, 0};
int main()
{
int Time = 0; // Duration
double Dt = 0; // Time steps
pointmass p0;
cerr << "Duration: ";
cin >> Time;
cerr << "Time steps: ";
cin >> Dt;
cerr << "Velocity of the point mass (vx,vy,vz)? ";
cin >> p0.v[0];
cin >> p0.v[1];
cin >> p0.v[2];
cerr << "Initial position of the point mass (x,y,z)? ";
cin >> p0.r[0];
cin >> p0.r[1];
cin >> p0.r[2];
for (double i = 0; i<Time; i+=Dt)
{
cout << i << setw(10);
for (int j = 0; j<=2; j++)
{
////////////position and velocity///////////
p0.r[j] = p0.r[j] + p0.v[j]*i + 0.5*G[j]*i*i;
p0.v[j] = p0.v[j] + G[j]*i;
///////////////////reflection/////////////////
if(p0.r[j] >= 250)
{
p0.r[j] = 500 - p0.r[j];
p0.v[j] = -p0.v[j];
}
else if(p0.r[j] <= 0)
{
p0.r[j] = -p0.r[j];
p0.v[j] = -p0.v[j];
}
//////////////////////////////////////////////
}
/////////////////////Output//////////////////
for(int j = 0; j<=2; j++)
{
cout << p0.r[j] << setw(10);
}
for(int j = 0; j<=2; j++)
{
cout << p0.v[j] << setw(10);
}
///////////////////////////////////////////////
cout << endl;
}
}

Answer 1

F = ma
a = F / m
a dt = F / m dt
a dt is acceleration over a fixed time - the change in velocity for that frame.
You are setting it to F / m i
it is that i which is wrong, as comments have suggested. It needs to be the duration of a frame, not the duration of the entire simulation so far.

Answer 2

I am a little concerned about the time loop along with other commenters - make sure that it represents an increment of time, not a growing duration.
Still, I think the main problem is you are changing the sign of all three components of velocity
on reflection.
That's not consistent with the laws of physics -conservation of linear momentum and energy - at the boundaries.
To see this, consider the case if your particle is moving in just the x-y plane (velocity in z is zero) and about to hit the wall at x= L.
The collision looks like this:
The force exerted on the point mass by the wall acts perpendicular to the wall. So there is no change in the momentum component of the particle parallel to the wall.
Applying conservation of linear momentum and kinetic energy, and assuming a perfectly elastic collision, you will find that
The component of velocity perpendicular to the wall DOES change sign
The component of velocity parallel to the wall DOES NOT change sign
In three dimensions, to have an accurate simulation, you have to work out the momentum components parallel and perpendicular to the wall on collision and code the resulting velocity changes.
In other words, this code:
///////////////////reflection/////////////////
if(p0.r[j] >= 250)
{
p0.r[j] = 500 - p0.r[j];
p0.v[j] = -p0.v[j];
}
else if(p0.r[j] <= 0)
{
p0.r[j] = -p0.r[j];
p0.v[j] = -p0.v[j];
}
//////////////////////////////////////////////
does not model the physics of reflection correctly. To fix it here is an outline of what to do:
Take the reflection checks out of the loop over x,y,z coordinates (but still within the time loop)
The collision condition for all six walls needs to be checked,
according to the direction of the normal vector to the wall.
For example for the right wall of the cube defined by X=250, 0<=Y<250, 0<=Z<250, the normal vector is in the negative X direction. For the left wall defined by X=0, 0<=Y<250, 0<=Z<250, the normal vector is in the positive X direction.
So on reflection from those two walls, the X component of velocity changes sign because it is normal (perpendicular) to the wall, but the Y and Z components do NOT change sign because they are parallel to the wall.
Apply similar considerations at the top and bottom wall (constant Y), and front and back wall (constant Z), of the cube -left as exercise to work out the normals to those surfaces.
Finally you shouldn't change sign of the position vector components on reflection, just the velocity vector. Instead recompute the next value of the position vector given the new velocity.

Answer 3

OK, so there are a few issues. The others have pointed out the need to use Dt rather than i for the integration step.
However, you are correct in stating that there is an issue with the reflection and energy conservation. I've added an explicit track of that below.
Note that the component wise computation of the reflection is actually fine other than the energy issue.
The problem was that during a reflection the acceleration due to gravity changes. In the case of the particle hitting the floor, it was acquiring kinetic energy equal to that it would have had if it had kept falling, but the new position had higher potential energy. So the energy would increase by exactly twice the potential energy difference between the floor and the new position. A bounce off the roof would have the opposite effect.
As noted below, once strategy would be to compute the actual time of reflection. However, actually working directly with energy is much simpler as well as more robust. However, please note although the the simple energy version below ensures that the speed and position are consistent, it actually does not have the correct position. For most purposes that may not actually matter. If you really need the correct position, I think we need to solve for the bounce time.
/* Start of the code */
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
struct pointmass
{
double m; // mass
double r[3]; // coordinates
double v[3]; // velocity
};
// Grav.constant
const double G[3] = { 0, -9.81, 0 };
int main()
{
// I've just changed the initial values to speed up unit testing; your code worked fine here.
int Time = 50; // Duration
double Dt = 1; // Time steps
pointmass p0;
p0.v[0] = 23;
p0.v[1] = 40;
p0.v[2] = 15;
p0.r[0] = 100;
p0.r[1] = 200;
p0.r[2] = 67;
for (double i = 0; i<Time; i += Dt)
{
cout << setw(10) << i << setw(10);
double energy = 0;
for (int j = 0; j <= 2; j++)
{
double oldR = p0.r[j];
double oldV = p0.v[j];
////////////position and velocity///////////
p0.r[j] = p0.r[j] + p0.v[j] * Dt + 0.5*G[j] * Dt*Dt;
p0.v[j] = p0.v[j] + G[j] * Dt;
///////////////////reflection/////////////////
if (G[j] == 0)
{
if (p0.r[j] >= 250)
{
p0.r[j] = 500 - p0.r[j];
p0.v[j] = -p0.v[j];
}
else if (p0.r[j] <= 0)
{
p0.r[j] = -p0.r[j];
p0.v[j] = -p0.v[j];
}
}
else
{
// Need to capture the fact that the acceleration switches direction relative to velocity half way through the timestep.
// Two approaches, either
// Try to compute the time of the bounce and work out the detail.
// OR
// Use conservation of energy to get the right speed - much easier!
if (p0.r[j] >= 250)
{
double energy = 0.5*p0.v[j] * p0.v[j] - G[j] * p0.r[j];
p0.r[j] = 500 - p0.r[j];
p0.v[j] = -sqrt(2 * (energy + G[j] * p0.r[j]));
}
else if (p0.r[j] <= 0)
{
double energy = 0.5*p0.v[j] * p0.v[j] - G[j] * p0.r[j];
p0.r[j] = -p0.r[j];
p0.v[j] = sqrt(2*(energy + G[j] * p0.r[j]));
}
}
energy += 0.5*p0.v[j] * p0.v[j] - G[j] * p0.r[j];
}
/////////////////////Output//////////////////
cout << energy << setw(10);
for (int j = 0; j <= 2; j++)
{
cout << p0.r[j] << setw(10);
}
for (int j = 0; j <= 2; j++)
{
cout << p0.v[j] << setw(10);
}
///////////////////////////////////////////////
cout << endl;
}
}