在仿真中实现碰撞响应

Question

我正在尝试在我正在创建的模拟中实现碰撞响应。 基本上，该程序模拟了一个球以一定的初始速度从 50 米的建筑物中抛出。

我不相信该程序会输出实际的碰撞时间值以及 x、y 和 vx、vy 的值。

这是程序：

 #include<stdio.h>
 #include<math.h>
 #include<stdlib.h>

 int main() {

     FILE *fp; 
     FILE *fr;

     //Declare and initialize all variables to be used
     float ax = 0, ay = 0, x = 0, y = 0, vx = 0, vy = 0; 
     float time = 0, deltaTime = .001; 
     float vyImpact = 0, vxImpact = 0, xImpact = 0; 

     float old_y = 0,  old_x = 0, old_vy = 0, old_vx = 0;
     float deltaTime2 = 0,  deltaTime3 = 0;

     int numBounces = 0;

     //Coefficient of Restitution; epsilon = ex = ey
     float ex = .5;
     float ey = .5;

     fr = fopen("input_data.txt", "rt"); //Open file for reading

     fp = fopen( "output_data.txt", "w" ); // Open file for writing

     if(fr == NULL){ printf("File not found");} //if text file is not in directory...

     if(fp == NULL){ printf("File not found");} //if text file is not in directory...

     fscanf(fr, "ax: %f ay: %f x: %f y: %f vx: %f vy: %f\n", &ax, &ay, &x, &y, &vx, &vy); 

     while (numBounces < 9) {

          //time = time + deltaTime
          time = time + deltaTime;

          //velocity[new] = velocity[old] + acc * deltaTime
          vx = vx + ax*deltaTime;
          vy = vy + ay*deltaTime;

          //position[new] = position[old] + velocity*deltaTime + .5*acc*(deltaTime)^2
          x = x + vx*deltaTime + (.5*ax*deltaTime*deltaTime);
          y = y + vy*deltaTime + (.5*ay*deltaTime*deltaTime);  

          fprintf(fp, "%f\t%f\t%f\t%f\t%f\t%f\t%f\t\n", ax, ay, x, y, vx, vy, time);

               //Collision occurs; implement collision response
              if (y < 0) {

                   //"Undo" values for y, x, and velocity
                   old_y = y - vy*deltaTime - (.5*ay*deltaTime*deltaTime); 
                   old_x = x - vx*deltaTime - (.5*ax*deltaTime*deltaTime); 
                   old_vy = vy - ay*deltaTime;
                   old_vx = vx - ax*deltaTime; 

                   //Calculate time of collision
                   deltaTime2 = (-old_y + sqrt((old_y*old_y) - 2*ay*old_y)) / (ay);
                   printf("Time of Collision = %f\n", time - deltaTime2); 

                   //Calculate velocity and x position at collsion
                   vyImpact = old_vy + ay*deltaTime2;
                   vxImpact = old_vx + ax*deltaTime2;
                   xImpact = old_x + old_vx*deltaTime2 + .5*ax*(deltaTime2*deltaTime2);

                   //Calculate new time for when ball bounces
                   deltaTime3 = deltaTime - deltaTime2;

                   //Calculate new x and y position and velocity for when ball bounces
                   x = xImpact + (ex)*vxImpact*deltaTime3 + .5*ax*(deltaTime3*deltaTime3);
                   y = 0 + (-ey)*vyImpact*deltaTime3 + .5*ay*(deltaTime3*deltaTime3);
                   vy = (-ey)*vyImpact + ay*deltaTime3;
                   vx = (ex)*vxImpact + ax*deltaTime3; 

                   numBounces++; 
                   printf("Number of Bounce(s) = %d\n", numBounces);


                   fprintf(fp, "%f\t%f\t%f\t%f\t%f\t%f\t%f\t\n", ax, ay, x, y, vx, vy, time);

               }
     }

     fclose(fp); //Close output file
     fclose(fr); //Close input file

     //system ("PAUSE"); 
     return 0;
  }

基本上，我正在尝试生成准确的值，以便我可以看到这个模拟应该是什么样子的图。我假设逻辑错误与物理学有关。但是由于我的物理知识有限，我无法看出到底是哪里出了问题。

这是示例输入： ax: 0 ay: -9.8 x: 0 y: 50 vx: 8.66 vy: 5

Answer 1

在我看来，您的问题可能在于您如何实现运动学方程。

//velocity[new] = velocity[old] + acc * deltaTime
vx = vx + ax*deltaTime;
vy = vy + ay*deltaTime;

//position[new] = position[old] + velocity*deltaTime + .5*acc*(deltaTime)^2
x = x + vx*deltaTime + (.5*ax*deltaTime*deltaTime);
y = y + vy*deltaTime + (.5*ay*deltaTime*deltaTime);

这里有两件事：您已经在 vx 和 vy 的方程中考虑了加速度，并且您使用的是求和而不是积分方程。不应包含 .5*ax*deltaTime*deltaTime 和 .5*ay*deltaTime*deltaTime。当基于速度方程的积分计算在总时间内由于恒定加速度而行进的距离时，使用方程 x= 0.5*a*t^2。由于您正在进行求和并且已经在速度方程中包含加速度，因此无需在位置方程中包含加速度。