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，该程序模拟一个球从一个50米的建筑物抛出一些初始速度。

我不相信程序正在输出碰撞时间的实际值以及x，y和vx，vy的值。

以下是该计划：

 #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有关。但由于我的物理知识有限，我无法看出究竟是什么问题。

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

Answer 1

在我看来，你的问题可能在于你如何实现运动学方程式。

//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：你已经在vx和vy的等式中考虑了加速度，而你使用的是求和而不是积分方程式。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由于总时间量的恒定加速度而行进的距离时，使用等式x = 0.5 * a * t ^ 2。当您进行求和并且已经包含速度方程中的加速度时，不需要在位置方程中包括加速度。

在仿真中实现碰撞响应

1 个答案: