Question

我正在尝试计算方程式描述的速度的时程： dV / dt = g −（C_d / m）* V ^ 2 。 g = 9.81，m = 1.0，C_d = 1.5。为此，我需要在c ++中创建一个程序，该程序使用Euler显式方法对方程进行数值求解。我试图以三种不同的步长delta_t = 0.05、0.1和0.2秒找到从t = 0到t = 1秒的速度。然后，您应该按照以下公式显示分析解决方案的百分比误差： V（t）= sqrt（（m * g）/ C_d）* tanh（sqrt（（g * C_d）/ m）* t ）。 我的问题是我不确定如何在不同的时间间隔内多次遍历欧拉方法。到目前为止，我已经解决了解析方程，但是不确定从这里开始。如果有人能帮助我指出正确的方向，将不胜感激。

#include <iomanip>
#include <cmath>
#include <math.h>

using namespace std;

int main() {

    double m = 1.0;     // units in [kg] 
    double g = 9.81;    // units in [m/s^2]
    double C_d = 1.5;   // units in [kg/m]  
    double t;           // units in [s]
    double v;           // units in [m/s]


    cout << "The velocity will be examined from the time t = 0 to t = 1 seconds." << endl;
    cout << "Please select either 0.05, 0.1, or 0.2 to be the time interval:" << endl;
    cin >> t;
    cout << "You have chosen the time interval of: " << t << " seconds." << endl;


    v = sqrt((m * g) / C_d) * tanh(sqrt((g * C_d) / m) * t);

    cout << "The velecity at a time of "<< t << " seconds is equal to: " << v << " m/s." << endl;


    return 0;

} ```

Answer 1

如果要以 A 的增量迭代 t ，并用每个 t 计算公式的结果，则可以编写一个for循环。

#include <iostream>

int main()
{
    double m = 1.0;     // units in [kg] 
    double g = 9.81;    // units in [m/s^2]
    double C_d = 1.5;   // units in [kg/m]

    std::cout << "The velocity will be examined from the time t = 0 to t = 1 seconds." << std::endl;
    std::cout << "Please select the time interval:" << std::endl;
    std::cout << "1: 0.05" << std::endl;
    std::cout << "2: 0.1" << std::endl;
    std::cout << "3: 0.2" << std::endl;

    double A = 0; // increment in for loop
    int x;
    std::cin >> x;
    switch (x) { // check what the input is equal to
        case 1: A = 0.05; break;
        case 2: A = 0.1; break;
        case 3: A = 0.2; break;
        default: std::cout << "Unknown option!" << std::endl; return 1;
    }
    std::cout << "You have chosen the time interval of: " << A << " seconds." << std::endl;

    std::cout << "Results of V(t):" << std::endl;
    // this initializes a variable t as 0, 
    //and while t is lower than or equal to 1, 
    //it will increment it by a and execute the logic within the scope of the loop.
    for (double t = 0; t < (1 + A); t += A) {
        std::cout << "at t = " << t << ": " << sqrt((m*g) / C_d) * tanh(sqrt((g*C_d) / m) * t) << std::endl;
    }

    return 0;
}

有关更多信息，请参见https://beginnersbook.com/2017/08/cpp-for-loop/。注意：我还在代码中引入了switch语句，以防止输入未知值。 https://beginnersbook.com/2017/08/cpp-switch-case/

使用欧拉方法创建一个C ++程序来求解运动方程

1 个答案: