How to solve the following question using the provided Runge-Kutta method in python

Question

The question body:

A skydiver of mass m in a vertical free fall experiences an aerodynamic drag force F=cy'² (' c times y prime square') where y is measured downward from the start of the fall, and y is a function of time ( y' denotes the derivative of y wrt time). The differential equation describing the fall is:

y''=g-(c/m)y'²

(where g = 9.80665 m/s^2; c = 0.2028 kg/m; m = 80 kg). And y(0)=y'(0)=0 as this is a free fall.

Task: The function must return the time of a fall of x meters, where x is the parameter of the function. The values of g , c and m are given below.

The Runge-Kutta function is defined as follows:

from numpy import *
def runge_kutta_4(F, x0, y0, x, h):
   '''
   Return y(x) given the following initial value problem:
   y' = F(x, y)
   y(x0) = y0 # initial conditions
   h is the increment of x used in integration
   F = [y'[0], y'[1], ..., y'[n-1]]
   y = [y[0], y[1], ..., y[n-1]]
   '''
   X = []
   Y = []
   X.append(x0)
   Y.append(y0)
   while x0 < x:
       k0 = F(x0, y0)
       k1 = F(x0 + h / 2.0, y0 + h / 2.0 * k0)
       k2 = F(x0 + h / 2.0, y0 + h / 2 * k1)
       k3 = F(x0 + h, y0 + h * k2)
       y0 = y0 + h / 6.0 * (k0 + 2 * k1 + 2.0 * k2 + k3)
       x0 += h
       X.append(x0)
       Y.append(y0)
   return array(X), array(Y)

And this is what I've done so far:

def prob_1_8(x)
    g = 9.80665  # m/s**2
    c = 0.2028  # kg/m
    m = 80  # kg

    def F(x, y):
        return array([
            y[1],
            g - (c / m) * ((y[1]) ** 2)
        ])

    X, Y = runge_kutta_4(F, 0, array([0, 0]), 5000, 1000)
    for i in range(len(X)):
        if X[i] == 5000:
            return Y[i]

However, when I tried to print prob_1_8(5000), the number looks ridiculous and it displayed:

RuntimeWarning: overflow encountered in double_scalars. 

According to the answer provided, I should get a value close to 84.8 when x=5000 . Can someone help me with this? I don't know what's the problem and how to fix it.

Answer 1

Please contemplate the function call of X, Y = runge_kutta_4(F, 0, array([0, 0]), 5000, 1000) . You are integrating over a time span of 5000 sec > 1 hour in steps of 1000 sec > 16 min . It is intuitively clear that this will be imprecise, as most of the acceleration will happen in the first 10 sec.

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Then the question is what exactly you are trying to filter out with the loop. Is it the speed after this time?

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The limit speed is where the right side is zero, at vmax=sqrt(g*m/c) = 62.1972 = 223.91 km/h , the claimed value of 84.8 can not be reached as a speed starting from rest. The fall time to a distance of x will be a little more than x/vmax , so you could use tmax = 100+x/vmax in T, Y = runge_kutta_4(F, t0, y0, tmax, 1) .

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Integrating in 1 sec time steps and looking for the speed after 5000 meters fall distance gives a result of 85 sec , distance 5013.33465614 m , speed 62.1972 m/s which is as to be expected close to the limit speed.

You can get a more precise time value by using (reverse) linear interpolation, then at time about 84.786 sec you reach distance 5000 m with a speed 62.1972 m/s . This again is compatible with the claimed result value, which now is a time, not a velocity.