by Paulo Marques, 2013/09/23

This notebook discusses the drag forces exerted on a body when traveling through air.

A falling body is subject to two forces: a downward force, $\vec{F_g}$, due to gravity, and an upward force $\vec{F_a}$, due to air resistance. Let's consider a body with mass $m$. Assuming an upward oriented YY axis, the following applies:

$$ F_g = - g \cdot m $$$$ F_a = - sgn(v) \cdot {1 \over 2} \rho C_D A v^2 $$In these formulas, $m$ is the mass of the body, $\rho$ is the density of air ($\approx 1.2 Kg/m^3$), $g$ is the accelaration of gravity ($\approx 9.8 m/s^2$), $v$ is the velocity of the body, $C_D$ is the drag coeficient of the body and $A$ is its cross-section area. $sgn(x)$ is the sign function that given a number returns +1 for positives and -1 for negatives. In the formula it makes the force always contrary to the movement of the body.

Thus, the resulting force $\vec{F}$ experienced by the body is:

$$ F = - g \cdot m - sgn(v) \cdot {1 \over 2} \rho C_D A v^2 $$Given that force is directly related to accelaration, and accelaration to velocity:

$$ v = \frac{dy}{dt} $$$$ F = m \cdot \frac{dv}{dt} $$substituting, we get this differencial equation:

$$ m \cdot \frac{dv}{dt} = - g \cdot m - sgn(v) \cdot {1 \over 2} \rho C_D A v^2 $$In fact, this can be expressed as a system of differential equations in the form ${\bf f}' = {\bf g}({\bf f}, t)$:

$$ \begin{cases} \frac{dy}{dt} = v \\\\ \frac{dv}{dt} = -g - sgn(v) \cdot {1 \over 2} {\rho \over m} C_D A v^2 \end{cases} $$So, let's simulate this with Python!

Let's start by importing some basic libraries:

In [1]:

```
%pylab inline
from scipy.integrate import odeint
from math import sqrt, atan
```

We now define the initial conditions and constants of the problem.

In [2]:

```
# Constants
g = 9.8 # Accelaration of gravity
p = 1.2 # Density of air
# Caracteristics of the problem
m = 0.100 # A 100 g ball
r = 0.10 # 10 cm radius
Cd = 0.5 # Drag coeficient for a small spherical object
y0 = 1000.0 # Initial height of the body (1000 m)
v0 = 10.0 # Initial velocity of the body (10 m/s^2, going up)
A = math.pi*r**2 # Cross-section area of the body
```

As said, let's define a system of ordinary differential equations in its normal form ${\bf f}' = {\bf g}({\bf f}, t)$. (In the code bellow we substitute $g()$ for $gm()$ so that it doesn't clash with the acceleration of gravity constant - $g$).

In [3]:

```
sgn = lambda x: math.copysign(1, x) # Auxiliary function to calculate the sign of a number
def gm(f, t):
(y, v) = f # Extract y and v (i.e., dy/dt) from the f mapping
dy_dt = v # The differential equations
dv_dt = -1.0*g - sgn(v)*(1./2.)*(p/m)*Cd*A*v**2
return [dy_dt, dv_dt] # Return the derivatives
```

Let's define the conditions to numerically solve the problem, including a time vector:

In [4]:

```
# Initial conditions (position and velocity)
start = [y0, v0]
# Time vector (from 0 to 5 secs)
tf = 5.0
t = linspace(0, tf, tf*100)
```

Now let's solve the equations numericaly and extract the corresponding $y(t)$ and $v(t)$:

In [5]:

```
f = odeint(gm, start, t)
y = f[:, 0]
v = f[:, 1]
```

Finally, we can plot the solution.

In [6]:

```
figure(figsize=(14, 6))
subplot(1, 2, 1, title='Velocity over time')
xlabel('Time (sec)')
ylabel('Velocity (m/sec)')
plot(t, v)
subplot(1, 2, 2, title='Height over time')
xlabel('Time (sec)')
ylabel('Height (m)')
plot(t, y)
```

Out[6]:

As you can see, the velocity starts at 10 $m/s^2$, with the ball going up. Its velocity starts decreasing, goes to zero at max height, and then becomes negative as the ball starts coming down. After a while it reaches its maximum speed: terminal velocity.

The theorerically terminal velocity, $V_t$, is:

$$ V_t = \sqrt{\frac{2 m g}{\rho A C_D}} $$Calculating it is easy:

In [7]:

```
vt = sqrt( (2.*m*g) / (p*A*Cd) )
vt
```

Out[7]:

Now, with our numerical simulation, the terminal velocity is:

In [8]:

```
# The terminal velocity
vt_numeric = abs(min(v))
vt_numeric
```

Out[8]:

Now, that's what I call pretty neat!

Copyright (C) 2013 Paulo Marques ([email protected])

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