This notebook discusses the drag forces exerted on a body when traveling through air. It's also done in Julia so that I can better understand the potencial of the language.

Image courtesy of WikiMedia.

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 Y 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 ${\bf f}' = {\bf g}({\bf f}, t)$:

$$ \begin{cases} \frac{dy}{dt} = v \\\\ \frac{dv}{dt} = -g - sgn(v) \cdot {1 \over 2} {\rho \over m} C_D A v^2 \end{cases} $$In [1]:

```
using ODE;
```

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

In [2]:

```
# Constants
const g = 9.8 # Accelaration of gravity
const p = 1.2 # Density of air
# Caracteristics of the problem
const m = 0.100 # A 100 g ball
const r = 0.10 # 10 cm radius
const Cd = 0.5 # Drag coeficient for a small spherical object
const y0 = 1000.0 # Initial height of the body (1000 m)
const v0 = 10.0 # Initial velocity of the body (10 m/s^2, going up)
const A = 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 - $g$). Note that $f$ is an $R^2$ function containing $y$ and $y'$ (and $v = y'$).

In [3]:

```
function gm(t, f)
(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 - sign(v)*(1./2.)*(p/m)*Cd*A*v^2.0
[dy_dt; dv_dt] # Return the derivatives
end;
```

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

In [4]:

```
# Initial conditions (position and velocity)
const start = [y0; v0]
# Time span (from 0 to 5 secs)
ts = [0.0; 5.0];
```

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

In [5]:

```
t, res = ode45(gm, start, ts)
y = map(x -> x[1], res)
v = map(x -> x[2], res);
```

Finally, we can plot the solution.

In [6]:

```
using PyPlot
fig, ax = subplots(1, 2, sharex=true, figsize=(10,4))
ax[1][:plot](t, v)
ax[1][:set_title]("Velocity over time");
ax[1][:set_xlabel]("Time (sec)")
ax[1][:set_ylabel]("Velocity (m/sec)")
ax[2][:plot](t, y)
ax[2][:set_title]("Height over time");
ax[2][:set_xlabel]("Time (sec)")
ax[2][:set_ylabel]("Height (m)");
```

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 enough:

In [7]:

```
vt = sqrt( (2.0*m*g) / (p*A*Cd) )
```

Out[7]:

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

In [8]:

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

Out[8]:

The two agree.