Tuesday, 13 September 2016

Some physical considerations on the dynamics of a skydiver

Recently, a friend went skydiving and, me being me, the first thing I could think about was making some physical considerations on his adventure :)

If you think of a falling object, at first you would think of it as falling with a constant acceleration of g. That is, you would neglect air drag. However, if you think of it, air drag is not exactly neglectable when describing the fall of a skydiver. If you neglect air drag, you would get an ever increasing speed which is not at all the case.

Let’s make some physical considerations:

We’ll assume that the only two forces acting on the skydiver are the force of gravity and the air drag. This is the resulting free body diagram:

Image 1

 

If we apply Newton’s Law, we obtain the following equation:

$$m\overrightarrow{a} = m\overrightarrow{g} + \overrightarrow{F}_{drag}$$

which could be simplified, considering the $y$ direction only

$$m\ddot{y} = mg - \frac{1}{2}\rho A C \dot{y}^2$$

where $rho$ is the air density, $A$ is the surface of the skydiver facing the stream of air, $C$ is the drag coefficient, $\ddot{y}$ is the acceleration on the $y$ axis and $\dot{y}$ is the speed on the $y$ axis.

Of course this is not an entirely accurate description of what is actually happening either. For instance, air density and drag coefficient are not constant (I’ll assume both to be constant though for simplicity). In particular, as far as the drag coefficient is concerned, we’ll assume that the skydiver is facing the earth and approximate them as a flat surface.

First of all, note that  acceleration is not constant. In fact, it gets smaller and smaller up until the drag force is equal to the force of gravity. At that point the acceleration is zero and our skydiver is falling at a constant speed. If we had neglected air friction, we would have reached a very different conclusion.

What terminal speed can we expect? Well, just set the acceleration to be zero and solve for the constant $\dot{y}_{eq}$:

$$\dot{y} _{eq}= \sqrt{\frac{2mg}{\rho AC}}$$

which evaluates to about 47.86 m/s (or about 172.29 km/h) when using the data showed further down, in the code. This result seems plausible and indeed it is very close to the actual value of about 180 km/h for a free falling skydiver.

What about the solution to this differential equation though? Using Wolfram Alpha one finds out the solution:

$$y = \frac{2m}{\rho AC}\ln \left [ {\cosh{\sqrt{ \left ( \frac{\rho ACg}{2m} \right )} t}} \right ]$$

which, by the way, would have been a little painful to find out by hand.

Using the power of Python we can find the speed by calculating (numerically) the derivative of the position, then we can use the first equation to find the acceleration, as follows:

$$ \ddot{y} = g - \frac{1}{2}\frac{\rho AC}{m}\dot{y}^2 $$

Using the values found on Wikipedia for the air density etc, this is the result of the simulation of a free fall of 20 seconds:

figure_1

As you can see, most of the action is happening around the first 10 seconds, then the acceleration goes almost to zero, the speed becomes constant and the position becomes proportional to time. By the way, another way of assessing that indeed speed becomes constant is by checking that $\lim_{t \rightarrow \infty  } \frac{y(t)}{t}$ is constant.

If we zoom in the first 5 seconds, we can get a better view of the transient phase

figure_2

Note how the simulation is consistent with the assumptions of initial acceleration of g, initial vertical speed equal to 0. Then note how the position starts out with a parabola-shaped curve but then tends to a line (since the acceleration tends to zero).

This model is not that accurate when increasing the mass of the diver. For instance when skydiving in tandem, two people, usually an instructor and a student, skydive together. In that case the total mass approximately doubles and the maximum speed achievable should be around 320 km/h. If you try to plug in the numbers for this scenario you get about 270 km/h which is short of about 50 km/h. Nevertheless, the model is quite accurate for modelling a single skydiver and could be refined for describing the tandem skydivers by taking into account other relevant details.

Below you can find the code I used for the simulation:

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