Skydiving Terminal Velocity

The Problem

I like interdependency among subjects, and this is the reason why I wrote this article. Some years ago, I saw an article standing that a skydiving object whose drag force is proportional to the square of the velocity will have its speed constant at some time. This proposition caught my attention. So I've decided to investigate it more.

Skydiving terminal velocity is an interesting question because it involves math and physics. The level of the calculus theory is not too easy neither too hard. It's an optimal point to achieve the flow state. We also need to understand some physics concepts to set up some boundaries conditions that will help us solve some equations.

Some days ago I remembered this problem and I decided to reproof it, this time writing this article. The questions we want to answer is:

When an object is in a skydiving fall, its velocity will be constant after enough time, considering that the drag force is proportional to the square of the speed.

Modeling

The resultant force, $\vec{F_R}$ , acting in a skydiver falling can be expressed by:

\tag{1} \vec{F_R} = \vec{F_g} + \vec{F_d}

where $\vec{F_g}$ is the gravitational force and, $\vec{F_d}$ , the drag force. Let's assume that the direction of the gravitational force is positive and that the magnitude of the drag force is proportional to the square of velocity, than we have:

\tag{2} m \frac{dv}{dt} = mg - kv^2

where $k$ is a constant, related to the drag coefficient, fluid density through which the object is falling, and the projected area of the object.

To solve $(2)$ , we can manipulate the equation to transform it to:

\tag{3} m\frac{dv}{dt} = mg - kv^2 \iff \frac{1}{\left(1 - \frac{k}{mg}v^2\right)}\frac{dv}{dt} = g

Considering $\beta^2 = \cfrac{mg}{k}$ , we can integrate $(3)$ in function of the time in a such way that on $t=0$ , the initial velocity is $v_0$ . Then, we have:

\tag{4} \displaystyle\int_{v_0}^{v} \frac{dv}{1 - \left(\frac{v}{\beta}\right)^2} = g\displaystyle\int_{0}^{t}dt = gt

Now we need to solve $(4)$ .

Hyperbolic Trigonometry

How could we solve $(4)$ ? We can use hyperbolic trigonometry to help us. Hyperbolic trigonometry uses the hyperbola $x^2 - y^2 = 1$ to determine its trigonometric identities, differently from the ordinary trigonometry, that uses the circle $x^2 + y^2 = 1$ .

By Hyperbolic_functions.svg: The original uploader was Marco Polo at English Wikipedia.derivative work: Jeandavid54 (talk) - Hyperbolic_functions.svg, Public Domain, Link

Before applying hyperbolic functions, we need to set up some physics boundaries. In our modeling, we're going to assume that the drag force never has its magnitude greater than the gravitational force. So we can assume, from $(2)$ , that:

m \frac{dv}{dt} = mg - kv^2 \geq 0 \iff v \leq \sqrt \frac{mg}{k} = \beta

Thus,

\tag{5} \frac{v}{\beta} \leq 1

The hyperbolic tangent, $\tanh$ , is defined by:

\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}

If we check its properties, we can see that $\tanh$ is a bijective function, which means that, for every $y \in \, \rbrack -1, 1\lbrack$ , must exist a $x \in \reals$ such as $\tanh x = y$ . If we consider the integral

\tag{6} \displaystyle\int \frac{dv}{1 - \left(\frac{v}{\beta}\right)^2},

we can assume, because of $(5)$ , that must exists a $x$ such that:

\tag{7} \tanh x = \cfrac{v}{\beta}

Deriving $(7)$ in function of $v$ , we have:

\tag{8} \frac{1}{\cosh^2 x} \frac{dx}{dv} = \frac{1}{\beta} \iff dv = \beta \frac{1}{\cosh^2 x} dx

Replacing $(7)$ and $(8)$ in $(6)$ , we have:

\displaystyle\int \frac{dv}{1 - \left(\frac{v}{\beta}\right)^2} = \beta \displaystyle\int \frac{1}{\cosh^2 x} \frac{1}{1 - \tanh^2 x}dx = \beta \displaystyle\int dx = \beta x

Then, considering $(7)$ , we have:

\tag{9} \displaystyle\int \frac{dv}{1 - \left(\frac{v}{\beta}\right)^2} = \beta x = \beta\tanh^{-1}\left(\frac{v}{\beta}\right)

Finally, applying integral limits in $(9)$ , we have:

\tag{10} \displaystyle\int_{v_0}^{v} \frac{dv}{1 - \left(\frac{v}{\beta}\right)^2} = \beta\tanh^{-1}\left(\frac{v}{\beta}\right) - \beta\tanh^{-1}\left(\frac{v_0}{\beta}\right)

Because of our assumption in $(5)$ , we have that $v_0$ won't have its value greater than $\beta$ .

The Terminal Velocity

If we substitute $(10)$ in $(4)$ and isolate $v$ , we have the value of the velocity in function of the time:

\tag{11} v = \beta\tanh \left(\sqrt\frac{kg}{m}t + \tanh^{-1}\left(\frac{v_0}{\beta}\right) \right)

Our question is, "What is the velocity when the time is sufficiently large?" To answer this question, let's see what happens when $t$ becomes large. Applying the limit on both sides of $(11)$ , we have:

\tag{12} v_t = \lim\limits_{t \rightarrow \infty }v = \beta \lim\limits_{t \rightarrow \infty}\tanh \left(c_1t + c_2\right) = \beta = \sqrt\frac{mg}{k}

where $v_t$ is the terminal velocity, and constants $c_1 = \sqrt\frac{kg}{m}$ and $c_2 = \tanh^{-1}\left(\frac{v_0}{\beta}\right)$ .

Physics Analysis

From $(12)$ we can conclude that after enough time, the velocity of the skydiver becomes constant, which was to be demonstrated. Furthermore, if the velocity is constant, the acceleration must be equal $0$ . If we take $(2)$ and assumes that $\cfrac{dv}{dt} = 0$ , then we have:

0 = mg - kv_t^2 \iff v_t = \sqrt\frac{mg}{k},

that is the same value we find on $(12)$ .

Lastly, if we check this "Speed skydiving" Wikipedia article, we can see that their definition of the terminal velocity is

v_t = \sqrt\frac{2mg}{\rho A C_d},

where $C_d$ is the drag coefficient, $\rho$ the fluid density through which the object is falling, and $A$ the projected area of the object. If we compare both terminal velocities equations, we conclude that our $k$ is defined by:

k = \frac{\rho A C_d}{2}.