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Skydiving Terminal Velocity

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.
#calculus
Articles, May 23, 2021 (changes)
Reading time: 5 minutes
Photo by Vincentiu Solomon on Unsplash

Photo by Vincentiu Solomon on Unsplash

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, FR\vec{F_R}, acting in a skydiver falling can be expressed by:

FR=Fg+Fd(1)\tag{1} \vec{F_R} = \vec{F_g} + \vec{F_d}

where Fg\vec{F_g} is the gravitational force and, Fd\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:

mdvdt=mgkv2(2)\tag{2} m \frac{dv}{dt} = mg - kv^2

where kk 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)(2), we can manipulate the equation to transform it to:

mdvdt=mgkv2    1(1kmgv2)dvdt=g(3)\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 β2=mgk\beta^2 = \cfrac{mg}{k}, we can integrate (3)(3) in function of the time in a such way that on t=0t=0, the initial velocity is v0v_0. Then, we have:

v0vdv1(vβ)2=g0tdt=gt(4)\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)(4).

Hyperbolic Trigonometry

How could we solve (4)(4)? We can use hyperbolic trigonometry to help us. Hyperbolic trigonometry uses the hyperbola x2y2=1x^2 - y^2 = 1 to determine its trigonometric identities, differently from the ordinary trigonometry, that uses the circle x2+y2=1x^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

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)(2), that:

mdvdt=mgkv20    vmgk=βm \frac{dv}{dt} = mg - kv^2 \geq 0 \iff v \leq \sqrt \frac{mg}{k} = \beta

Thus,

vβ1(5)\tag{5} \frac{v}{\beta} \leq 1

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

tanhx=sinhxcoshx=exexex+ex\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\tanh is a bijective function, which means that, for every y]1,1[y \in \, \rbrack -1, 1\lbrack, must exist a xRx \in \reals such as tanhx=y\tanh x = y. If we consider the integral

dv1(vβ)2,(6)\tag{6} \displaystyle\int \frac{dv}{1 - \left(\frac{v}{\beta}\right)^2},

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

tanhx=vβ(7)\tag{7} \tanh x = \cfrac{v}{\beta}

Deriving (7)(7) in function of vv, we have:

1cosh2xdxdv=1β    dv=β1cosh2xdx(8)\tag{8} \frac{1}{\cosh^2 x} \frac{dx}{dv} = \frac{1}{\beta} \iff dv = \beta \frac{1}{\cosh^2 x} dx

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

dv1(vβ)2=β1cosh2x11tanh2xdx=βdx=βx\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)(7), we have:

dv1(vβ)2=βx=βtanh1(vβ)(9)\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)(9), we have:

v0vdv1(vβ)2=βtanh1(vβ)βtanh1(v0β)(10)\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)(5), we have that v0v_0 won't have its value greater than β\beta.

The Terminal Velocity

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

v=βtanh(kgmt+tanh1(v0β))(11)\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 tt becomes large. Applying the limit on both sides of (11)(11), we have:

vt=limtv=βlimttanh(c1t+c2)=β=mgk(12)\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 vtv_t is the terminal velocity, and constants c1=kgmc_1 = \sqrt\frac{kg}{m} and c2=tanh1(v0β)c_2 = \tanh^{-1}\left(\frac{v_0}{\beta}\right).

Physics Analysis

From (12)(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 00. If we take (2)(2) and assumes that dvdt=0\cfrac{dv}{dt} = 0, then we have:

0=mgkvt2    vt=mgk,0 = mg - kv_t^2 \iff v_t = \sqrt\frac{mg}{k},

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

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

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

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

k=ρACd2.k = \frac{\rho A C_d}{2}.
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