The Physics
Opus in profectus

Aerodynamic Drag

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practice problem 1

Two related questions…
  1. Determine the drag coefficient of a 75 kg skydiver with a projected area of 0.33 m2 and a terminal velocity of 60 m/s.
  2. By how much would the skydiver need to reduce his project area so as to double his terminal velocity? How would he accomplish this?


  1. Terminal velocity for a falling object occurs when the drag on the object equals its weight.
    R =  W
    ½ρCAvt2 =  mg
    Solve for projected area, substitute values, and compute. (The density of air is in this book somewhere.)
    C =  2mg
    C =  2(75 kg)(9.8 m/s2)
    (1.207 kg/m3)(0.33 m2)(55 m/s)2
    C = 1.22  
    This agrees with the range of values stated in the table on the discussion page of this topic.
  2. Solve for terminal velocity this time, instead of drag coefficient.
    vt = √  2mg
    Don't plug in any numbers, just look at the way terminal velocity is related to projected area. Projected area is in the denominator, under a radical sign. That means terminal velocity is inversely proportional to the square root of projected area. That means the skydiver would have to reduce his projected area to one quarter of its original value.
    vt ∝ √  1  ⇒  2vt ∝ √  1
    A ¼A
    The skydiver can do this by changing his orientation from horizontal to vertical — from spread eagle to head first.

practice problem 2

Write something different.


Answer it.

practice problem 3

Determine the velocity of a falling body as a function of time when…
  1. drag is directly porpotional to speed and
  2. drag is proportional to the square of speed.


  1. Start with Newton's second law of motion. Identify the relevant forces. Weight (W) pulls the skydiver down. Drag (R) pushes her back up. We'll make down be the positive direction, since that's where the skydiver's headed.
    Σ F =  ma
    W − R =  ma
    mg − bv =  ma

    Acceleration is the rate of change of velocity with time.

    mg − bv = m  dv

    This is a first order differential equation — equation because there's an equals sign, differential because it contains two infinitesimals (dt and dv), and first order because the infinitesimals are not raised to anything higher than the first power. This kind of equation is best solved by separation of variables. Place all the time terms on one side (a rather lonely side) and all the velocity terms on the other side (a somewhat crowded side).

    dt =  m  dv
    mg − bv

    Integrate both sides. Time starts at zero and ends at some later time. Velocity starts at zero and ends at some later velocity. Note that the symbols t and v are doing double duty in the equation. They are acting as both variables and upper limits.

    t   v  

    dt = 
    m  dv
    mg − bv
    0   0    

    Complete the integral…

    t   v

     = −  m
    ln|mg − bv|
    0   0

    Evaluate over the limits…

    (t − 0) = −  m  (ln |mg − bv| − ln |mg|)


    −  b  t = ln    mg − bv  
    m mg
    −  b  t = ln    1 −  b  v  
    m mg

    Remember, we're trying to make velocity into a function of time. We need to get v out of the logarithm. Raise both sides of the equation into a power of e.

    ebt / m = 1 −  b  v

    Finish things off with a little bit of algebra…

    v =  mg  (1 − ebt / m )

    Let's check the limits of this equation to see if they make sense. This function returns the value v = 0 when t = 0.

    v(0) =  mg  (1 − eb0 / m )
    v(0) =  mg  (1 − 1) = 0
    v(0) =  0  

    This agrees with our condition that the skydiver wasn't moving at first.

    This function always increases, but it never quite reaches a final value. It approaches v = mg/b as we get closer and closer to t = ∞.

    v(∞) =  mg  (1 − eb∞ / m )
    v(∞) =  mg (1 − 0)
    v(∞) =  mg  

    This is our terminal velocity. We'd get the same thing if we set drag equal to weight and solved algebraically for speed.

    One last test. What happens to our function if we let b = 0? What happens if we get rid of drag?

    v(b = 0) =  mg  (1 − e−0t / m ) = 
    v(b = 0) =  mg (1 − 1) = mg 
    v(b = 0) = mg  0  = uh oh!

    What is the limit of zero divided by zero? To answer that question, we'll use a little trick called L'Hôpital's rule — named for the French mathematician Guillaume de l'Hôpital (1661–1704). The ratio of two limits that approach zero is the same as the ratio of the limits of their first derivatives. (There's a bit more to this rule, but I'll leave it to your math teacher to fill in the details.)

    lim ƒ(x)  = lim df/dx
    g(x) dg/dx

    For our problem, where the limiting variable is b, we'll let…

    ƒ(b) =  (1 − ebt / m )  ⇒ df  =  t  ebt / m
    db m
    g(b) =  b  ⇒ dg  = 1

    So now we need to take the limit of this instead…

    v(b = 0) =  mg t  ebt / m  
    1 m  
    v(b = 0) =  gt  
    v(b = 0) =  oh yeah!  

    When we get rid of drag, we get back the velocity–time equation for uniform acceleration — in other words, free fall…

    v = gt

    The solution makes sense.

  2. Repeat the previous approach using a drag that is proportional to speed squared.
    Σ F =  ma
    W − R =  ma
    mg − bv2 =  ma

    Rearrange into a first order differential equation…

    mg − bv2 = m  dv

    Separate the variables…

    dt =  m  dv
    mg − bv2

    Integrate both sides.

    t   v  

    dt = 
    m  dv
    mg − bv2
    0   0    

    Let me be honest here. I have no idea what to do next. I had to consult an expert. Apparently, the solution is…

    t = √  m  tanh−1
    v √  b
    bg mg
    v = √  mg  tanh
    t √  bg
    b m

    Have you ever seen anything that looks like this before? Did you notice the odd function? That is not a typo. I did not accidentally drop an "h" into "tan" (short for tangent). I really meant to write "tanh". Let me digress briefly.

    The normal trig functions (sine, cosine, tangent, etc.) are ratios of the sides of a right triangle. A standard way to imagine this triangle is with one vertex at the center of a circle. In this configuration, the hypotenuse of our standard triangle is the radius of the circle in which the triangle is embedded. Take that radius and sweep it around the circumference. Watch the ratios evolve. This is the traditional way to graph them —as functions of an angle swept around a circle.

    Well now, who says we have to do it this way? Who says we have to use a circle? Well now, I would if you asked me, since I tend to relate angles with fractions of a circle, but I'm just a physicist. Mathematicians on the other hand are some clever people. One of them actually had the temerity to respond to the question, "Why a circle?" with another, more interesting question, "Why not a hyperbola?" And the rest was history.

    There are trig functions defined on a circle and trig functions defined on a hyperbola. The former are called sine (sin), cosine (cos), tangent (tan), and so on (cot, sec, and csc). The latter are called hyperbolic sine (sinh), hyperbolic cosine (cosh), hyperbolic tangent (tanh), and hyperbolic so on (coth, sech, and csch). Here's what the mathematicians have determined these functions look like (where x is the "angle" measured in radians).

    sinh (x) =  e+x − ex
    cosh (x) =  e+x + ex
    tanh (x) =  e+x − ex
    e+x + ex

    So what does that mean for us? Make the following replacement in the definition of tanh.

    x = t √  bg

    Look at the beautiful pile of symbols we get.


      e t √  bg  − e − t √  bg  

    v = √  mg m m
    b e t √  bg  + e − t √  bg
      m m

    Isn't that something? Astoundingly, this little mathematical diversion turns out to be useful. Hyperbolic trig functions are more than just an intellectual challenge. They wind up being the solution to some real world problems.

    Let's not test the limits of this equation. There are so many terms to keep track of. It wouldn't be much fun. Instead, let's taste and compare. Graph the two solutions side by side.

    Both solutions are exponential at their core. The speed squared model approaches its terminal velocity faster than the directly proportional model. This makes sense, since squaring the speed makes the drag increase more quickly. Whether the terminal velocities would differ is another matter. The graphs above assume that both models generate the same constant of proportionality. Since these are entirely hypothetical models, no comparison can really be made.

practice problem 4

This tab-delimited text file consists of just the power and top speed data for 122 cars tested by Road & Track magazine in 1998. Use the data in this file and your favorite analysis software to determine the model that best describes aerodynamic drag for automobiles; that is, determine the value of the power n in the generalized drag equation…

R = −bvn


My favorite data analysis software gave me the following coefficients when I graphed P vs. vmax and did a power curve fit.

It looks like aerodynamic drag for cars is proportional to the square of speed. Bernoulli's equation tells us that drag is proportional to the square of speed and I see a power that's approximately 2 in the curve fit above. Isn't life grand when everything behaves exactly as you expect it to.

Halt! Proceed no further with this logic. It is wrong, wrong, wrong, wrong, wrong, wrong, wrong. The graph above shows the relationship between power and speed, not drag force and speed. The question now is how to make the jump from what we have to what we need.

Begin with the definition of work and play around with it a bit.

P =  W  =  Fs  = Fv
t t

Replace the nonspecific force F with power law form of aerodynamic drag.

P = Fv = bvn v = bvn + 1

The curve fit that I did gave me one greater than the power in the general relation. This means that the aerodynamic drag on automobiles is proportional to speed not speed squared. Bernoulli's law does not apply for some reason. I don't know how to reconcile this discrepancy. I suspect that automotive engineers use…

R = ½ρCAv2

for drag calculations, not the computationally simple, but physically unrealistic…

R = −bv

but then, I don't know any automotive engineers.