When reading about a harmonic oscillator, the most common form of damping is proportional to the velocity.

$$F_D = c \;\dot{x}$$

Without giving much thought to viscous damping, one example I always thought of is the following. Taking my hand outside a car with velocity v. As the velocity increases the drag also increases.

However, I also know that the aerodynamic drag of a vehicle is proportional to the square of the velocity. I.e. :

$$F_A = \frac{1}{2} C_D \rho A (\dot{x})^2$$

So, either my example does not fall under viscous damping, or there is something fundamental that I don't understand.

I would very much appreciate your insight on this.


1 Answer 1


The equation $F_{\textrm{A}} = C_{\textrm{D}}A\rho\dot{x}^2/2$ doesn't tell you that drag is proportional to the square of velocity. In itself, that equation doesn't tell you anything at all about how drag depends on velocity, because that equation is just a definition of the drag coefficient $C_{\textrm{D}}$. What tells you how drag depends on velocity is the combination of:

  • the insight that (for all mutually geometrically similar objects in incompressible Newtonian fluids) the value of $C_{\textrm{D}}$ depends on other variables (including velocity) only through the combination of variables known as the Reynolds number; and
  • the graph showing how the value of $C_{\textrm{D}}$ depends on Reynolds number.

At small values of the Reynolds number, $C_{\textrm{D}}$ is inversely proportional to the square root of Reynolds number (Landau, Lifshitz and Pitaevskii, 1987, Fluid mechanics, Butterworth Heinemann). Since the definition of Reynolds number has a single factor of velocity in its numerator, this means that, in this regime, drag is proportional to the $3/2$ power of velocity. While this still isn't the form known in dynamical systems theory as "viscous damping", it's closer to it than would be the case if drag were proportional to the square of velocity. On the other hand, as the value of the Reynolds number tends to infinity, $C_{\textrm{D}}$ tends to a non-zero constant. In this regime, drag really is proportional to the square of velocity, and drag is definitely not of a viscous-damping form.

If you're interested in the micromechanics of why $C_{\textrm{D}}$ is inversely proportional to Reynolds number for small Reynolds numbers, but tends to a non-zero constant as the Reynolds number tends to infinity, there's a nifty explanation in Faber (1995, Fluid dynamics for physicists, Cambridge University Press), under the headings "skin friction" and "boundary layer separation".

  • $\begingroup$ In case anyone saw this answer in the first twelve hours or so after I posted it, and has now come back and is confused as to why it's become a substantively different answer, it's because I spotted that I'd made an error in the exponent of the relationship between drag coefficient and Reynolds number at small Reynolds numbers, and made a few rounds of edits to correct it. $\endgroup$ Oct 20, 2020 at 10:20

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