June 10, 2026. Some quantitative meditations upon seeing the wind whip up a dust devil and hearing it whisper through leaves and moan through branches.

Dust devils

I recently observed a dust devil (or small whirlwind) in an empty cul-de-sac on a cool day of gusty wind and later rain. The dust devil was the on the order of the size of the cul-de-sac, say $L \sim 10 \text{ m}$ across, and moving at around $u = 1 \text{ m/s}$. The kinematic viscosity of air at $T = 15^\circ$ and standard pressure is around $\nu \approx 1.48 \times 10^{-5} \text{ m}^2/\text{s}$. This is not really estimable from dimensional analysis, since the simplest dimensionally consistent formula

\[\nu \sim uL \sim 10 \text{ m}^2/\text{s}\]

is off by a factor of $10^5$. What we are missing is a dimensionless quantity called the Reynolds number, the ratio of kinetic to viscous forces:

\[\text{Re} = \frac{uL}{\nu} \approx 6.8 \times 10^4.\]

Reynolds number above $\simeq 4 \times 10^3$ indicates kinetic forces dominate and we enter the regime of turbulent flow. In turbulence, laminar sheets of the fluid (in this case air) break into eddies of all sizes, the largest being eddies on the size of the macroscopic length scale (like our cul-de-sac) all the way down to the microsopic or Kolmogorov scale $\eta$ where viscous forces once again take over and disperse the vortices. The Kolmogorov scale obeys

\[\eta = \left(\frac{\nu^3}{\epsilon}\right)^{1/4}\]

where $\epsilon$ is a somewhat obscure quantity, namely, the average rate at which “turbulence kinetic energy” is lost per unit mass. The goal I set myself: determine the size of the microscopic vortices the cul-de-sac dust devil would dissipate itself into. The only missing quantity here is $\epsilon$, but rather than look it up, I figured I should estimate it from dimensional analysis. For a large-scale eddy, the relevant quantities are $u$, $L$, and density $\rho$, i.e. kinetic quantities rather than viscosity. The rate at which energy is dissipated per unit mass has dimensions

\[\frac{\text{energy}}{\text{time}\cdot \text{mass}} = \frac{ML^2/T^2}{TM} = \frac{L^2}{T^3}.\]

The only combination of relevant quantities with this dimension is

\[\epsilon \sim \frac{u^3}{L}.\]

Plugging in our numbers gives $\epsilon \sim 0.1 \text{ m}^2/\text{s}^3$, consistent with measurements of air on a day of mild wind. Plugging this into the Kolmogorov microscale equation gives

\[\eta = \left(\frac{\nu^3}{\epsilon}\right)^{1/3} \sim \left(\frac{L\nu^3}{u^3}\right)^{1/4} \sim\left(\frac{10 \cdot (1.5 \times 10^{-5})^3}{1^3}\right)^{1/4} \text{ m} \approx 0.43 \text{ mm}.\]

This is consistent with a typical Kolmogorov length scale of $\eta \sim 1 \text{ mm}$, which in turn, has implications for the preferential formation of raindrops of that size! So there’s a nice throughline from the wind that blew the dust devil to the rain that accompanied later.

Whispering trees

On that note, trees make a nice whispering sound when the wind passes through them. We can actually work out the base frequency it sussurates at! We basically model it as the natural resonance of the leaf, considered as a tiny cantilever attached to the tree. If $k$ is the stiffness and $m$ the leaf mass, then the natural frequency is given by

\[f^* = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{k}{A\rho}},\]

where $A$ is the area and $\rho$ the areal mass density. The stiffness, in turn, is determined by the cantilever stiffness equation:

\[k = \frac{3 E I}{\ell^3}\]

where $\ell$ is the length of the leaf, $E$ its Young modulus, and $I = A^2/4\pi$ its areal moment of inertia. Putting it together, we find

\[f^* = \sqrt{\frac{3Ew}{16\pi^3\rho\ell^2}}.\]

For gum leaves, we take average length $\ell \sim 10 \text{ cm}$ and width $w\sim 4 \text{ cm}$ (determined by measuring a few specimens), modulus of elasticity $E \sim \times 10^8 \text{ Pa}$, and density $\rho \sim 0.2 \text{ kg/m}^2$. Plugging in the numbers yields

\[f^* \sim 3500 \text{ Hz}.\]

This is consistent with the spectrum of a whisper (see e.g. this doctoral thesis, Figure 3.5), which is usually perceived in the $10^3$–$10^4 \text{ Hz}$ range. (Technically, there is a lognormal probability distribution obeying Taylor’s power law over leaf size and hence frequency. But this complicates the analysis without shifting the result.) Thus, we have a nice first-principles account of why trees whisper when the wind blows!

From whispers to moans

If wind blowing through tree leaves whispers, wind blowing over branches tends to moan. Presumably, this is not due to the bending moment. So what causes it? The answer takes us back to the dust devils. When wind blows around an object, it forms a cavity in the wake at sufficiently high ($\text{Re} > 90$) Reynolds number. That wake gets filled with turbulent eddies in a process called the von Kármán vortex street. For a wind of speed $u \sim 10 \text{ km/h} \approx 7 \text{ m/s}$ and a branch of of width $L \sim 0.02 \text{ m}$, the Reynolds number associated with the flow is

\[\text{Re}_L = \frac{uL}{\nu} = \frac{3.5 \times 0.02}{1.48 \times 10^{-5}} \approx 5 \times 10^3 \gg 90,\]

using our value for $\nu$ from above. We are will into the vortex shedding regime. The frequency of vortex shedding $f$ is encoded by the Strouhal number

\[\text{St} = \frac{fL}{u}.\]

To determine $f$, we note the relation between Strouhal and Reynolds number:

\[\text{St} = 0.198\left( 1 - \frac{19.7}{\text{Re}_L}\right) \approx 0.2.\]

This gives us the vortex shedding frequency

\[f = \frac{u\text{St}}{L} = \frac{0.2 \times 3.5}{0.02} = 35 \text{ Hz},\]

a low, haunting, and barely audible moan. Smaller objects and stronger winds will produce higher frequency moans, which is consistent with what we hear. That completes our whirlwind tour of wind-induced phenomena!