How far can you shout upwind?

That sound should be blown back by a high wind does not at first sight appear to be unreasonable. Sound is known to travel forward through or on the air; and if the air is itself in motion, moving backwards, it will carry the sound with it, and so retard its forward motion — just as the current of a river retards the motion of ships moving up the stream. A little consideration, however, serves to show that the effect of wind on sound cannot be explained in this way.
–Osborne Reynolds, “On the Refraction of Sound by the Atmosphere,” 1874

One of the joys of being a child (and/or a physicist) is pointing at everyday phenomena and asking why they occur. As a child, I assumed the naive hypothesis described by Reynolds was the correct explanation for the difficulty of shouting upwind. However, a brief treatment by physics reveals a considerably different mechanism: the vertical velocity gradient of wind speed causes the sound to refract upwards into the atmosphere, away from a listener on the ground.

Denote the speed of sound in air as \(c\) and the windspeed as \(v\). Due to the presence of a boundary layer — the no-slip boundary condition dictates zero wind velocity at the ground — the velocity is really a function of height. For simplicity, we take \(v(z) = \alpha z\) for proportionality constant \(\alpha\) and distance \(z\) from the ground. By definition, the resulting index of refraction \(n\) as a function of height is $$\begin{aligned} n_\pm(z) &= \frac{c}{c \pm \alpha z}, \end{aligned}$$ where \(n_+\) corresponds to downwind and \(n_-\) to upwind refraction. We know that the path of a sound ray is then given by Snell’s law, $$\begin{aligned} n_\pm(z) \sin \theta(z) &= n_\pm(z+\Delta z) \sin \theta(z + \Delta z). \end{aligned}$$ We could solve this in the limit of \(\Delta z \to 0\), or we can swashbuckle our way to the answer. Since the index of refraction is changing continuously with \(z\), we know the path of the sound ray must curve; the lowest-order parameter describing this curve is its radius of curvature. The only length scale provided by parameters \(c\) and \(\alpha\) is \(c/\alpha\). Hence, to a first approximation, the sound ray should bend in a circle with radius \(c/\alpha\). If downwind, the index of refraction decreases with height, causing the ray to curve downwards towards the ground; if upwind, the ray curves upwards into the sky. This is the underlying mechanism beneath the difficulties of shouting upwind.

For our edification, we can check this guess by solving the differential equation. Substituting \(n_\pm(z)\) and taking \(\Delta z \to 0\) with initial height \(z_0\) gives $$\begin{aligned} [c \pm \alpha (z+\Delta z)] \sin \theta(z) &= [c \pm \alpha z] \sin \theta(z + \Delta z)\\ \frac{\sin \theta(z + \Delta z)}{\sin \theta(z)} &= \frac{1 \pm (z+\Delta z)\alpha/c}{1 \pm z\alpha/c}\\ \frac{d\theta}{dz} \cot\theta(z) &= \frac{1}{z\pm c/\alpha}\\ \int d\theta~\cot\theta &= \int \frac{dz}{z\pm c/\alpha}\\ \sin \theta(z) &= \left(\frac{\sin \theta(z_0)}{z_0 \pm \frac{c}{\alpha}}\right) \left(z \pm \frac{c}{\alpha}\right), \end{aligned}$$ i.e., Snell’s law describes an exactly circular path with radius \(R=c/\alpha\) curving either upwards or downwards as we predicted.

We can plug in a few numbers. On a windy day, the wind blows at perhaps \(v = 5~\mathrm{m~s^{-1}}\) at the height of a (short) human (1 meter). This is a fairly intuitive guess: the breeze roughly disappears when running downwind at \(\sim 5~\mathrm{m~s^{-1}}\). However, it is bold to assume that the wind speed increases by \(5~\mathrm{m~s^{-1}}\) for every meter increase in the altitude; at 1 km, the wind would be traveling at 5 km/s! Our linear wind profile is only accurate near the ground; it must increase more slowly after some altitude, likely as a power law. For instance, on the top of the Empire State Building (400 meters), it can get very windy, but it remains possible to walk. Linearly interpolating to a wind speed of 15 m/s at the height of the observation deck, we can guess \(\alpha \sim 0.04~\mathrm{s^{-1}}\). Taking the speed of sound at \(343~\mathrm{m~s^{-1}}\), this gives a refraction radius of around \(8000\) meters.

If a person — fine, a hobbit — were to shout upwind from a height of one meter, a spherical wavefront would emerge from the (apparently angry) hobbit, and each ray would then refract in an approximately circular path. Part of the lower half of the sphere would reflect off the ground as well, but the reflected rays would also refract upwards. In fact, since the wind speed gradient is larger near the ground, they should refract especially quickly. Consequently, we will assume that once the shout refracts to a height of two meters, the ray will miss a recipient at the height of one meter. (Note that the common \(1/r^2\) falloff doesn’t apply due to the refraction, so missing the recipient by a meter doesn’t mean that the sound will be about as loud as on a still day.) Solving the displacement \(x^2 + (8000 - 1)^2 = 8000^2\) gives a horizontal distance \(x\) of 100 meters upwind before the shout is inaudible. Even if we assumed that the sound must be displaced by a distance of 10 meters, we would obtain a maximum distance \(x \sim 400\) meters; hence, we may be satisfied that our estimate is robust to order of magnitude. In practice, the noise of the gusts wind may also make it more difficult to hear a shout, biasing our estimate towards the lower end of the \(O(100)\)-meter range.

Historical aside

One aspect of our analysis might strike the reader as especially odd: we never directly accounted for the volume of the shout. Due to the refraction, we expect that a sound of any volume should become inaudible. (In reality, due to stochasticity, the probability of a shout leaking through to the listener increases with volume.) Indeed, historical evidence suggests that even the loudest of sounds — an entire Civil War battle — can refract away. We quote directly from [1].

At Iuka, Mississippi, on Sept. 20, 1862, Major General Ulysses S. Grant formulated a plan typical of the day, with sounds of battle acting as a trigger for troop movements. His plan, if successful, would have brought about the defeat of one of the primary Confederate armies (under Sterling Price). Grant’s plan called for forces under Brigadier General William Rosecrans to come upon Iuka (where Price’s men were based) from the south. The remainder of Grant’s men, under Major General Edward Ord, were to wait four miles north of Iuka until the sounds of engagement between Rosecrans’s forces and those of Price were heard. Late in the afternoon, Ord and Grant saw smoke rising from Iuka but heard nothing and assumed that Price was burning supplies to prevent their capture. Rosecrans and Price had actually been engaged for over two hours, but by the time couriers notified Grant it was too late: Price had slipped out between the two Union armies and avoided the intended pincer movement. The culprit in this case was a strong wind blowing from north to south that had placed Grant and Ord in an acoustic shadow as sounds of battle were refracted upward over their heads.

Additionally, the shouting downwind analysis is considerably trickier due to the repeated reflection of sound off the ground. As sound rays refract in circular arcs bending towards the ground instead of towards the sky (unlike the upwind case), sound “bounces” along the ground and can propagate large distances. During this process, it creates silent zones (acoustic shadows) where the bounce is high in the sky and arcing once again towards the ground. Again, we reference [1] for a dramatic historical example.

This effect can cause audibility of sounds of battle at unusually long distances: at times, the battle of Gettysburg was inaudible ten miles away, while it was heard clearly at times in Pittsburgh, 160 miles (250 km) to the west.

Experimental results

In progress, currently awaiting a windy day.


[1] Ross, Charles D. “Blending history with physics: Acoustic refraction.” The Physics Teacher 38.4 (2000): 208-209.