How far can a shout travel?

Gabriel Golfetti's answer assumes no dissipation. In reality, atmospheric attenuation is quite important for this calculation. According to Engineering Acoustics/Outdoor Sound Propagation: Attenuation by atmospheric absorption (Wikibooks), dissipation in the atmosphere exponentially decreases the sound's intensity with distance, which leads to a linear reduction in the loudness of the sound in dB. Therefore, the loudness of the sound will actually be


where $a$ is the attenuation coefficient in dB/m. The table below gives the attenuation coefficient as a function of frequency and relative humidity for air at 20 degrees Celsius:

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For air at a pressure of 1 atm and sound at a frequency of 1 kHz (which is around the peak of the human vocal spectrum), for most values of relative humidity the attenuation coefficient is approximately $a\approx 1\;\text{dB}/100\;\text{m}$. So our equation for the loudness becomes


Solving for $L=-9\;\text{dB}$ gives

$$r\approx 2\;\text{km}$$

which is drastically reduced from the original answer. Changing the attenuation coefficient by a factor of two (which is approximately how much it varies at that frequency for non-dry air) changes the maximum distance by a factor of 2, so the proper answer, accounting for this uncertainty, is a few km.

To answer this, we need to estimate the level of sound that a shout creates near its source. Since I have no idea what that value is, I googled it: around 88dB at 0.3m away (

For the human voice, the minimum hearing threshold is around -9dB (, and such we can now estimate this distance.

The sound intensity $I$ varies with distance $r$ as $$I\propto r^{-2}.$$

As the intensity is related to the pressure $p$ as $$I\propto p^2,$$ we can say the sound pressure goes as

$$p\propto r^{-1}.$$

Since sound level is given by

$$L=20\log\left(\frac{p}{p_0}\right)\mathrm{dB},$$ for a reference pressure $p_0$ that I can't recall the value of right now, we can say that the sound level of the shout goes as

$$L=88\mathrm{dB}-20\log\left(\frac{r}{0.3\mathrm m}\right)\mathrm{dB}.$$

As such, we need to find $r$ such that this becomes around -9dB. Solving this we get $$r=21\mathrm{km}.$$

Note that this result does not take into account reflections on the surface of the Earth or dissipation. As such, the tower should be much higher that the value we found for $r$.


As @probably_someone commented, taking dissipation into account is not that difficult. We just need to add an attenuation of 1dB per 100m, which turns our equation of sound level into

$$L=88\mathrm{dB}-\left(20\log\left(\frac{r}{0.3\mathrm m}\right)+\frac{r}{100\mathrm m}\right)\mathrm{dB}.$$

This equation can be solved numerically, and gives us the value of $r$ as

$$r=2\mathrm{km},$$ which is quite smaller than our original estimate without dissipation.