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I was wondering, if a person climbs a tower and shouts at the top of his lungs how much distance would the sound travel ? Would it reach someone 1km far? Assuming that there are no tall buildings in the way and the wind is still.

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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

$$L=88\;\text{dB}-20\log_{10}\left(\frac{r}{0.3\;\text{m}}\right)-ar$$

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:

enter image description here

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

$$L=88\;\text{dB}-20\log_{10}\left(\frac{r}{0.3\;\text{m}}\right)-\frac{r}{100\;\text{m}}$$

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.

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    $\begingroup$ Wow, you beat me to it haha $\endgroup$ Jul 6, 2018 at 15:35
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    $\begingroup$ @archaic If you look at that chart, the attenuation coefficient varies pretty widely with frequency. What this means is that different parts of the human vocal spectrum will be attenuated at different rates. Lower-frequency sounds tend to carry further (have less attenuation) than high-frequency sounds, as you can see. The further away you get from the source, the more high frequencies will be suppressed. But that's not the whole picture, either; the ear's nonlinear response prefers high-frequency sounds. The overall effect is a distortion of sound over long distances. $\endgroup$ Jul 6, 2018 at 17:06
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    $\begingroup$ @archaic This distortion means that conveying understandable meaning over distances more than several hundred meters (a significant fraction of the maximum distance) is probably difficult. One way to get around this is to use forms of communication that don't depend on pitch modulation to convey meaning. Such forms are common in older military tactics; For long distances, low-frequency drums carry very well with distance, and can communicate just based on a rhythm. For shorter distances, or cases in which there is a substantial low-frequency background, high-frequency drums and whistles work. $\endgroup$ Jul 6, 2018 at 17:12
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    $\begingroup$ @archaic And no, this result does not take into account the details of the terrain or the height above ground. As such, this is still a very rough, order-of-magnitude approximation. $\endgroup$ Jul 6, 2018 at 17:15
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    $\begingroup$ @archaic In a word, yes, but it's difficult to say how much, because at that point the problem becomes highly situational. I'm pretty confident that the corrections should be less than an order of magnitude in most cases, as I have successfully yelled at someone across a football field (roughly 100 m) before. Also, not all of the corrections necessarily decrease the range -- for example, reflection off of the ground will roughly double the sound intensity (less if there's interference), and make the sound travel something like 30 percent farther (less than an order of magnitude). $\endgroup$ Jul 6, 2018 at 17:34
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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 (https://www.engineeringtoolbox.com/voice-level-d_938.html).

For the human voice, the minimum hearing threshold is around -9dB (https://en.m.wikipedia.org/wiki/Absolute_threshold_of_hearing), 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 $r$.

EDIT

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.

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    $\begingroup$ Atmospheric attenuation plays quite a large role in propagation. For sounds around 1 kHz (which is around the peak of the human voice) at atmospheric pressure and intermediate humidity, you should get an attenuation of about 1 dB per 100 m (en.wikibooks.org/wiki/Engineering_Acoustics/…), which will significantly change the answer. $\endgroup$ Jul 6, 2018 at 15:14
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    $\begingroup$ The attenuation should be linear in $r$, not constant. $\endgroup$ Jul 6, 2018 at 15:26
  • $\begingroup$ Oh right. Damn, that makes the equation quite hard to solve then. $\endgroup$ Jul 6, 2018 at 15:27
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    $\begingroup$ ctd ... We could not actually see his house from near ours, however. On many occasions that he did so, he could not only be heard, but even (at least in good conditions) more-or-less understood. If he'd been a little higher up I'd expect it may well have carried a little further. I'd guess that 2km would be just about plausible in more ideal circumstances. $\endgroup$
    – Glen_b
    Jul 7, 2018 at 1:50
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    $\begingroup$ I have witnessed a woman in Greece ordering bread from the baker by shouting across the valley, at a distance of about 1km. It was clearly a daily routine. $\endgroup$ Jul 7, 2018 at 8:34
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This question is answered by the Guinness World Records.

The normal intelligible outdoor range of the male human voice in still air is 180 m (590 ft 6.6 in). The silbo, the whistled language of the Spanish-speaking inhabitants of the Canary Island of La Gomera, is intelligible under ideal conditions at 8 km (5 miles). There is a recorded case, under optimal acoustic conditions, of the human voice being detectable at a distance of 17 km (10.5 miles) across still water at night.

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You can probably get much louder than 88dB with a shout especially at 0.3m. You'd only have to raise your voice a bit to get that loud as in a teacher in a classroom. I managed to get 104dBA across the room with my best Tarzan level wail but as far as shouting intelligible words 100 dBA at 1 meter in an anechoic chamber is probably a good round ballpark figure, although it's possible to get a bit louder than this.

A sound level of 0dB will be inaudible outdoors. Even the background noises in your ear can easily drown it (blood flow and any mild tinnitus).

Probably 20dB is about your limit and if you live in a city maybe 40-50 dB is needed to understand the words. The fan in your laptop is about 35dB and you clicking keys about 45dB. On inverse square law alone you could be heard a kilometre away at 40 dB. I have yet to look up the relationship between air absorption and frequency which drops exponentially with distance rather than simply squaring and affecting the high frequencies far more than the bass.

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All these answers assume linear behavior. A few other things should be pointed out.

  1. For very high amplitude sounds, large volume, the ordinary wave equation is not valid. So there might be more to the story based on solutions, or approximate solutions, to the full non-linear equations. Not sure this would greatly increase the distance but it may change some of the quantities used to approximate the answers. There is a famous account of filed artillery exercises during the U.S. Civil War (I think), where soldiers herd shots fired before the shouted command to fire. This can be explained with non-linear wave propagation as the very high amplitude shock traveled with a higher effective speed (or just speed). Of course as amplitude decreases the remaining wave will behave as an ordinary wave.

  2. The atmospheric attenuation is frequency dependent, and the non-linear propagation is also frequency dependent. A shout is to some extent percussive, e.g. a short burst. Each frequency will (a) travel at different speeds in the non-linear regime, and (b) dampen differently due to attenuation. With this in mind I wonder if what is picked up at 2km could even be correlated with the original source. There will be information loss.

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In all these discussions, you must take into account the fact that the world outdoors does not exist in silence. There is a background noise floor at speech frequencies which is usually too faint to hear unless you are paying close attention to it. Shouts which are within 3dB of the local noise floor will be masked by it, and hence inaudible to your ears.

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Atmosperic refraction of sound in temperature inversion conditions. This was discovered in XIX century.See Osborne Reynolds publications.This makes sounds arrive further, even with increase in expected dB, due to constructive interference: direct wave added to refracted wave and reflected wave, in still waters.

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