I'm trying to figure out if you have a crowd of people, how big does the crowd have to be so that when you add more people at the edges it doesn't make a difference in decibel levels anymore?

In a hypothetical situation where we pack a group of screaming people in a grid, what would be the limit on the sound level that we can go up to? For example, can we pack 2 million screaming people, and it'll be louder than 1 million screaming people?

Assuming we arrange people in a perfect grid, and each person stands on 0.66 square meters, is there a point at which we can't add any more screaming people and the combined sound levels wouldn't increase? We know that at some point sound waves may cancel out due to phase interference, but what effect would this have on the ultimate loudness we can achieve, considering that frequencies will vary?

Additionally, is there a maximum decibel (dB) level that we should consider when arguing about this hypothetical scenario? And if yes, how would practical real-world factors such as distance and perception of loudness play a role in that?

Some theories suggest that the limit of dB levels could be infinite in theory, while others state that there might be a limit at 194 dB, when low-pressure regions are completely empty, and no further increase in loudness is possible. How does this ultimate loudness limit, if it exists, influence the scenario of packing a crowd of screaming people into a grid?

  • $\begingroup$ "in a grid" What kind of a grid? Like are you imagining an effectively two-dimensional grid of people on an effectively infinite flat plane? Or a three dimensional grid of people? On the surface of the earth? Or what? $\endgroup$
    – hft
    Commented Jun 28, 2023 at 16:42
  • $\begingroup$ "Some theories suggest that..." Which ones? $\endgroup$
    – hft
    Commented Jun 28, 2023 at 16:48
  • $\begingroup$ "is there a point at which we can't add any more screaming people" Why would we not be able to add any more screaming people? You mean because the earth is round or what? $\endgroup$
    – hft
    Commented Jun 28, 2023 at 16:51
  • $\begingroup$ At what point are you measuring the sound level? The center of the crowd? The edge? Some distance from the edge? Somewhere else? $\endgroup$ Commented Jun 28, 2023 at 16:59
  • $\begingroup$ @MichaelSeifert The center of the crowd. $\endgroup$
    – macco
    Commented Jun 28, 2023 at 18:56

1 Answer 1


Assuming we arrange people in a perfect grid, and each person stands on 0.66 square meters...

Assuming you are interested in the sound intensity at the origin in a two dimensional infinite flat grid and you have people whose individual power is $P_0$ standing at grid locations other than the origin, then the intensity at the origin is: $$ I = \sum_{(i,j) \neq (0,0)} \frac{P_0}{4\pi a^2(i^2 + j^2)}\;, $$ where $a^2$ is 0.66 square meters. (And where we have assumed that our two-dimensional grid of people is embedded in our usual three-dimensional flat space.)

Clearly $I$ increases as you increase the number of people. It increases "slowly" but increases nevertheless.

Whether or not the sum increases without limit depends on the dimensionality of the grid.

For a two-dimensional grid, you can look at the behavior far from the origin in the continuum limit. In this limit assume the intensity at the origin due to a patch of area $\delta A$ is proportional to $\frac{\delta A P_0}{a^2 r^2}$. Then: $$ I \to I_{\text far} \sim \int_{\text far} \frac{\delta A P_0}{a^2 r^2} $$ $$ \sim \int_{\text far} \frac{r dr d\theta P_0}{a^2 r^2}\;. $$

In this two-dimensional case we have the "far" contribution as $$ I_{\text far}\sim\frac{2\pi P_0}{a^2} \int_{r=R}^\infty \frac{dr}{r}\;, $$ where $R$ is some large distance (large enough that we are fine with the continuum approximately). The integral diverges logarithmically due to the upper limit of integration being infinity. The integral diverges worse for a three-dimensional grid. The integral doesn't diverge for a one-dimensional grid (e.g., a line of people embedded in three-dimensional flat space).

Per the comments, OP now also wants to ask a similar question about coherent audio sources.

Note first, however, that a bunch of people screaming do not produce coherent audio. There is no clear mechanism by which a bunch of people screaming would coordinate the waves coming out of all their mouths.

But anyways. Let's assume the audio sources are can all be specified at an exact wavelength $\lambda$ and phase offset $\phi$. In this case the amplitude at the origin of the entire grid of coherent screamers is: $$ A = \alpha_0\sum_{(i,j)\neq (0,0)}\frac{e^{i\left(\frac{2\pi a}{\lambda} \sqrt{i^2 + j^2}\right)+i\phi_{ij}}}{a\sqrt{i^2 + j^2}} $$ and the intensity is: $$ I = |A|^2 = |\alpha_0|^2 \sum_{(i,j)\neq (0,0)} \sum_{(k,\ell)\neq (0,0)} \frac{e^{i\left(\frac{2\pi a}{\lambda} (\sqrt{i^2 + j^2} - \sqrt{k^2+\ell^2}\right)}}{a^2\sqrt{i^2 + j^2}\sqrt{k^2+\ell^2}} e^{i(\phi_{ij}-\phi_{kl})}\;. $$

The average intensity (averaging over the site-specific phase) is: $$ <I> = |\alpha_0|^2 \sum_{(i,j)\neq (0,0)} \sum_{(k,\ell)\neq (0,0)} \frac{e^{i\left(\frac{2\pi a}{\lambda} (\sqrt{i^2 + j^2} - \sqrt{k^2+\ell^2}\right)}}{a^2\sqrt{i^2 + j^2}\sqrt{k^2+\ell^2}} <e^{i(\phi_{ij}-\phi_{kl})}>\;, $$ where we recover the incoherent result when $<e^{i(\phi_{ij}-\phi_{kl})}>=\delta_{(i,j),(k,\ell)}$ and we see that $|\alpha_0|^2\sim P_0$.

OTOH, if the site-specific phases are the same (e.g., all $\phi=0$) then: $$ I = |A|^2 = {\left| \alpha_0\sum_{(i,j)\neq (0,0)}\frac{e^{i\left(\frac{2\pi a}{\lambda} \sqrt{i^2 + j^2}\right)}}{a\sqrt{i^2 + j^2}} \right|}^2 $$

  • $\begingroup$ In the two dimensional grid the integral diverges logarithmically to infinity, but what if we try to make it a (tiny bit) more realistic scenario and also take into account that some sound waves will cancel out. Does it still diverge? $\endgroup$
    – macco
    Commented Jun 28, 2023 at 18:55
  • 2
    $\begingroup$ Why would the sound waves "cancel out"? Why would the audio sources be at all coherent? A bunch of people standing around screaming don't have any clear method to make all their sound waves coherent... $\endgroup$
    – hft
    Commented Jun 28, 2023 at 19:32
  • $\begingroup$ If you have coherent sources then the calculation is different and the phase of the waves' phases need to be considered. It's also a fairly straightforward calculation and I don't think the answer changes much other than some form factor, but I haven't worked it out exactly. Is this what you really want to ask? Or are you really interested in "people screaming"? If you want to know about coherent sources then modify your question post. $\endgroup$
    – hft
    Commented Jun 28, 2023 at 19:35
  • $\begingroup$ ah ok, I assumed if many people would scream at the same time then there would be a lot of overlapping frequencies which are bound to cancel eachother at some points. But maybe that's not how it works? $\endgroup$
    – macco
    Commented Jun 28, 2023 at 23:36
  • $\begingroup$ Thanks! I think practically there is an upper limit on how many people you can place and it won't get louder, e.g., if you take 194 decibels to be the loudest a sound can reach in the atmosphere, because at some point the fluctuations in air pressure are so large that the low pressure regions hit zero pressure—a vacuum—and you can't get any lower than that. $\endgroup$
    – macco
    Commented Jun 28, 2023 at 23:42

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