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When computing the noise level x amount of meters away from a singer who is singing outdoors, should one rely on the standard formula

$I=\frac{P}{4πr^2}$

(and then convert 'I' into decibels)

or rather on the formula $I=\frac{P}{2πr^2}$, given that the sound waves cannot spread downwards, and are thus forming a "half-sphere" only?

My confusion stems from the fact that I remember calculating the radio waves emitted by an antenna using the formula $I=\frac{P}{2πr^2}$, based on the fact that the antenna was located on the ground, meaning that the waves formed a semi-sphere only. This was confirmed by my textbook.

However, that same textbook uses $I=\frac{P}{4πr^2}$ for the scenario given above, which does not seem to make sense.

Can anybody help out?

Thank you!

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These examples are idealized.

For instance, the example with the antenna appears to be based on the assumption that it has a hemispherical radiation diagram, i.e., has uniform radiation above the ground, which is not possible.

The example with the singer appears to be based on the assumption that the sound is radiated uniformly in all directions (isotropic radiator), which is theoretically possible, but not very feasible (for instance, all sound energy directed toward the ground would have to be fully absorbed).

If we accept these unrealistic assumptions, the formulas look fine.

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    $\begingroup$ Thank you for the reply! But in the case of the singer: would you say it is more realistic to assume a sphere or a semiphere? And in case the answer is a "sphere": is there really any logical reason why the assumption for the antenna was a "hemisphere"? I.e. is there a physical reason that explains why in one case the book opted for assuming a sphere and in the other a hemisphere? Is it - in relative terms - more realistic to assume one of these two scenarios in one case than in the other? $\endgroup$ – Pregunto Sep 15 '18 at 14:50
  • $\begingroup$ @Pregunto To me, these two examples are similar, once all the simplifications have been accepted. $\endgroup$ – V.F. Sep 15 '18 at 15:05
  • $\begingroup$ And would you yourself recommend the hemisphere or the sphere-model in these two cases? $\endgroup$ – Pregunto Sep 15 '18 at 20:55
  • $\begingroup$ @Pregunto I would recommend nither for the reasons outlined before, but if I had to choose, it would be hemisphere. $\endgroup$ – V.F. Sep 15 '18 at 21:07
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I believe V.F. provided very nice reasoning. Despite the fact though, I believe it would be a bit instructive to see why your textbook uses two different approaches to those two cases.

Since Radio Frequencies (RF) cover a humongous range one would have to clarify a bit on the range of interest. For this reason, and the fact that my field is Acoustics, I will cover the audio first and then try to expand on the RF.

Whether someone uses a half-sphere or full-sphere solid-angle of radiation depends primarily on the wavelength (and frequency through the $c = \lambda f$ equation). For wavelengths way longer than the distance of the source to the closest reflective boundary (the ground in this case), the phase difference between the radial component pointing "upwards" (or in general, not reflected) and the reflected energy is so small that you get almost perfect coupling (i.e. constructive interference) irrespective of the angle of observation.

If we tried to quantify that, we would say that the phase offset due to the travel distance from the source to the boundary and back has to be small (in the limiting case it should be zero). In this way, the two components will be in phase. This can be somewhat be given by

$$L < \frac{\lambda}{2n}$$

where $L$ is the distance from the source to the closest (reflective) boundary, $\lambda$ is the wavelength and $n$ is an arbitrary number that most probably depends on how accurate you want to be. Ex. by setting $n = 10$ your reflected and "originally emitted" waves will have a phase difference of $\frac{\lambda}{10}$ for a frequency that corresponds to this wavelength (as mentioned above, depends on phase speed $c$). Obviously, $n$ has to be as big as possible (being infinite in the limiting case, for constant $\lambda$).

Now, regarding the textbook approach. I will have to speculate about some things but I believe I'll manage to provide a basic explanation on which you could build. For "normal" conditions the speed of sound is considered to be $c \approx 343 m/s$. This means that for a person of average height (let's say $1.75 m$) the limit where the distance between the source (mouth) and ground reaches $\frac{\lambda}{2}$, for which you have complete cancellation due to the reflection is $f \approx 98 Hz$. If nothing else, this means that even for the frequency of $f = 90 Hz$, you can't use the $2 \pi$ solid angle radiation approximation! Considering that the human voice (even for some baritone or bass singers) rarely goes below $100 Hz$ (for more information on that you could refer to "Communication Acoustics" by Pulkki and Karjalainen) then it is reasonable to not use this model. So, considering which model is left to use as a first approximation, you would realise that the corresponding $4 \pi$ solid-angle radiation one is what is left (of course you could go harder on yourself and use a more accurate but more complex one).

In addition to that, the higher you go in frequency, up to a certain limit, the more accurate this model will be. This does not mean that this will be a very accurate model and this frequency limit is where your mouth starts to exhibit considerable directivity characteristics where a spherical radiation model is not accurate at all (even the hemispherical would not be useful in this case).

The main factors here are the increase in the directivity of the source (mouth) and the reflections by the ground. The phenomenon is kind of self-contradicting in the sense that, while the directivity increases you will get less reflected energy from the ground making the $4 \pi$ model more accurate, but the higher the directivity the less spherical your radiation will be invalidating the model! Now, to help the situation a little bit, attenuation/absorption from the ground also strikes in after some frequency (this depends on the kind of ground - soil, gravel, etc. For more info see "Ground, Terrain and Structure Effects on Sound Propagation" report by Hannah Lindsay).

To conclude this part, the main reason that your textbook uses a $4 \pi$ solid-angle of radiation is that the $2 \pi$ model is even worse.

On the other hand, RF antennas can very well accommodate wavelengths in the range of $1 km - 10m$. While they can be in the range of lower wavelengths( higher frequencies), I believe that your textbook refers to this (or even higher wavelength) range. Following our initial approach, you can easily see that the $2 \pi$ radiation model covers very well the phenomenon for installation at a height in the range of about $1 m$.

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