<p>This sounds like a class exercise, but nevertheless:</p>
<p>The answer to your question is in the transmission at interfaces.
When arriving at an interface part of the sound will be reflected and the rest will pass the interface.</p>
<p>The amount of sound that transfers an interface depends on the acoustic impedance (product of density and sound speed) of the materials (water,metal,air) and the angle of arrival.</p>
<p>For the three materials we may for example have</p>
<ul>
<li>Steel: <span class="math-container">$Z=ρ_sv_s = 46.5 \times 10^6$</span> kg/m<span class="math-container">$^2$</span>s</li>
<li>Water: <span class="math-container">$Z=ρ_sv_s = 1.5  \times 10^6$</span> kg/m<span class="math-container">$^2$</span>s</li>
<li>Air:   <span class="math-container">$Z=ρ_sv_s = 420$</span> kg/m<span class="math-container">$^2$</span>s</li>
</ul>
<p>For simplicity lets ignore the arrival angle and assume perpendicular incidence.</p>
<p>If <span class="math-container">$A_i, A_r, A_t$</span> are the sound aplitudes of incomming, reflected and transitted sound amplitudes, respectively, then the formulas for reflection and transmission coefficients are</p>
<ul>
<li>Reflection <span class="math-container">$R_{12} =A_r/A_i= (Z_2 - Z_1)/(Z_1+Z_2)$</span></li>
<li>Transmission <span class="math-container">$T_{12} =A_t/A_i= 2Z_2/(Z_1+Z_2)$</span></li>
</ul>
<p>incoming and reflected sounds are in medium 1, and transmitted sound is in medium 2</p>
<p>So, we have for the transmission of the sound the following cases</p>
<ul>
<li>water-steel: <span class="math-container">$T_{ws}=1.94$</span></li>
<li>steel-water: <span class="math-container">$T_{sw}=0.063$</span></li>
<li>steel-air: <span class="math-container">$T_{sa}=1.8 \times 10^{-5}$</span></li>
<li>air-steel: <span class="math-container">$T_{as}=2 $</span></li>
</ul>
<p>Ignoring multiple reflections within the double hull and assuming that frequency is high enough, i.e. steel is thick enough and gap between double hull is wide enough for sound to propagate (no tunnel effect) then we may combine all transmission factors by multiplication.</p>
<ul>
<li>case 1: water-steel-water-steel-air <span class="math-container">$T = T_{ws}T_{sw}T_{ws}T_{sa} =   4.2 \times 10^{-6} = -108 \mathtt{dB}$</span></li>
<li>case 2:water-steel-air-steel-air <span class="math-container">$T = T_{ws}T_{sa}T_{as}T_{sa} =   1.3 \times 10^{-9} = -178 \mathtt{dB}$</span></li>
</ul>
<p>The sound pressure within the submergible is therefore significantly  lower than in the surrounding water.</p>
<p>Observation: in case of lower frequencies, where wavelength is much larger than the hull thickness, then effectively only the water to air transmissivity counts (hull is becomming nearly transparent).</p>
<p>Note: Cases that deviate from the simplifying assumption, obviously require detailed acoustic modelling.</p>

CC BY-SA 4.0
2023-11-04T20:02:11.497Z
Clarification: this is story research, not academics. Consequently, I'm not terribly interested in mathematical precision. That being said, if lower frequencies make the multi-layer transition better than higher frequencies, am I right to guess that to an observer inside the sub, the whale song and/or sonar ping would sound lower-pitched than it would in the water? If the sound source was, say, approximately 8 km away from the sub, would either sound even be audible? I just looked up dB levels for whale song and military sonar: less than 200 and more than 200, respectively, if that helps.
rccapps
CC BY-SA 4.0
2023-11-05T07:07:25.277Z
I keep my answer nevertheless, and the numbers indicate that it very difficult to hear (unaided) outside natural sound within subs. But, I have no personal experience, so I cannot be certain. I edited my answer to provide also dB attenuation.
WMXZ
CC BY-SA 4.0
2023-11-03T09:19:35.717Z
2023-11-05T07:11:48.710Z
2023-11-05T07:11:48.710Z
WMXZ
<p>To be more specific: assuming an inner and outer hull, with the in-between gap partially serving as a ballast tank, would the transition from water to metal to air/water to metal to air:</p>
<p>a) amplify or mute the sounds, and</p>
<p>b) affect the range of perceived frequencies.</p>
<p>I understand that even the &quot;best&quot; human ears cannot detect the <em>full</em> range of whale song, but would the submersible enclosure alter the frequencies perceived <em>within</em> the range of human hearing?</p>
<p>Full disclosure: I'm just as interested in how the environment within a submersible affects the perception of <em>artificial</em> sounds (e.g. broadband sonar &quot;pings&quot;). But I'm hoping/assuming the answer would be more or less the same for biologically-generated whale song and mechanically/electronically-generated sonar &quot;chirps&quot;; if anyone has real-world experience in this area - i.e. being on a sub and hearing whale song and/or active sonar - feel free to chime in.</p>

<underwater><echolocation><cetacean>
How does cetacean echolocation sound inside a submersible?
# How does cetacean echolocation sound inside a submersible?

(Asked by: username_0 on 2023-11-02)

To be more specific: assuming an inner and outer hull, with the in-between gap partially serving as a ballast tank, would the transition from water to metal to air/water to metal to air:

a) amplify or mute the sounds, and

b) affect the range of perceived frequencies.

I understand that even the "best" human ears cannot detect the *full* range of whale song, but would the submersible enclosure alter the frequencies perceived *within* the range of human hearing?

Full disclosure: I'm just as interested in how the environment within a submersible affects the perception of *artificial* sounds (e.g. broadband sonar "pings"). But I'm hoping/assuming the answer would be more or less the same for biologically-generated whale song and mechanically/electronically-generated sonar "chirps"; if anyone has real-world experience in this area - i.e. being on a sub and hearing whale song and/or active sonar - feel free to chime in.

**Tags:** underwater, echolocation, cetacean


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## 1 answers

### Answer by username_1 on 2023-11-03. Score: 2

This sounds like a class exercise, but nevertheless:

The answer to your question is in the transmission at interfaces.
When arriving at an interface part of the sound will be reflected and the rest will pass the interface.

The amount of sound that transfers an interface depends on the acoustic impedance (product of density and sound speed) of the materials (water,metal,air) and the angle of arrival.

For the three materials we may for example have

* Steel: $Z=ρ\_sv\_s = 46.5 \times 10^6$ kg/m$^2$s
* Water: $Z=ρ\_sv\_s = 1.5 \times 10^6$ kg/m$^2$s
* Air: $Z=ρ\_sv\_s = 420$ kg/m$^2$s

For simplicity lets ignore the arrival angle and assume perpendicular incidence.

If $A\_i, A\_r, A\_t$ are the sound aplitudes of incomming, reflected and transitted sound amplitudes, respectively, then the formulas for reflection and transmission coefficients are

* Reflection $R\_{12} =A\_r/A\_i= (Z\_2 - Z\_1)/(Z\_1+Z\_2)$
* Transmission $T\_{12} =A\_t/A\_i= 2Z\_2/(Z\_1+Z\_2)$

incoming and reflected sounds are in medium 1, and transmitted sound is in medium 2

So, we have for the transmission of the sound the following cases

* water-steel: $T\_{ws}=1.94$
* steel-water: $T\_{sw}=0.063$
* steel-air: $T\_{sa}=1.8 \times 10^{-5}$
* air-steel: $T\_{as}=2 $

Ignoring multiple reflections within the double hull and assuming that frequency is high enough, i.e. steel is thick enough and gap between double hull is wide enough for sound to propagate (no tunnel effect) then we may combine all transmission factors by multiplication.

* case 1: water-steel-water-steel-air $T = T\_{ws}T\_{sw}T\_{ws}T\_{sa} = 4.2 \times 10^{-6} = -108 \mathtt{dB}$
* case 2:water-steel-air-steel-air $T = T\_{ws}T\_{sa}T\_{as}T\_{sa} = 1.3 \times 10^{-9} = -178 \mathtt{dB}$

The sound pressure within the submergible is therefore significantly lower than in the surrounding water.

Observation: in case of lower frequencies, where wavelength is much larger than the hull thickness, then effectively only the water to air transmissivity counts (hull is becomming nearly transparent).

Note: Cases that deviate from the simplifying assumption, obviously require detailed acoustic modelling.

> **Comments:**
> **username_0**: Clarification: this is story research, not academics. Consequently, I'm not terribly interested in mathematical precision. That being said, if lower frequencies make the multi-layer transition better than higher frequencies, am I right to guess that to an observer inside the sub, the whale song and/or sonar ping would sound lower-pitched than it would in the water? If the sound source was, say, approximately 8 km away from the sub, would either sound even be audible? I just looked up dB levels for whale song and military sonar: less than 200 and more than 200, respectively, if that helps.  
> **username_1**: I keep my answer nevertheless, and the numbers indicate that it very difficult to hear (unaided) outside natural sound within subs. But, I have no personal experience, so I cannot be certain. I edited my answer to provide also dB attenuation.  

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