Answers.Body: <p>The simple answer is not a lot, the more complicated answer is it depends on angle.</p>
  <p>Before going further, it is important to note that while sound pressure is sufficient to describe acoustics in one media, sound intensity is the ultimate measure of how &quot;loud&quot; a sound is. There is a good discussion of this on the <a href="https://dosits.org/science/sounds-in-the-sea/how-does-sound-in-air-differ-from-sound-in-water/" rel="nofollow noreferrer">DOSITS website</a>. I made this mistake earlier, despite having been repeatedly warned about this issue by my teachers in the past.</p>
  <p>The sound intensity of a plane wave is the product of the acoustic pressure and the acoustic velocity (Eq. numbers from Computational ocean acoustics):</p>
  <p>(1.12) I = p <strong>v</strong> / 2</p>
  <p>for a plane wave, the velocity is related to the pressure by</p>
  <p>(2.20) p = rho c <strong>v</strong></p>
  <p>where the product of rho and c is the <a href="https://en.wikipedia.org/wiki/Acoustic_impedance" rel="nofollow noreferrer">acoustic impedance</a> of media. The symbol Z is introduced for the acoustic intensity. This leads to the definition of acoustic intensity for a plane wave</p>
  <p>(Sec. 1.3.3.1) I = p**2 / Z</p>
  <p>So, when comparing sound pressures in the same media this issue of changing acoustic intensity can be ignored, and there is instead a pressure reference level for the media. For historical reasons the reference level is 20 micropascal in air and 1 micropascal in water, which makes comparison of levels between these media a little tricky.</p>
  <p>Besides the difference in dB reference levels, however, there is an important physical difference between the intensity of waves with the same pressure in the two media. According to the last equation, the same pressure level in water (Z ~= 1.5e6 rayl) represents a very different acoustic intensity than in air (Z ~= 450 rayl).</p>
  <p>Now for the transmission problem: A plane wave will have an angle a1 in the air, and angle a2 in the water. For angle a1, the angle a2 is given by Snell's law:</p>
  <p>sin(a1) / sin(a2) = c1 / c2</p>
  <p>the pressure amplitude of the transmitted wave in the second media is :</p>
  <p>(1.57) T = (2 Z2 / sin(a2)) / (Z2 / sin(a2) + Z1 / sin(a1))</p>
  <p>*Edit: As pointed out by @WMXZ, the transmission coefficient at normal incidence for air to water is ~2, while the transmission coefficient from water to air is ~0. Both transmission coefficients of exactly 0 and 2 indicate perfect reflections, produced by the ideal pressure release and rigid boundary boundaries, respectively.</p>
  <p>For the ideal pressure release surface, a plane wave of 0 amplitude is transmitted into the other media. In the rigid boundary case (Z2 &gt;&gt; Z1), a plane wave with double amplitude is transmitted into the media, but according to the Eq. in Sec. 1.3.3.1, the transmitted intensity is also 0.</p>
  
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Answers.Body: <p>For vertical incidence and sound moving from medium 1 to medium 2
  we have for the transmission coefficient</p>
  <p>T=2 * (rho2 * c2)/(rho2 * c2 + rho1 * c1)</p>
  <p>rho is density
  {take sin to 90 deg in formula given by @neilrichards)}</p>
  <p>then for air to water the transmission coefficient becomes close to 2
  and for water to air the transmission coefficient becomes about 4.5 e-4</p>
  <p>sound generated in water reflects easily on the surface, and sound generated in air transits easily into water. In fact detecting acoustically the overflight of an helicopter or an aircraft can been done.</p>
  <p>Note however, that this does not mean that the sound intensity is the same.
  The formula hold for sound pressure values and for the same pressure value, the sound intensity in water is about 3710 times lower than in air.</p>
  <ul>
  <li><p>Edit: One consequence of the acoustic impedance mismatch between air and water is that sound detection may be impacted. If your define your detector as intensity ratio, then you must consider the acoustic impedance mismatch (and not only pressure squared) between sound in air
  and in water.</p>
  </li>
  <li><p>comment: A technical consequence is that sound sensors must be generated such that no impedance mismatch occurs. Microphones do not work well underwater, hydrophones do not work well in air. They both work in the other domain, but not so good.</p>
  </li>
  </ul>
  
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Answers.Text: I edited my previous answer below due to your helpful comments. I think it is misleading to say "sound generated in air transits easily into water". The same pressure amplitude in air represents significantly less acoustic intensity in water. Overflight of a helicopter is very loud and low frequency compared to many acoustic sources of practical interest.
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Body: <p>How would one go about calculating the amount of sound transmitted from air to water?</p>
  <p>From experience recording bats over water surfaces I suspect a lot of sound is reflected back up into the air. However, can someone tell me broadly how much of the sound makes it into the water?</p>
  <p>I tried looking into this and found one somewhat deep-sea oriented <a href="https://www.acoustics.asn.au/conference_proceedings/AAS2002/AAS2002/PDF/AUTHOR/AC020127.PDF" rel="noreferrer">report</a>, which didn't bring me far.</p>
  <p>I'm looking to understand how much of the sound (say the received level at 1m below surface) will go into the water from the call of a Daubenton's bat flying 30-40 cm above the water surface. Let's assume a source level of 100 dB SPL @1 m (20uPa ref).</p>
  
Tags: <sound-propagation><modelling>
Title: extent of sound transmission from air to water
ThreadText: # extent of sound transmission from air to water
  
  (Asked by: username_0 on 2022-07-12)
  
  How would one go about calculating the amount of sound transmitted from air to water?
  
  From experience recording bats over water surfaces I suspect a lot of sound is reflected back up into the air. However, can someone tell me broadly how much of the sound makes it into the water?
  
  I tried looking into this and found one somewhat deep-sea oriented [report](https://www.acoustics.asn.au/conference_proceedings/AAS2002/AAS2002/PDF/AUTHOR/AC020127.PDF), which didn't bring me far.
  
  I'm looking to understand how much of the sound (say the received level at 1m below surface) will go into the water from the call of a Daubenton's bat flying 30-40 cm above the water surface. Let's assume a source level of 100 dB SPL @1 m (20uPa ref).
  
  **Tags:** sound-propagation, modelling
  
  
  ---
  
  ## 2 answers
  
  ### Answer by username_2 on 2022-07-12. Score: 11 | Accepted answer
  
  For vertical incidence and sound moving from medium 1 to medium 2
  we have for the transmission coefficient
  
  T=2 \* (rho2 \* c2)/(rho2 \* c2 + rho1 \* c1)
  
  rho is density
  {take sin to 90 deg in formula given by @neilrichards)}
  
  then for air to water the transmission coefficient becomes close to 2
  and for water to air the transmission coefficient becomes about 4.5 e-4
  
  sound generated in water reflects easily on the surface, and sound generated in air transits easily into water. In fact detecting acoustically the overflight of an helicopter or an aircraft can been done.
  
  Note however, that this does not mean that the sound intensity is the same.
  The formula hold for sound pressure values and for the same pressure value, the sound intensity in water is about 3710 times lower than in air.
  
  * Edit: One consequence of the acoustic impedance mismatch between air and water is that sound detection may be impacted. If your define your detector as intensity ratio, then you must consider the acoustic impedance mismatch (and not only pressure squared) between sound in air
    and in water.
  * comment: A technical consequence is that sound sensors must be generated such that no impedance mismatch occurs. Microphones do not work well underwater, hydrophones do not work well in air. They both work in the other domain, but not so good.
  
  > **Comments:**
  > **username_1**: I edited my previous answer below due to your helpful comments. I think it is misleading to say "sound generated in air transits easily into water". The same pressure amplitude in air represents significantly less acoustic intensity in water. Overflight of a helicopter is very loud and low frequency compared to many acoustic sources of practical interest.  
  
  ---
  
  ### Answer by username_1 on 2022-07-12. Score: 8
  
  The simple answer is not a lot, the more complicated answer is it depends on angle.
  
  Before going further, it is important to note that while sound pressure is sufficient to describe acoustics in one media, sound intensity is the ultimate measure of how "loud" a sound is. There is a good discussion of this on the [DOSITS website](https://dosits.org/science/sounds-in-the-sea/how-does-sound-in-air-differ-from-sound-in-water/). I made this mistake earlier, despite having been repeatedly warned about this issue by my teachers in the past.
  
  The sound intensity of a plane wave is the product of the acoustic pressure and the acoustic velocity (Eq. numbers from Computational ocean acoustics):
  
  (1.12) I = p **v** / 2
  
  for a plane wave, the velocity is related to the pressure by
  
  (2.20) p = rho c **v**
  
  where the product of rho and c is the [acoustic impedance](https://en.wikipedia.org/wiki/Acoustic_impedance) of media. The symbol Z is introduced for the acoustic intensity. This leads to the definition of acoustic intensity for a plane wave
  
  (Sec. 1.3.3.1) I = p\*\*2 / Z
  
  So, when comparing sound pressures in the same media this issue of changing acoustic intensity can be ignored, and there is instead a pressure reference level for the media. For historical reasons the reference level is 20 micropascal in air and 1 micropascal in water, which makes comparison of levels between these media a little tricky.
  
  Besides the difference in dB reference levels, however, there is an important physical difference between the intensity of waves with the same pressure in the two media. According to the last equation, the same pressure level in water (Z ~= 1.5e6 rayl) represents a very different acoustic intensity than in air (Z ~= 450 rayl).
  
  Now for the transmission problem: A plane wave will have an angle a1 in the air, and angle a2 in the water. For angle a1, the angle a2 is given by Snell's law:
  
  sin(a1) / sin(a2) = c1 / c2
  
  the pressure amplitude of the transmitted wave in the second media is :
  
  (1.57) T = (2 Z2 / sin(a2)) / (Z2 / sin(a2) + Z1 / sin(a1))
  
  \*Edit: As pointed out by @username_2, the transmission coefficient at normal incidence for air to water is ~2, while the transmission coefficient from water to air is ~0. Both transmission coefficients of exactly 0 and 2 indicate perfect reflections, produced by the ideal pressure release and rigid boundary boundaries, respectively.
  
  For the ideal pressure release surface, a plane wave of 0 amplitude is transmitted into the other media. In the rigid boundary case (Z2 >> Z1), a plane wave with double amplitude is transmitted into the media, but according to the Eq. in Sec. 1.3.3.1, the transmitted intensity is also 0.
  
  
  ---
  
  