In the BBC Science in Action podcast the acoustic power of a 1 millisecond "click" from a whale was said to be 230 dB. The related BBC article mentions this number and elaborates:

The blue whale is not the loudest animal on Earth, despite what you may have learned in school. While its calls are claimed to be louder than a jet engine at take-off, clocking in at an impressive 188 decibels (dB), the sperm whale is actually louder: its communicative clicks have been measured at 230 dB.

If I use the following equation from here;

$$I(dB)=10\log_{10}\frac{I}{I_0} $$

rearrange it,

$$I= I_0 \ 10^{I(dB)/10} $$

and plug in the value of 10-12 watts/m2 given there and use a nominal area of 1 m2, I get 1023-12 = 1011 watts/m2! So, if the duration of the "click" is 1 millisecond, that's 100 MJ of energy. It's not surprising then that it can also use acoustic pulses to disable or stun potential prey.

But the article goes on to say:

This record raises another important point about loudness. Decibels in water are not equivalent to decibels in air. "Water is denser than air, so sound travels through it differently – the speed of sound is different," says bio-acoustics expert James Windmill from the University of Strathclyde in the UK, who discovered the water boatman's remarkable call.

Roughly, to convert from dB in water to dB in air, you have to subtract [around] 61 dB from the reported sound level."

I have a feeling this is somehow related to sound pressure vs intensity; force/area versus power/area. Water is roughly 1000x as dense as air.

Question: I am not sure if I should be using this 61 dB offset in my calculation above. Is the energy of the "click" still 100 MJ?

  • $\begingroup$ You are using 230dB on the power formula, when it's really meant to be used with the amplitude formula, i.e. 20dB is a change of a factor of 10, not 10dB. That should put things into the right order of magnitude. :-) $\endgroup$
    – CuriousOne
    Aug 5 '16 at 2:38
  • $\begingroup$ @ChrisWhite - Ah, I assumed 1 square meter but didn't mention it - I'll fix that. Thanks! $\endgroup$
    – uhoh
    Aug 5 '16 at 2:46
  • $\begingroup$ Loudness is usually specified in $dB_{Amplitude}$, so that's twice the number of $dB_{Intensity}$. $\endgroup$
    – CuriousOne
    Aug 5 '16 at 2:50
  • $\begingroup$ @CuriousOne this is acoustic power, not loudness. 10 is correct here. However that means that the 61dB should be 30.5dB which looks like the ratio of the densities., My question is, do I need to apply that "water to air" correction to my power (and energy) estimate, or is that only for loudness considerations? $\endgroup$
    – uhoh
    Aug 5 '16 at 2:51
  • 1
    $\begingroup$ Water and air don't even use the same reference level. 0 dBSPL in air is 20 μPa. 0 dBSPL underwater is 1 μPa. $\endgroup$
    – endolith
    Mar 23 '21 at 21:57

First, I went into great detail on the propagation of sound in different media in the following answer: https://physics.stackexchange.com/a/266046/59023.


Intensity (or specifically sound intensity) of a linear sound wave is related to sound pressure, $P$, through: $$ I = \frac{ P^{2} }{ \rho_{o} \ C_{s} } $$ where $\rho_{o}$ is the mass density and $C_{s}$ is the speed of sound in the medium. One can look up the properties of water to find that $\rho_{o}$ ~ 999.972 kg/m3 and $C_{s}$ ~ 1484 m/s. We can also look up the reference pressure level for water (or at NOAA) finding $P_{H2O}$ ~ 1 $\mu$Pa (compared to $P_{air}$ ~ 20 $\mu$Pa) at 1 meter from source. This corresponds to a reference intensity of $I_{o} \sim 6.74 \times 10^{-19}$ W/m2 for water.

We can then use: $$ I = I_{o} \ 10^{L/10} $$ where $I$ is intensity (in W/m2) and $L$ is intensity (in dB). If we use the intensity I show above and $L$ ~ 230 dB, then $I \sim 6.74 \times 10^{4}$ W/m2 under water.

If we use your 1 m2 area and $\Delta t$ ~ 1 ms, then the total energy would be ~67.4 J.


I am not sure if I should be using this 61 dB offset in my calculation above. Is the energy of the "click" still 100 MJ?

No, it is much lower at ~67.4 J. The $\sim 10^{11}$ W you found is rather extreme when you consider that it corresponds to peak electrical power consumption of France. Though whales are rather enormous mammals and a ~10,000 calorie diet does correspond to $\sim 10^{7}$ J of energy (i.e., diet of an Olympic athlete, which is probably much less than a sperm whale), that would still be a lot of power to produce.

  • $\begingroup$ So I think the answer should perhaps include "Yes, you are right. You do have to subtract the 61 dB offset"? That would drop 10${}^{11}$ watts to 10${}^{5}$ watts, and for 1 millisecond that would be 10${}^{2}$ Joules, the same as your value. Thank you for working the problem through clearly! Also I like your linked answer as well $\endgroup$
    – uhoh
    Oct 5 '16 at 0:39
  • $\begingroup$ Thanks again for your answer, I've just used it here. $\endgroup$
    – uhoh
    Oct 22 '18 at 11:00
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    $\begingroup$ @uhoh - No worries. The linked answer took a lot of work to dig up all those reference numbers and then extrapolate things properly. Once I had that answer figured out, this one was relatively straight forward to address. $\endgroup$ Oct 22 '18 at 20:11
  • 1
    $\begingroup$ P_air is 20 μPa, not 10 $\endgroup$
    – endolith
    Mar 23 '21 at 21:58
  • 1
    $\begingroup$ @endolith - Ah good catch, thanks! $\endgroup$ Mar 23 '21 at 22:25

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