<p>@MichaelLadegaard's answer is perfectly correct, so I try here a short version of it.</p>
<p>The Term &quot;195-224 dB re 1 μPa at 1 m&quot; (note I added a 1 before &quot;μPa&quot; (micro Pascal) to be more precise for underwater situations) is simply a short notation for:</p>
<p>&quot;Sound source generates at 1 m a sound pressure between 10^(195/20) μPa  and 10^(224/20) μPa&quot;.</p>
<p>Observation 1: Pascal (Pa) measures sound pressure with 1 Pa = 1N/m^2
( atmospheric pressure at sea level (1 atm = 101.325 kPa)</p>
<p>We know that 10^(195/20)= 5.62x10^9 and 10^(224/20)= 1.58x10^11, and considering that 1 μPa = 10^(-6) Pa, the term in question may be translated to</p>
<p>&quot;Sound source generates at 1 m a sound pressure between 5620 Pa and 15800 Pa&quot;.</p>
<p>Observation 2: sound pressure is rarely measured at 1 m, but always back calculated to 1 m using appropriate spreading models.</p>

CC BY-SA 4.0
2024-11-22T18:39:27.153Z
2024-11-22T18:39:27.153Z
WMXZ
<p>I'll try. First a little background:</p>
<p>Sound Intensity = sound_pressure * particle_velocity</p>
<p>Sound intensity on a decibel scale = 10*log10( Intensity_measured / Intensity_reference)</p>
<p>Because intensity is quantified relative to a reference intensity, and because of the relation between sound_pressure and particle_velocity, there are things in the equation above that can be substituted or cancel out, so you can in fact estimate the sound intensity (in a free field, i.e. &quot;far&quot; away from the sound source and reflecting surfaces) from the pressure alone:</p>
<p>Sound intensity level (or sound pressure level) = <strong>10</strong>*log10( pressure_measured^2 / pressure_reference^2)</p>
<p>That can also be written this way:
Sound pressure level (SPL) = <strong>20</strong>*log10( pressure_measured / pressure_reference)</p>
<p>In water, the standard reference pressure is 1 µPa and 20 µPa in air.</p>
<p>So when you have you equation from above, and if you want to know the absolute pressure (in Pascal instead of dB re 1 µPa), you can insert your value in the equation above:</p>
<p>SPL = 224 dB re 1 µPa = 20 * log10( pressure / 1 µPa)</p>
<p>pressure = 10^( 224 / 20 ) * 1 µPa =  1.58 * 10^11 µPa = 15,800 Pascal</p>
<p>So you can see that sometimes, the µPa and Pascal values become very large and slightly complicated to do computations with at the top of your head. But with a bit of practice, it becomes quite easy with decibels. For example, 195 dB re µPa is approximately 30 dB lower than the 224 dB re µPa from the SPL range you provided. If you are used to working with decibels then you might know that a 30 dB difference is approximately a 30x difference on a linear scale, so 195 dB re 1 µPa approximately equals the 15,800 Pascals from above divided by 30 = 5,270 Pascals, which is very close to the actual value of 5,012 Pascals. Decibels are also useful because you get a feel for whether differences between values are significant or not. A difference of ~270 Pascal may sound like a lot, but it depends on what you compare it to. Relative to 1 Pascal, yes (the difference is 20<em>log10(270/1) = 49 dB), but relative to 5000 Pascals, no, as 20</em>log10(5270/5000) is &lt;1 dB difference.</p>
<p>Rules of thumb:</p>
<ul>
<li><p>6 dB difference is a 2x difference in sound pressure</p>
</li>
<li><p>10 dB difference is a ~3x difference in sound pressure</p>
</li>
<li><p>20 dB difference is a 10x difference in sound pressure</p>
</li>
</ul>
<p>-- so a difference of 30 dB = 20 dB + 10 dB = 10 * 3 = 30 times difference on a linear scale.</p>
<p>-- a 60 dB difference in sound level is the same as 20 dB + 20 dB + 20 dB = 10 * 10  * 10 = 1000 times difference in sound pressure</p>
<p>NOTE: Because intensity ratios are computed as <strong>10</strong><em>log10(intensity ratio) and pressure ratios using <strong>20</strong></em>log10(pressure ratio), a 60 dB (1000 times) difference in sound pressure is equal to a 60 dB (1,000,000 times) difference in intensity, or energy.</p>
<p>The &quot;at&quot; (which is often written @) from your question relates to the number being a &quot;source level&quot;. That is to say that if you measure a received level at a certain range from the sound source, then you can back-calculate what the level would be if you were to measure it 1 m away from the sound source by estimating the transmission loss, which you then add to your received level.</p>
<p>I hope that helped a bit.</p>

CC BY-SA 4.0
2024-11-22T11:07:52.483Z
2024-11-22T12:00:21.663Z
2024-11-22T12:00:21.663Z
Michael Ladegaard
<p>I'm reading a journal article on wind and vessel noise in the sound fields experienced by southern killer whales, and I have struggled with a lot of technical jargon in the methods and materials section. However, I came across this &quot;195-224 dB re μPa at 1 m,&quot; the only part I can decipher from it is at 1 meter part. Can someone please break it down into layman's terms for me?</p>

<noise><echolocation><noise-impact><source-level>
Can someone explain 195-224 dB re μPa at 1 m to me?
# Can someone explain 195-224 dB re μPa at 1 m to me?

(Asked by: username_0 on 2024-11-22)

I'm reading a journal article on wind and vessel noise in the sound fields experienced by southern killer whales, and I have struggled with a lot of technical jargon in the methods and materials section. However, I came across this "195-224 dB re μPa at 1 m," the only part I can decipher from it is at 1 meter part. Can someone please break it down into layman's terms for me?

**Tags:** noise, echolocation, noise-impact, source-level


---

## 2 answers

### Answer by username_2 on 2024-11-22. Score: 4 | Accepted answer

I'll try. First a little background:

Sound Intensity = sound\_pressure \* particle\_velocity

Sound intensity on a decibel scale = 10\*log10( Intensity\_measured / Intensity\_reference)

Because intensity is quantified relative to a reference intensity, and because of the relation between sound\_pressure and particle\_velocity, there are things in the equation above that can be substituted or cancel out, so you can in fact estimate the sound intensity (in a free field, i.e. "far" away from the sound source and reflecting surfaces) from the pressure alone:

Sound intensity level (or sound pressure level) = **10**\*log10( pressure\_measured^2 / pressure\_reference^2)

That can also be written this way:
Sound pressure level (SPL) = **20**\*log10( pressure\_measured / pressure\_reference)

In water, the standard reference pressure is 1 µPa and 20 µPa in air.

So when you have you equation from above, and if you want to know the absolute pressure (in Pascal instead of dB re 1 µPa), you can insert your value in the equation above:

SPL = 224 dB re 1 µPa = 20 \* log10( pressure / 1 µPa)

pressure = 10^( 224 / 20 ) \* 1 µPa = 1.58 \* 10^11 µPa = 15,800 Pascal

So you can see that sometimes, the µPa and Pascal values become very large and slightly complicated to do computations with at the top of your head. But with a bit of practice, it becomes quite easy with decibels. For example, 195 dB re µPa is approximately 30 dB lower than the 224 dB re µPa from the SPL range you provided. If you are used to working with decibels then you might know that a 30 dB difference is approximately a 30x difference on a linear scale, so 195 dB re 1 µPa approximately equals the 15,800 Pascals from above divided by 30 = 5,270 Pascals, which is very close to the actual value of 5,012 Pascals. Decibels are also useful because you get a feel for whether differences between values are significant or not. A difference of ~270 Pascal may sound like a lot, but it depends on what you compare it to. Relative to 1 Pascal, yes (the difference is 20*log10(270/1) = 49 dB), but relative to 5000 Pascals, no, as 20*log10(5270/5000) is <1 dB difference.

Rules of thumb:

* 6 dB difference is a 2x difference in sound pressure
* 10 dB difference is a ~3x difference in sound pressure
* 20 dB difference is a 10x difference in sound pressure

-- so a difference of 30 dB = 20 dB + 10 dB = 10 \* 3 = 30 times difference on a linear scale.

-- a 60 dB difference in sound level is the same as 20 dB + 20 dB + 20 dB = 10 \* 10 \* 10 = 1000 times difference in sound pressure

NOTE: Because intensity ratios are computed as **10***log10(intensity ratio) and pressure ratios using **20***log10(pressure ratio), a 60 dB (1000 times) difference in sound pressure is equal to a 60 dB (1,000,000 times) difference in intensity, or energy.

The "at" (which is often written @) from your question relates to the number being a "source level". That is to say that if you measure a received level at a certain range from the sound source, then you can back-calculate what the level would be if you were to measure it 1 m away from the sound source by estimating the transmission loss, which you then add to your received level.

I hope that helped a bit.


---

### Answer by username_1 on 2024-11-22. Score: 2

@username_2's answer is perfectly correct, so I try here a short version of it.

The Term "195-224 dB re 1 μPa at 1 m" (note I added a 1 before "μPa" (micro Pascal) to be more precise for underwater situations) is simply a short notation for:

"Sound source generates at 1 m a sound pressure between 10^(195/20) μPa and 10^(224/20) μPa".

Observation 1: Pascal (Pa) measures sound pressure with 1 Pa = 1N/m^2
( atmospheric pressure at sea level (1 atm = 101.325 kPa)

We know that 10^(195/20)= 5.62x10^9 and 10^(224/20)= 1.58x10^11, and considering that 1 μPa = 10^(-6) Pa, the term in question may be translated to

"Sound source generates at 1 m a sound pressure between 5620 Pa and 15800 Pa".

Observation 2: sound pressure is rarely measured at 1 m, but always back calculated to 1 m using appropriate spreading models.


---

