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I know of events that are happening about 45 KM away from me which are said to be 210 or 213 dB at 75 meters distance from multiple sources. I think that I can hear them, so I did the obvious:

$$ \frac{I_2}{I_1} = \left( \frac{d_1}{d_2} \right)^2$$

$$ I_2 = I_1 * \left( \frac{d_1}{d_2} \right)^2\text{ dB}$$

For my values:

$$ I_2 = 213 \left( \frac{75}{45000} \right)^2\text{ dB}$$

$$ I_2 = 213 \left( \frac{1}{600} \right)^2\text{ dB}$$

$$ I_2 = \frac{213}{360000}\text{ dB}$$

$$ I_2 \approx 0.6 \cdot 10^{-3}\text{ dB}$$

Well, 0.6 e-3 dB I should not be able to hear!

However, going by the rule of thumb "twice the distance, minus three decibels":

$$ 75 * 2^x = 45000 $$

$$ 2^x = 600 $$

$$ x = \log_2(600) $$

$$ x \approx 9.2 $$

$$ 213 - 3 \cdot 9.2 = 213 - 27.6 = 185.4 $$

The value of 184 dB seems too loud, and additionally does not agree with the previous method.

Lastly I tried tool for estimating Sound Levels With the Inverse Square Law from the usually-terrific HyperPhysics website. This tool gave a value of 157 decibels! Another on-line tool gives the same result.

Which result should I trust?

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2  
Can't answer your questions, but I can say that the inverse square law can't be expected to be valid between two points at ground level. The existence of the ground, terrain, trees, structures, etc. will screw it up. –  garyp Jul 25 at 13:46
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Also, the temperature gradient in the vertical direction is known to affect the propagation of sound waves significantly. –  auxsvr Jul 25 at 13:59
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I thought it was half the intensity, minus three decibels. –  BMS Jul 25 at 15:22
    
@BMS: Yes, it looks like you are correct. Thank you. –  dotancohen Jul 25 at 15:27
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@ChrisWhite: the first reference gives the intensity of a 1000 lb bomb at 250 feet which is around 75 m. –  Edward Jul 25 at 15:40

3 Answers 3

up vote 10 down vote accepted

The problem is with your first calculation and also with the somewhat misleading equation that you've found. It's true that $$\frac{I_2}{I_1}=\left(\frac{d_1}{d_2}\right)^2$$ but units are important here. In that formula, $I_1$ and $I_2$ would properly be expressed as power values. To compute with decibels, which are logarithmic quantities, one would instead use $$I_1 + 10\log\left(\frac{d_1}{d_2}\right)^2 = I_2$$ or equivalently, $$I_1 + 20\log\left(\frac{d_1}{d_2}\right) = I_2$$, where $I_1$ and $I_2$ are decibels and $d_1$ and $d_2$ are in identical linear units (feet or meters, for example).

With your particular numbers we get $$\begin{eqnarray} I_2 &=& 213\text{ dB} + 20 \log\left(\frac{75}{45000}\right) \\ &=& 213\text{ dB} + 20\log\left(\frac{1}{600}\right) \\ &\approx& 213\text{ dB} + 20(-2.78) \\ &\approx& 213\text{ dB} - 55.56 \\ &\approx&157.4\text{ dB} \end{eqnarray}$$

Estimating manually

You've correctly remembered that -3dB is half the power. That is, $$\frac{1}{2}P = -3\text{dB}$$. Another easily remembered fact is $$\frac{1}{10}P = -10\text{dB}$$. Both are very commonly used in engineering for rough estimations. So in this case, because it's an inverse square law, we have $$\begin{eqnarray} \left(\frac{75}{45000}\right)^2 &=& \frac{1}{600^2} \\ &=& \frac{1}{360000} \\ &\approx& \frac{1}{400000} \\ &\approx& \frac{1}{2^2\cdot 10^5} \\ &\approx& -6\text{dB} - 50\text{dB} \\ &\approx& -56\text{dB} \end{eqnarray}$$ So this would give $213\text{dB} - 56\text{dB} = 157\text{dB}$

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Thank you Edward. Considering then the facts as you've stated them, a 1 ton bomb dropped 45 KM away is as loud as an NHRA Dragster right next to someone? That just does not seem realistic! –  dotancohen Jul 25 at 15:22
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@dotancohen: It isn't realistic. On the surface of the earth, propagation of sound is affected by many other factors, including the frequencies of the sound, attenuation due to objects between you and the sound, multipath effects from echos and many others. You may find this wikibook of interest. –  Edward Jul 25 at 15:34
    
That wikibook is terrific, thank you! –  dotancohen Jul 25 at 16:25
    
I see now where my confusion was. Thank you for showing the maths explained, and especially for the comment which explains why the perceived value to my ears is so much lower that what the maths make us expect. –  dotancohen Jul 25 at 20:05

The formula you use does not make any sense - you don't measure intensity of sound in dB, but the logarithm of the intensity, so you cannot multiply by the distance ratio squared.

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Thank you. The online information that I've been able to locate has been loose in their interpretations and strict in their expectations of the level of knowledge of the reader. Thank you for stating it so clearly! –  dotancohen Jul 25 at 15:20

I'll piece together some of what's been said in answers and comments in a different light.


With acoustics, it pays to be very careful with units. A sound wave has a pressure $p$, and this corresponds to what I'll call it's intensity $I$. Intensity goes as the square of pressure: $$ \frac{I}{I_0} = \left(\frac{p}{p_0}\right)^2. $$ Here $I_0$ is the threshold of human hearing, corresponding to pressure $p_0 = 20\ \mathrm{μPa}$. Intensity also falls off as the square of distance: $$ \frac{I_1}{I_2} = \left(\frac{d_2}{d_1}\right)^2. $$

Now loudness $L$ is measured logarithmically with respect to intensity or pressure. In particular, $$ L = 10\ \mathrm{dB} \log_{10}\left(\frac{I}{I_0}\right). $$ Thus the change in loudness is $$ L_1 - L_2 = 10\ \mathrm{dB} \log_{10}\left(\frac{I_1}{I_2}\right) = 20\ \mathrm{dB} \log_{10}\left(\frac{d_2}{d_1}\right). $$ Going from $75\ \mathrm{m}$ to $45\ \mathrm{km}$ does indeed induce a loudness change of $56\ \mathrm{dB}$, all else being equal.


One catch is that sound has problems traveling long distances near the ground. Things get in the way, and its propagation is complicated. I won't attempt to model that here, but it probably means we're overestimating the sound far from the source.


Another point important for this particular problem is that different measurements will fold in different types of energy. In particular, this chart says a $1$-ton bomb at $250$ feet comes in at $213\ \mathrm{dB}$, including all power such as wind. The sound (pressure wave) is only $176\ \mathrm{dB}$ at the same distance.


Another catch is that human perception doesn't quite follow the loudness scale. Really one should correct for our ears being more attuned to certain frequencies than others. One way to do this is ask, "How loud will a pure $1\ \mathrm{kHz}$ sound need to be to sound as loud to our ears?" The result is the value $L_N$ in phons. Since we don't have any frequency data, I'll just skip this step, but in reality low rumblings don't sound as loud as their raw loudness $L$ values imply. For very low frequencies the difference can be dramatic.

Even once different frequencies are accounted for, our hearing still doesn't quite jibe with decibels. A better measurement is done in sones, leading to a perceived loudness $N$. The formula is $$ N = \left(10^{L_N/(10\ \mathrm{dB})-4}\right)^{0.30103}. $$

At $75\ \mathrm{m}$ we have $L_1 = 176\ \mathrm{dB}$, so we assume $L_{N,1} = 176\ \mathrm{dB}$, meaning $N_1 = 12{,}400\ \mathrm{sone}$. At $45\ \mathrm{km}$ we have $L_2 = L_{N,2} = 120\ \mathrm{dB}$, meaning $N_2 = 256\ \mathrm{sone}$. That puts it at the threshold of hearing damage even for short exposures, but remember the actual value is probably much lower due to all the sound being in low frequencies.

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Thank you. You certain understand how to phrase the ideas into terms that are easier to digest, and to address the points of confusion. –  dotancohen Jul 25 at 20:03
    
@ChrisWhite: you've mentioned phons and sones but have neglected scones which is defined as "the maximum sound level that a typical adult human can be subjected to during the preparation of breakfast tea and still plausibly pretend to be asleep." :) –  Edward Jul 25 at 20:13

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