by Neil Bauman, Ph.D.

October 29, 2016

A person asked,

How do you calculate the difference in sound intensity in decibels between any two sound intensities. For example, how do you calculate the increase in sound intensity between 0 dB and 15 dB or between 52 and 94 dB?

There is a mathematical relationship between decibels (dB) and sound intensities. It works like this. Each 10 dB increase results in a **10-fold** increase in sound **intensity** which we **perceive** as a **2-fold** increase in sound **volume**.

Thus, from 0 dB to 10 dB there is a 10-fold increase in sound intensity, just as there is from 10 dB to 20 dB or from 34 dB to 44 dB.

Note: Sound **intensity** is the **energy** (power) needed to produce a given level of sound. Don’t confuse sound intensity (the amount of energy needed to produce a given level of sound) with sound **volume** (the level at which we **perceive** the resulting sound.)

The table below shows the increase in sound intensity between 0 dB and each of the values listed.

Decibel

Value Increase in Sound Intensity Perceived Increase in Volume

0 dB

10 dB 10 times the sound intensity 2 times as loud

20 dB 100 (10 x 10) 4 (2 x 2)

30 dB 1,000 (10 x 10 x 10) etc. 8 (2 x 2 x 2) etc.

40 dB 10,000 16

50 dB 100,000 32

60 dB 1,000,000 64

70 dB 10,000,000 128

80 dB 100,000,000 256

90 dB 1,000,000,000 512

100 dB 10,000,000,000 1024

110 dB 100,000,000,000 2048

120 dB 1,000,000,000,000 4096

As you can see, these numbers quickly get large. For example, if you had a 120 dB loss at a certain frequency, in order to hear a sound at that frequency, it would have to be 1 trillion times as intense (it would require 1 trillion times the energy to produce it) as needed for a person who had “perfect” hearing (and thus could hear it at an intensity of 0 dB).

Note this well. Since our ears **perceive** sound logarithmically, we do not perceive a sound of 120 dB as being 1 trillion times louder than a sound of 0 dB. Rather, we perceive it as about 4,000 times louder.

Now that we have a little background, we are ready to proceed with the details of how to calculate the differences in sound intensities and relate them to decibel values.

Unfortunately, far too often people assume that there is a simple linear interpolation between any two decibel values. Thus, since there is a 10-fold increase between 10 dB and 20 dB in sound intensity, they assume the increase at the half-way point (15 dB in this case) is a 5-fold increase.

If you assumed this, you would be wrong. Even hearing health care professionals that should know don’t always get this right.

The reason you can’t just simply interpolate between two decibel values is because we are not working with linear numbers, but with logarithmic numbers. This means there is a logarithmic relationship between such values, not a linear relationship.

The formula for calculating the increase in sound intensity between two decibel values is:

x-fold increase in sound intensity = 10 ^{(ending dB value – starting dB value)/10}

Therefore, to find the increase in sound intensity between 10 dB and 15 dB, you simply subtract the higher dB value from the lower value and divide the result by 10 to get the exponent. Calculating (15 – 10)/10 gives you an exponent of 0.5. Raising 10 to the 0.5 power (10 ^{0.5}) gives 3.162. Thus, the intensity increase between 10 dB and 15 dB is 3.162-fold.

In like manner, to calculate the difference in sound intensity between 52 dB and 94 dB, just follow the same procedure and use the same formula. (94-52)/10 gives an exponent of 4.2. 10 ^{4.2} = 15,848.9. Thus, the intensity increase between 52 dB and 94 dB is 15,848.9-fold. To put it another way, it takes 15,848.9 times as much energy to produce a sound of 94 dB than to produce a sound of 52 dB.

It’s easy to check your work to be sure you are in the right ball park. You know the difference you are working with is 42 dB. You already know that for a 40 dB increase, the intensity value is 10,000 times higher (10 x 10 x 10 x 10) and that for a 50 db increase, the value would be 100,000 times higher (10 x 10 x 10 x 10 x 10). (See above table.) So your answer must lie somewhere between these two values, and sure enough, it does.

To make things simple, in case you don’t have a fancy calculator*, here is a table to help you.

dB Difference x-fold Multiplier

1 1.259

2 1.584

3 1.995

4 2.512

5 3.162

6 3.981

7 5.011

8 6.309

9 7.943

10 10.000

In order to use this table, just take the multiplier figures for values between 1 and 10 and then move the decimal point to the right one place for each whole 10 dB difference.

Thus, if you want to find the difference in sound intensity between 3 dB and 9 dB, and since the value is less than 10 dB, just read off the value from the table for a 6 dB difference, namely 3.981. Thus for a 6 dB increase, there is a 3.981-fold increase in intensity.

If you want to find the sound intensity increase between 52 and 94 dB, you subtract 52 from the 94 to get 42 dB. Take the units figure (2) and from the table for a 2 dB difference, you see the multiplier is 1.584. Now to get your final answer, move the decimal to the right by the value of the tens figure (4) and you have a 15,840-fold increase in intensity. (If the decibel difference is larger than 100, then use the tens and hundreds figures. Thus if the difference was 124 dB, you’d move the decimal to the right by 12 decimal places.) That’s how simple it is.

And if you ever want to calculate how much louder you **perceive** one sound as compared to another you can do it by using the following formula.

perceived x-fold volume increase = 2^{ (ending dB value – starting dB value)/10}

Therefore, to find the perceived increase in sound volume between 10 dB and 15 dB, you simply subtract the higher dB value from the lower value and divide the result by 10 to get the exponent—(15 – 10)/10 gives you an exponent of 0.5. (So far, everything is the same as for calculating intensity differences. Now comes the change—you use base 2 rather than base 10.) Raising 2 to the 0.5 power (2^{ 0.5}) gives 1.4. Thus you would perceive the sound as being 1.4 times louder.

In like manner, to calculate the difference in perceived sound volume between 52 dB and 94 dB, just follow the same procedure and use the formula. (94-52)/10 gives an exponent of 4.2. 2^{ 4.2} = 18.4 times louder.

Note: Perceived volume varies from person to person so the calculated results may not agree with any given person’s subjective results, but it certainly puts you in the right ball park.

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* Note: if you have an iPhone, you have a fancy built-in calculator. Swipe up from the bottom and you’ll see it there with your flashlight, timer and camera. When you hold your iPhone vertically you have a simple calculator. Turn your phone on its side and it automatically switches to a fancy scientific calculator where you have the 10^{x} and x^{y} functions.

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