Sound at a distance

Goal

Understand computations involving sound over a distance including delays and level.

Distance

distance=rate\times time

d=rt

I like to remember this formula as the word "dirt" with out the "i" so "drt".

Imagine you are in a hallway and you are going to the end it. As you start walking you start at position 0. Maybe in 1 second you read door 1. Then the next second you reach door 2, and so on until you reach the end. In order to get to the end of the hallway you had to change your position over time. This changing position position over time described your rate or speed at which you moved. If we know your speed and how long you moved we can deduce how far you went.

If we know any of the two quantities we can find the third. Maybe we know how fast you will move and the distance and want to know how long it will take. In the cases we will be doing we will assume a constant rate. To deal with more complex cases with a changing rate we would require calculus.

Sound at a Distance

To calculate how loud a certain sound is at a given distance we should recall a couple basic facts..

Speed of sound is
$1130\frac{ft}{s} \ or\ 343\ \frac{m}{s}$
$d=rt$
Distance is a comparison from where you want to end to where you started from, so if the reference is a distance compared to the source then use 1 for the reference. You can think of it as "you" are starting out at your location which is the level "it is" at your spot, in other words, "1" is used to represent "all 1 of it is there". Then you compare it to the distance change.

This means if you have any two: rate (speed), time, or distance, then you can find the third one. The application is very straightforward.

Example

If I yell “HI” to you when you are 80 meters away at 80 dB SPL, how long will it take to reach you?

Show Answer

We need the distance equation:

d=rt

We know the distance is

80m

. Since we are using meters we will need to use the meter version of the speed of sound

343\ \frac{m}{s}

Have have the rate and the distance so solve for time and use that as our equation:

\frac{d}{r} =t

Now we sub in the known values:

\frac{80m}{343\frac{m}{s}} =\boxed{.23\ s}

Level at a distance

We have seen how to find distance or time given we know one of them and the speed of sound. Now we can look at what the level difference would be at different distances. Since distance follows the inverse square law it will use the 20 log rule so doubling or halving distance will increase or decrease the level by 6 dB SPL.

The equation to describe sound a distance then will be the ratio of two distances. A reference distance and the measured distance. This ratio will describe the level change between the two distances. It is not the level itself but the change in level!

dB_{SPL} =20\log\left(\frac{distance_{measured}}{distance_{reference}}\right)

The log for ratios greater than 1 will yield a positive number, ratios less than 1 will yield a negative number. Ratios cannot be negative so we don't need to worry about negative numbers in the log. We wish to correlate the ratios with distance. For ratios greater than 1 the signal should get lower because the distance grew and if the ratio is less than one the signal should get higher because the distance shrank. Therefore we add a minus sign to the front of this ratio to reflect this.

dB_{SPL} =-20\log\left(\frac{distance_{measured}}{distance_{reference}}\right)

Now if we know the initial level we can combine it with the level change and find the new level at the desired distance.

dB_{SPL} =inital\ level-20\log\left(\frac{distance_{measured}}{distance_{reference}}\right)

When determining the reference location is it often obvious by whether or not the ratio should be bigger than 1. You can also think of it as the location your "starting at" as the reference and the location you are "going to" as the measured value. If you are starting at your location (or the mic's location) then the value is 1 so that the ratio works out as the measured distance to 1.

NOTE that the level here assumes free field conditions! Often other factors influence the level so this will be a ball park answer in practice.

Example

If I yell “HI” to you when you are 80 meters away at 80 dB SPL, what will the level be at your location?

Show Answer

We need the distance equation:

dB_{SPL} =inital\ level-20\log\left(\frac{distance_{measured}}{distance_{reference}}\right)

Our reference location is our location so it is 1. The ending location is 80 meters. We are given an initial level at 80 dB SPL. If we weren't then we would use 0 dB SPL.

Now we sub in the known values:

dB_{SPL} =80-20\log\left(\frac{80}{1}\right) \approx \boxed{41.94\ dB_{SPL}}

NOTE: If you moved right next to me the ratio would become 1/1 which when put in the log will give 0 resulting in only the initial value.

Example

Suppose I am playing clarinet and you are 25 feet away. From where I am it's 75 dB SPL. What level is it where you are?

Show Answer

We need the distance equation:

dB_{SPL} =inital\ level-20\log\left(\frac{distance_{measured}}{distance_{reference}}\right)

Our reference location is our location so it is 1. The ending location is 25 feet. We are given an initial level at 75 dB SPL.

Now we sub in the known values:

dB_{SPL} =70-20\log\left(\frac{25}{1}\right) \approx \boxed{47.04\ dB_{SPL}}

Example

Suppose I am playing a clarinet and you are 25 feet away. From where you are its 30dB SPL. How loud is it where I am?

Show Answer

We need the distance equation:

dB_{SPL} =inital\ level-20\log\left(\frac{distance_{measured}}{distance_{reference}}\right)

This time the reference location is where you are, which is 25 feet away. The measured location is 1 because that is where we are interested in getting to.

Now we sub in the known values:

dB_{SPL} =30-20\log\left(\frac{1}{25}\right) \approx \boxed{57.96\ dB_{SPL}}

Rounding and converting to milliseconds

Often questions will ask for a number to be rounded to a particular value or for the answer to be in milliseconds rather than seconds.

Rounding

Rounding is a way of cutting of a number to an approximation that is close to the more exact value. Rounding does so by specifying a place value to round to.

The way it works is you look at the place value 1 spot to the left of the specified spot. If that number is a 5 or more you round it up. If it is a 4 or 0 you round it down. Rounding up means you add 1 the specified place and make the rest of the spots to the left 0. Rounding down means you leave the number in the specified place alone and make all the spots to the left of it 0.

Say I have the following number 100.011 and I ask you to round it to the nearest hundredth. So you find the decimal in the hundredth place which is 1 and look one spot the left which is also 1. So the new number becomes: 100.010 or 100.01

Suppose it was 100.015 then it would round to 100.02

If you had the number 100.5 and I said to round to the nearest whole number you would look at the ones place. The place to the left of the ones place is the 10ths. It has a 5 so you round up to 101.0 or 101.

Suppose now I asked you to round 100.9543 to the nearest tenths. In this case the 5 would cause it to round up and so we would add 1 to 9. This means the 100.9543 would actually increase to 101.0000 or 101.

Milliseconds to seconds

Milliseconds has the prefix milli meaning 1000, so there are 1000 milliseconds in a second! This gives us a ratio by which we can use to convert. If we wish to go from seconds to milliseconds just multiply by 1000. If we wish to go from milliseconds to seconds just divide by 1000. It is that simple! Just remember to use s for seconds and ms for milliseconds!

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