Sigh.. thought I knew this:

Speaker system rated 130 dB SPL at 1 meter (Mackie SA1521).

As we move away from the speaker, SPL drops by inverse square law.

Uh... 6dB each 2x distance? or was it 3 dB POWER each 2x?

Darn! I need to know this quick. THANKS!

Example: at 2,4,8,16 (What's the SPL at 16 meters?)

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### [quote=David French]... In 16 meters, the sound should lose abo

[quote=David French]... In 16 meters, the sound should lose about 35 dB , so the answer should be 106 dB. ....

Thanks David!

OK, 'Pressure' is like 'voltage' and 2:1 voltage is 6 dB. So going from 1 meter to 16 meters is 2,4,8,16 so 4 * 6 = 24 dB and 130 - 24 = 106 dB and... Sure. I shoulda known that!

And now I have the equation in my notebook.

Or, for a person in the audience 16 meters from this speaker, it's, technically speaking, plenty loud.

### Hi David, Terry, Sorry to jump in. I was browsing RO a bit. Da

Hi David, Terry,

Sorry to jump in. I was browsing RO a bit.

David shows the standard published method.

Another manner (which is where the formula comes from) is just calculating the surface the sound is spread over.

Lp = Lw-10*log(S) [S= surface in m2]

If one analyses the inverse square law one notices that this is noting else than a surface calculation of a sphere with a radius r which represents the distance to the source, representing a dimionless point source (center point of sphere).

So the formula for e.g. 16 m = -10*log(S) where S is the surface of the sphere = 4*pi*r^2 = 3216.99 m2

-10 log (3216.99) = -35.07 dB

For 1 m S = 12.57 m2

-10 log (12.57) = -10.99 dB

So the difference = -24.08 dB

You can also directly calculate versus the ratio of both spheres

3216.99/12.57 = 256

So the difference is -10*log(256) = -24.08 dB.

:? While this seems more complicated than the inverse square law it has it advantages to go to the basics of this inverse square law.

:roll: Why?

As David said: the inverse square law assumes a free field, meaning a point source without dimensions radiating as a sphere.

But once you are at a short distance of a huge large source this formula isn't valid anymore (in front of a radiating wall sound hardly decays). A.g. an array of speakers at a short distance does not behave as a point source. A large isolating enclosure around technical equipment does not behave as a point source.

Also understanding this factor S relationship makes it easier to adjust the formula to real live circumstances.

Assume you have a speaker in a very long very reflective room. The back wall absorbs perfectly. You measure at distance x.

How much is this noise decayed at that distance. There are numerous reflections via the side walls and wall behind the speaker.

Well that's easy: rather than using the inverse square law you calculate the area over which the sound is spread, which is about the section of the room. e.g. W 4.5 m x H 2.8 m = 12.6 m2

So your average Lp will be Lw - 10*log(12.6) = -11 dB (assuming no absorption on reflection paths.

Summary: You need to take the 10 * log of the surface the sound is spread over.

More you can find in the "Noise Decay" Word document on my site.

http://www.acoustics-noise.com

Or you can see this animation I once made:

http://home.planet.nl/%7Edesart/noisedec/NoiseDecayMovie.html

The formula on this animation may look complicated, but when one should set the measures of the source to 0 it becomes exactly as the inverse square law (dimensionless point source radiating as a sphere) and all the non-sphere factors in the formula can be removed.

One notices that at a longer distance from the source (relative to source measures) the propagation indeed start looking as a sphere making the inverse square law about correct, because the influence of the other factors becomes insignificant.

The last (very light grey colored) factor in the animation takes the diffuse field in a room into account but can be ignored if you only want to know the decay in function of distance.

Warm regards

(y)

This is meant as an addition, not correction of David's perfect reply.

### Eric, THANKS! for the excellent review of the basis for calculat

Eric, THANKS! for the excellent review of the basis for calculating sound level VS distance VS "The Real World".

I will be teaching some high school kids about sound, and basics of audio equipment and recording and mixing, later this year. I'd like to use parts of your explanation if that's OK. I recall the 'surface of the sphere' from High School Physics. That was, um, 2004-1957= 47 years ago, just to prove I can do some math(s).

You have done a great job of sharing your ideas and work on your website.

If we cross paths sometime, I hope to buy you a beer bigger than those shown in your post.

8-)

### TerryKing wrote: If we cross paths sometime, I hope to buy you

TerryKing wrote:

If we cross paths sometime, I hope to buy you a beer bigger than those shown in your post.

8-)

Terry,

Thanks

:lol: LOL

Sorry the RO beers only come in one size. Close your eyes and they grow infinite.

But I gladly share a real one.

Just simple. If the radius of a sphere doubles (= doubling distance) then the surface quadruples (x4).

10*log (4) = 6.0206....

That's where the 6 dB's per doubling of distance comes from. The sound energy is spread over a 4 times larger surface, so diminishes with a factor 4 per surface unit (expressed in dB - 6 dB)

## A handy equation for the dB of loss over a distance is 20 * log(

A handy equation for the dB of loss over a distance is 20 * log( distance in meters) + 11. If it's rated 130 dB SPL at 1 meter, then at the source it should be 141 dB. In 16 meters, the sound should lose about 35 dB , so the answer should be 106 dB. This is a free field equation, so the real dB might be a bit more depending on the surroundings.