[8thnerve]

Can anyone quickly calculate the distance that a spherically propogated sound wave would take to reduce by 60db?  How about a cylindrically propogated one?

[beat]

you may find something useful here:

http://www.audiodiycentral.com/resource/pdf/nflawp.pdf

[JohnR]

In free space, 6 dB drop with doubling of distance. So if you started at one meter, 60 dB would be 2 to the power of 10 = 1024 meters. So you would have to be rather a long way up in the air for the free space assumption to hold

Air absorption will have its effect too. Don't know the math for that, but it is frequency dependent.

[Occam]

and for the cylyndrical wave (line source as opposed to point source) you've got 3db attenuation with each doubling of distance, IN THE "Near Field". This would correspond to 2^20 meters, a bit over 3,000,000 feet.

But sadly, your line source is not infinitely long, and @ 60db attenuation, you moved to the 'far field'  which behaves like a point source (inverse square law) so you'd be there with near John's number, close to 3,000 feet.

8th,
Do you have any idea what ratio 60db represents?