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Natural Science Forum / Physics / Acoustics / August 2007Range of sound
Hi everyone, Hopefully someone can help me. Is there an equation that will determine how far a sound can be heard in certain atmospheric conditions? The source of the sound is that of a firework (4th July, or Guy Fawkes Night, as you prefer); atmopsheric temperature is about freezing point, the only unknown is humidity. Thanks in advance! Paul -- http://www.paullee.com > Hopefully someone can help me. Is there an equation that will > determine how far a sound can be heard in certain atmospheric > conditions? The source of the sound is that of a firework (4th July, > or Guy Fawkes Night, as you prefer); atmopsheric temperature is about > freezing point, the only unknown is humidity. Inversions (warm air aloft; 100m and higher) are the key. In the daytime the converse occurs; the sun heats the ground and thus the air in contact with it, causing refracrion upward so that popagation beyond the line of sight is hopeless. At night, especially more than 2 hours after sunset, the true inversions occur. The ground cools via infrared radiation into space through the transparent atmosphere. Sound is bent back toward the surface, "channeling" for quite some distance. I now believe that this propagation mode provides only 2D divergence, ergo only 3dB per distance doubled, whence much longer distances will occur. As a first approximation, change "inverse square law" having 6dB/dd into 2 dB/dd, to represent night sound propagation. For surface sounds, first diverge sound as 6dB/dd for the first 100m of travel, then convert to 3dB/dd for the remainder of the journey. Since the fireworks explode at about 100m altitude, the sound is nicely injected itnto the inversion channel, thus it is 3dB/dd all the way. (This propagation mode routinely occurs in the oceans. Whales use it daily..) Angelo Campanella Hi Angelo, In this case, the sound paths shall be curved instead of straight lines. Which distance (curved or direct) shall we use in your approximation (2 dB/dd) ? Thanks. CSL "angelo Campanella" <a.campanella@att.net> ???????:YZ2pi.9893$iX3.2720@bgtnsc05-news.ops.worldnet.att.net... >> Hopefully someone can help me. Is there an equation that will >> determine how far a sound can be heard in certain atmospheric [quoted text clipped - 26 lines] > > Angelo Campanella > In this case, the sound paths shall be curved instead of > straight lines. Which distance (curved or direct) shall we use in your > approximation (2 dB/dd) ? The curvature is so slight that I never consider it. Map it out... Ang. C. On 7/23/07 10:53 PM, in article Utgpi.363811$p47.188757@bgtnsc04-news.ops.worldnet.att.net, "angelo Campanella" <a.campanella@att.net> wrote: >> In this case, the sound paths shall be curved instead of >> straight lines. Which distance (curved or direct) shall we use in your [quoted text clipped - 3 lines] > > Ang. C. If you take into account the speed of sound, the curved path taken by sound takes the least time to travel between the endpoints. Fermat's principle. Actually, the time taken to cover the real path can be any extremum rather than a minimum. Bill  SignatureIraq: About three Virginia Techs a month > On 7/23/07 10:53 PM, in article > Utgpi.363811$p47.188757@bgtnsc04-news.ops.worldnet.att.net, "angelo > Campanella" <a.campanella@att.net> wrote: >>>In this case, the sound paths shall be curved instead of >>>straight lines. Which distance (curved or direct) shall we use in your >>>approximation (2 dB/dd) ? Correction; 3dB/dd >>The curvature is so slight that I never consider it. Map it out... > If you take into account the speed of sound, the curved path taken by sound > takes the least time to travel between the endpoints. Fermat's principle. > Actually, the time taken to cover the real path can be any extremum rather > than a minimum. We generally "hear" mostly the one wave, or ray, that has the least attenuation that has been refracted to our position. Other waves or rays will exist at other elevations; we aren't there to hear them. A strange possibility exists: For a distant line source (e.g. an interstate highway) the level reduction in a uniform atmosphere is 3dB/dd (cylindrical divergence. When an inversion occurs at night (no wind), where the second dimension is lost, there is 0 dB/dd! That is, line source noise will not be attenuated with distance beyond the nearby point where cylindrical divergence ceases. The only attenuation that does occur is due to ground absorption and air absorption. I sometimes muse that this is the source of the anomalous "hum" that has been reported from time to time in recent decades. The low frequency content, somewhere between 40Hz and 80Hz arises from the pure tone exhaust fundamental of many vehicles combined with up and down doppler that can cause frequent random undulation ("QSB" in radio ham lingo). When there is a wind, the direction favored for propagation becomes polarized to the extent that directly downwind, the least attenuated propagation occurs, so we hear that segment of the freeway OK. But waves or rays coming from either lateral segments of the highway will have more attenuation, so the overall scenario falls back toward a point source... maybe 1.5dB/dd plus ground and air absorption. Doppler would be diminished. Ang. C. Well, the reason why I'm writing is because of the following: http://home.earthlink.net/~dnitzer/4HaasEaton/Rockets2.html The conditions that night were warm (early-mid 80s, quite humid, moon visible, wind direction and speed unknown). I'm wondering what would have been heard in freezing conditions, close (maybe 5 miles off) to a huge ice field (water at freezing, air slightly warmer), no wind, and at least some humidity. TIA! Paul > > On 7/23/07 10:53 PM, in article > > Utgpi.363811$p47.188...@bgtnsc04-news.ops.worldnet.att.net, "angelo [quoted text clipped - 37 lines] > > Ang. C. Nice Comments Ang. C.. i always wondered about the 6db/dd on doubling the distance. if there is 100 db at 1 meter what woudl be sound level at 10 and 100 meters. enlighten me. v > i always wondered about the 6db/dd on doubling the distance. if there > is 100 db at 1 meter what woudl be sound level at 10 and 100 meters. > enlighten me. 10x = -20 dB, etc. Ang. C. |
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