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jfarch - 04 Mar 2004 23:34 GMT

Hi all,
I have a project to install several high powered, full range
speakers to warn approaching vessels 300 feet away! My question is
does anyone have any idea of an equalizing curve for this application?
Also, what angle would be best considering water temperature,
thermals, etc?

Thanks,
Jerry Arch

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Noral Stewart - 04 Mar 2004 23:42 GMT

> Hi all,
> I have a project to install several high powered, full range
[quoted text clipped - 5 lines]
> Thanks,
> Jerry Arch

You said only 300 feet, but then asked about "thermals" etc. At such close
distances, atmospheric effects are not very significant. I suspect you are
actually dealing with a longer distance.

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Angelo Campanella - 05 Mar 2004 04:35 GMT

> I have a project to install several high powered, full range
> speakers to warn approaching vessels 300 feet away! My question is
> does anyone have any idea of an equalizing curve for this application?
> Also, what angle would be best considering water temperature,
> thermals, etc?

    The absorption of sound by air, humidity being a modifier, is well
known. I can FAX you a graph if you wish. Generally, no EQ is needed
below 1 kHz, while at 4 kHz, the absorption becomes considerable.
Perhaps someone here can provide an algorithm.

    A few numbers: at 5 kHz, if the humidity is greater than 50%, the
absorption of sound at that frequency is less than o.01 dB/foot
(100'=>1dB; 300'=>3dB). At 10 kHz, the rate is four times that. The
worst humidity value is 15% where 5 kHz absorption peaks at o.04 dB/foot
and for 10 kHz, o.1 dB/foot. Over water, I would expect the humidity to
usually be over 50%, IMHO.

    I wonder why you are concerned about EQ'ing. Is it that you want to
project voice warnings (as we have all over Ohio for Tornado warnings)?
The audio I hear is really hoaky.. Barely intelligible (about 600'
away) when the horn is pointing my way, and hopeless in any other
direction. Part of the problem is that the horn here is too low.. so
that sound scattering and shielding by intervening houses make it a mess.

    Another problem is "finite amplitude distortion", or harmonic content
in audio terms. Here, when the horn throat SPL goes over about 160 dB
SPL, the pressure waves are so intense that shock waves can develop (the
unusual extreme). But more likely, the voice timbre becomes harsh, and
make one think that the horn driver is going nonlinear. But its the air
that is going nonlinear.

    I suggest that you position the speakers as high as practical so that
not only all obstacles will be cleared from the sound path, but also so
that the incidence (downward) angle is sufficiently steep the the
refraction from thermal gradients that may form in the air at high noon
will still not deflect the sound over the boats. This has been the
principle of all bell towers, for millennia!

    But over water, your installation should do real well, especially on a
calm day.

    Angelo Campanella

--------- www.CampanellaAcoustics.com ---------

"I have simply studied carefully whatever I've undertaken, and tried to
hold a reserve that would carry me through." - Charles A. Lindbergh.

"As for background noise level; 35 dBA is a good classroom; 45 dBA is a
sound masking system!" - Anthony K. Hoover

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Peter Larsen - 05 Mar 2004 11:58 GMT

> > I have a project to install several high powered, full range
> > speakers to warn approaching vessels 300 feet away! My question is
[quoted text clipped - 6 lines]
> below 1 kHz, while at 4 kHz, the absorption becomes considerable.
> Perhaps someone here can provide an algorithm.

+ 3 dB pr. octave above 1 kHz comes to mind as a good first attempt,
shelving so that no more than 10 dB boost is applied makes sense. No
reason to try to get it "right" because "right" varies too much with
humidity anyway.

Next question is how low a frequency range that is required, I would
suggest a second order acoustic roll-off below 240 Hz. There is a jap
hifi factory making very large high efficiency compression drivers and
horns, I think their gear could do it and do it very very well. I don't
have the url on this machine .... they have westrex compatible
spareparts too as I recall it.

> A few numbers: at 5 kHz, if the humidity is greater than 50%, the
> absorption of sound at that frequency is less than o.01 dB/foot
[quoted text clipped - 36 lines]
> "As for background noise level; 35 dBA is a good classroom; 45 dBA is a
> sound masking system!" - Anthony K. Hoover

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Eric Desart - 05 Mar 2004 20:13 GMT

A good on-line calculator
http://www.measure.demon.co.uk/Acoustics_Software/iso9613.html

I copied this from:
From: Desart Eric
Subject: Re: Open air absortion coeff.
Newsgroups: alt.sci.physics.acoustics
Date: 2001-08-30 11:21:53 PST

But in the original message dated 2001-80-30 I left some typos
I was informed about those typos by:
G.C. Duyckinck Dörner at Sept 27, 2001 for which my thanks.
At the time I forgot about them.
The algorithms below should be correct now.
Anyhow you can test them versus the above on-line calculator
note: I added some minor adjustments in the algorithms (more accurate)

ISO 9613-1 : 1993 (E)
NOISE ABSORPTION BY AIR

as = a * s [dB] total absorption at distance s

Pt = Pi * exp(-x * as) [Pa]

x = 1/(10 * log((exp(1))^2) = ca 0.1151 (value in norm, formula E. Desart)

Delta Lt = 10 * log( Pi^2 / Pt^2 ) = as [dB]

a = 8.686 * f ^2 * ((1.84 * 10^-11 * (Pa / Pr)^-1 * (T / To)^(1/2)) + y)
[dB/m]

y = (T / To)^(-5/2) * (0.01275 * exp(-2239.1 / T) * (frO + f ^2 / frO)^-1 +
z)

z = 0.1068 * exp(-3352 / T) * (frN + f ^2 / frN)^-1

frO = (Pa / Pr) * (24 + 4.04 * 10^4 * h * ((0.02 + h) / (0.391 + h)))

frN = (Pa / Pr)*(T / To)^(-1/2) * (9 + 280 * h * exp(-4.170 * ((T /
To)^(-1/3)-1)))

h = hr * ((Psat / Pr) / (Pa / Pr)) = hr * (Psat / Pa)

Psat = Pr * 10^(-6.8346 * (To1/T)^1.261 + 4.6151)

a pure-tone sound attenuation coefficient, in dB/m,
for atmospheric absorption
s distance in m through which the sounds propagates
Pi initial sound pressure amplitude, in Pa
Pt sound pressure amplitude, in Pa
Pa ambient atmospheric pressure in kPa
Pr reference ambient atmospheric pressure: 101.325 kPa
Psat saturation vapor pressure ca equals:
International Meteorological Tables WMO-No.188 TP94
World Meteorological Organization - Geneva Switzerland
T ambient atmospheric temperature in K
To reference temperature in K: 293.15 K (20 °C)
To1 triple-point isotherm temp: 273.16 K = 273.15 + 0.01 K (0.01 °C)
h molar concentration of water vapor, as a percentage
hr relative humidity as a percentage
f frequency
frO oxygen relaxation frequency
frN nitrogen relaxation frequency
x Just a help factor to shorten formula, improvement on norm by
Eric Desart
y Just a help factor to shorten formula
z Just a help factor to shorten formula

Hopes this helps

Kind regards
Eric

> > > I have a project to install several high powered, full range
> > > speakers to warn approaching vessels 300 feet away! My question is
[quoted text clipped - 59 lines]
> > "As for background noise level; 35 dBA is a good classroom; 45 dBA is a
> > sound masking system!" - Anthony K. Hoover

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Eric Desart - 05 Mar 2004 21:21 GMT

> But in the original message dated 2001-80-30 I left some typos

Is there a 80th month? I need a Keyboard without build-in typos.....
Guilty me...

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Natural Science Forum / Physics / Acoustics / March 2004

Paging over water