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3dB(A) = doubling of noise?
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#1 Posted : 20 July 2006 10:34:00(UTC)
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Posted By Juliet Fennell
Can anyone help? Im convinved that an increase of 3dB(A)is a doubling in noise while a colleague is convinced its 10dB(A) Anyone know the answer?
Thanks, Juliet
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#2 Posted : 20 July 2006 10:51:00(UTC)
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Posted By Martyn Hendrie
The Bell (or decibel scale) is logarithmic (i am going to make no attempt to explain logarithms) but the affect is that every increase of 10 decibels is an increase of 10 times the sound energy.

Working backwards from that a 3dB increase doubles the sound energy.
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#3 Posted : 20 July 2006 10:52:00(UTC)
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Posted By Diane Thomason
You are right, it's 3 (i.e. increase of 3dB(A) means that the sound pressure level is doubled.
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#4 Posted : 20 July 2006 10:54:00(UTC)
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Posted By Alan Hoskins
Hi Juliet,

An increase of 3 dBA is a doubling of sound pressure level.

An increase of 10 dBA is required for a sound to be perceived as being twice as loud.

Alan
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#5 Posted : 20 July 2006 10:59:00(UTC)
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Posted By Juliet Fennell
Thanks guys.
juliet
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#6 Posted : 20 July 2006 11:40:00(UTC)
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Posted By IT
It is actually called the rules of 3

I.E

1 machine 85 dba
2 machine both emitting 85 dba is represented as 88 dba

make your friend pay for the drinks
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#7 Posted : 20 July 2006 12:38:00(UTC)
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Posted By Charley Farley-Trelawney
Juliet

Alexander Graham Bell founded the concept of decibels and formulated a logarithmic range based on 10. "Deci" refers to the base 10 log scale, and "bel" refers to Alexander Bell. Each 10-dB increase represents a tenfold increase in sound intensity. In addition, a 10-dB increase is perceived as more or less doubling-up on loudness.
I believe therefore, your friend may well be referring to this equation; as other responders have explained in detail I will not at the risk of becoming boring repeat any of those replies.
Best wishes
CFT
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