Super Soundproofing Community Forum
Soundproofing Forum Topics => Other Soundproofing Questions => Topic started by: Ed on October 05, 2012, 01:07:08 PM

We are considering installing soundproofing in our home, but would first like to determine if it will make enough of a difference.
I would like to know how you can calculate the effective reduction in decibels if a supplier claims their product reduces the noise level by 50% or 70%, etc.
Because the decibel scale is logarithmic, I am unsure how this works and I assume it is not just a simple matter of reducing any given decibel reading by the claimed percentage reduction to obtain the new decibel figure.
For example, if I know traffic passing our house generates a decibel reading of 80dB and the soundproofing claims to reduce that noise level by 50%, then I assume the reduced decibel reading would not be 40dB?
Is there a formula that can be used to calculate the resulting decibel figure after a percentage reduction has been applied to a decibel reading?
Thanks for any help.

Well, it's a subjective thing... A 10 dB reduction will seem like 50% less noise. A 20 dB reduction would be 1/2*1/2 as much noise  75% less. 40 dB reduction would be (1/2)^4, or 1/16 as much noise, for a reduction of 93%.
The logarithmic scale only really applies to the reduction in sound energy. An 80 dB noise has 10,000 times more energy than a 40 dB one.

Thanks for the reply, that was very helpful.