### Author Topic: Decibel Readings  (Read 2527 times)

#### Ed

• Guest « 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.

#### johnbergstromslc

• Guest « Reply #1 on: October 06, 2012, 01:02:19 AM »
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.

#### Ed

• Guest 