6dB is half of the sound pressure. You don't need to explain logs to me. Commercial firearm suppressors typically have reductions of 30 to 40 decibels. Spring air guns generate a lot of sound from the vibration of the rifle. You can't use a suppressor for that. The air gun silencers are pretty ineffective because there isn't a lot of super sonic gas to slow down. If the pellets are super sonic, that's what you're going to hear anyway and a suppressor can't do anything about the super sonic crack.

You sure about that? I have been teaching radar repair to the military for 25 years and when talking about power levels a 3dB gain is twice as much power and a 3db reduction (-3dB) is half as much power. -10dB is 1/10 as much power, -20dB is 1/100 as much power, -30dB is 1/1000 as much power, etc... I am not trying to bust your chops, just wondering if sound dB measurements are different then power level db measurements.

Positive. Sound dB are 20log calculations instead of a 3dB change for 10log calculations meaning a 6dB change is a doubling or a half. dB = 20*log(RMS pressure / RMS reference pressure) 0dB = 20 * log(1) 6dB = 20 * log(y*1) => 10^0.3 = y = 2 0dB = 10 * log(1) 3dB = 10 * log(y*1) => 10^0.3 = y = 2 Finding the dB of a power like you said absolutely is 3dB for a factor of two, but things like sound pressure and voltage use the 20log equation and a 6dB change gives the factor of two. If you square the RMS pressure and RMS reference (this makes it sound intensity or sound power per unit area) then you would use the 10log equation, but you use the 20log equation when dealing with sound pressure.