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The Range
NFA & Class III Discussion
Air gun silencer
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<blockquote data-quote="338Shooter" data-source="post: 1005232" data-attributes="member: 3449"><p>Positive. <img src="/images/smilies/wink.png" class="smilie" loading="lazy" alt=";)" title="Wink ;)" data-shortname=";)" /></p><p></p><p>Sound dB are 20log calculations instead of a 3dB change for 10log calculations meaning a 6dB change is a doubling or a half.</p><p></p><p>dB = 20*log(RMS pressure / RMS reference pressure)</p><p></p><p>0dB = 20 * log(1) </p><p>6dB = 20 * log(y*1) => 10^0.3 = y = 2</p><p></p><p>0dB = 10 * log(1)</p><p>3dB = 10 * log(y*1) => 10^0.3 = y = 2</p><p></p><p>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.</p><p></p><p>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.</p></blockquote><p></p>
[QUOTE="338Shooter, post: 1005232, member: 3449"] 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. [/QUOTE]
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