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increasing power?
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<blockquote data-quote="squeak9798" data-source="post: 1243687" data-attributes="member: 555320"><p>The problem with that "equation" is that it tries to pinpoint an actual sound pressure level; 149db (for example), which makes it totally useless.</p><p></p><p>If you want to use <em>any</em> equation, this is the one to use:</p><p></p><p>10*log(power1/power2) = db</p><p></p><p>10*log(2/1) = 3.0103db</p><p></p><p>10*log(500/250) = 3.0103db</p><p></p><p>So, *in theory* you should gain 3.0103db by doubling your power. But, like I said....this doesn't account for things like power/Bl compression, etc etc, so it's only theoretical.</p></blockquote><p></p>
[QUOTE="squeak9798, post: 1243687, member: 555320"] The problem with that "equation" is that it tries to pinpoint an actual sound pressure level; 149db (for example), which makes it totally useless. If you want to use [I]any[/I] equation, this is the one to use: 10*log(power1/power2) = db 10*log(2/1) = 3.0103db 10*log(500/250) = 3.0103db So, *in theory* you should gain 3.0103db by doubling your power. But, like I said....this doesn't account for things like power/Bl compression, etc etc, so it's only theoretical. [/QUOTE]
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