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<blockquote data-quote="shredder2" data-source="post: 8657922" data-attributes="member: 674047"><p>You realize that sound is measured and felt differently than can be expressed to a layman... it's measured on a logarithmic scale and in a perfect world doubling cone and power will net 6db. So that +10db increase above is HUGE.</p><p></p><p>Here's something to chew on...</p><p></p><p>t's a typically over complicated explanation. The fact is that for any two sources, regardless of size or input power, the radiated sound pressure adds vectorial. If the sources are closely spaced (correlated or spaced much less than a wave length) then they are in phase at all points in space and the sound pressure just adds everywhere. So if they each radiate a sound pressure of 1 independently then they sum to 2. Acostic power is the square of the pressure so power goes from 1 squared = 1, to 2 squared = 4. In dB, 10 log(4/1) = 6dB. So two sources which radiate the same sound pressure radiate 6dB more acoustic power when correlated. Now the electrical power into the sources is just the sum of the power delivered into each. Just add the power. If the drivers receive the same power it's 1 +1 = 2 which in dB is 3dB more power. So the two sources radiate 6dB more acoustic power with a 3dB increase in input power resulting in a 3dB increase in efficiency. If you make the drivers identical fine. But its the same if one driver is a 6" woofer and the second is a 10". And it carries over to drivers with different efficiency too. Lets say you have driver 1 which radiates a sound pressure of 1 with 1 watt in. Driver 2 radiates a sound pressure of 1.3 with an input power of 0.8 watt. Assume they are closely space, the net sound pressure of both is 2.3 for a 7.23 dB increase in radiated acoustic power, relative to driver 1 alone.. The input power is 1.8 watt or 2.55 dB greater than 1 watt. So relative to the first driver the system radiates 7.23 dB more acoustic power with 2.55 more input power or a gain in efficiency of 4.68 dB compared to the first driver. Or you can use a more appropriate reference for sound pressure if you like and have a more standard measure of the increase in efficiency. you can go through the arguments from first principles if you like, but it just make things less simple.</p><p></p><p><strong>Now I got a headache... that type of explanation is why I just learned modeling software.</strong></p><p></p><p><strong>Then there's the louspeaker design cookbook</strong></p><p></p><p><strong><a href="https://www.parts-express.com/loudspeaker-design-cookbook-7th-edition-book--500-035" target="_blank">https://www.parts-express.com/loudspeaker-design-cookbook-7th-edition-book--500-035</a></strong></p><p></p><p><strong>Another Cliffhanger ^^^^.... snore</strong></p><p></p><p><strong>Good luck... I'm out</strong></p></blockquote><p></p>
[QUOTE="shredder2, post: 8657922, member: 674047"] You realize that sound is measured and felt differently than can be expressed to a layman... it's measured on a logarithmic scale and in a perfect world doubling cone and power will net 6db. So that +10db increase above is HUGE. Here's something to chew on... t's a typically over complicated explanation. The fact is that for any two sources, regardless of size or input power, the radiated sound pressure adds vectorial. If the sources are closely spaced (correlated or spaced much less than a wave length) then they are in phase at all points in space and the sound pressure just adds everywhere. So if they each radiate a sound pressure of 1 independently then they sum to 2. Acostic power is the square of the pressure so power goes from 1 squared = 1, to 2 squared = 4. In dB, 10 log(4/1) = 6dB. So two sources which radiate the same sound pressure radiate 6dB more acoustic power when correlated. Now the electrical power into the sources is just the sum of the power delivered into each. Just add the power. If the drivers receive the same power it's 1 +1 = 2 which in dB is 3dB more power. So the two sources radiate 6dB more acoustic power with a 3dB increase in input power resulting in a 3dB increase in efficiency. If you make the drivers identical fine. But its the same if one driver is a 6" woofer and the second is a 10". And it carries over to drivers with different efficiency too. Lets say you have driver 1 which radiates a sound pressure of 1 with 1 watt in. Driver 2 radiates a sound pressure of 1.3 with an input power of 0.8 watt. Assume they are closely space, the net sound pressure of both is 2.3 for a 7.23 dB increase in radiated acoustic power, relative to driver 1 alone.. The input power is 1.8 watt or 2.55 dB greater than 1 watt. So relative to the first driver the system radiates 7.23 dB more acoustic power with 2.55 more input power or a gain in efficiency of 4.68 dB compared to the first driver. Or you can use a more appropriate reference for sound pressure if you like and have a more standard measure of the increase in efficiency. you can go through the arguments from first principles if you like, but it just make things less simple. [B]Now I got a headache... that type of explanation is why I just learned modeling software.[/B] [B]Then there's the louspeaker design cookbook[/B] [B][URL="https://www.parts-express.com/loudspeaker-design-cookbook-7th-edition-book--500-035"]https://www.parts-express.com/loudspeaker-design-cookbook-7th-edition-book--500-035[/URL][/B] [B]Another Cliffhanger ^^^^.... snore[/B] [B]Good luck... I'm out[/B] [/QUOTE]
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