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Old 4th July 2009   #7
spm_gl
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Joined: Apr 2009
Location: Spreewald, near Berlin / Germany
Posts: 316

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Okay, got it. "Schallschutz & Raumakustik in der Praxis" (German books are usually better) tells me, the equivalent absorption area depends on volume and frequency. Goes from 0.2 to 3.0. There's an equation too, but no equation editor :-)
Edit: found a simplified formula: A = 6*10^-5 Vfk [m²] with V volume, f frequency, k "positioning factor"
k=1 for wall, k=2 for edge, k=3 for corner. (That's 6 times (10 to the power of -5))
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