How much absorbing material should I add to change the reverberation time?

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    jfairbanks

    Your sheet lists “total surface area of hall = 4,365 ft²” and “α of walls = 0.04”.

    If you (incorrectly, I think) assume all 4,365 ft² have α=0.04, then:

    𝐴 = 4365 (0.04) = 174.6 sabins

    which would predict: RT= 569.3/174.6 = 3.26sec and NOT 0.73 sec.

    So the provided 0.73 s RT implies the room already has much more absorption than α=0.04 over 4,365 ft² would provide (meaning: other surfaces/contents are more absorptive, or the “surface area” number is not the total enclosure area, or α=0.04 applies only to part of the room, etc.).  

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