Preserve high frequencies with plastic plates
Posted: Sat Mar 05, 2016 8:50 am
Hi,
I'm currently helping someone improving his room acoustics. He is deep into Hifi and does not want to use an eq in his signal path. OK so I try to give tips that preserve as much high frequency energy as possible. My question: do you have any experience using thin plastic plates to completely cover absorbers? A foil can't be used for areas that size because of mechanical and optical demands. The idea is to have the least possible weight that reflects high frequencies and needs no other cover to protect it (that would absorb high frequencies again).
What we already did:
I made a measurement and told him to build edge absorbers (mineral wool) for a start. We measured again and the result is as expected. But now we need to cover them with something. In addition the whole rest of the backwall (~12m^2) will be covered with 20cm mineral wool (plus 20cm air gap) and also needs coverage. I try to get him to a RT60 of 0,4sec. According to my calculations we will need even more than that as well.
Actual RT60: As you can see on the pic above we already reach the 0,4sec at 10k. I think it would be best to have a cover that reflects above ~6k Hz. There is this "Bergschersches Massengesetz" (I don't know the english word, should be something like "Mass-law of Berger") that is used to calculate the soundinsulation of walls. I don't know if that is accurate for that light-weight plates but if yes I would need 0,05 kg/m^2...that sound more like a foil than a plate.
Long story short, I don't have any real experience on that. That's all just numbers out of books.
So my main question I guess is...what would you try to do? Any thoughts on that?
Thanks
Andy
It's maybe not necessary to know for a general answer on how to preserve high frequencies but maybe it's good to know the context. This pic shows what I want to achieve. It's the equivalent absorption area. The gray lines mark my target area, the blue one is the actual state (with those edge absorbers) and the red lines mark the difference.
An "Equivalent absorption area" of 10m2 means I need 10m2 of an absorber with absorption factor 1. But I can also use 20m2 of absorbers with absorption factor 0.5. It's simply the product of the area and the alpha of the absorber.
I'm currently helping someone improving his room acoustics. He is deep into Hifi and does not want to use an eq in his signal path. OK so I try to give tips that preserve as much high frequency energy as possible. My question: do you have any experience using thin plastic plates to completely cover absorbers? A foil can't be used for areas that size because of mechanical and optical demands. The idea is to have the least possible weight that reflects high frequencies and needs no other cover to protect it (that would absorb high frequencies again).
What we already did:
I made a measurement and told him to build edge absorbers (mineral wool) for a start. We measured again and the result is as expected. But now we need to cover them with something. In addition the whole rest of the backwall (~12m^2) will be covered with 20cm mineral wool (plus 20cm air gap) and also needs coverage. I try to get him to a RT60 of 0,4sec. According to my calculations we will need even more than that as well.
Actual RT60: As you can see on the pic above we already reach the 0,4sec at 10k. I think it would be best to have a cover that reflects above ~6k Hz. There is this "Bergschersches Massengesetz" (I don't know the english word, should be something like "Mass-law of Berger") that is used to calculate the soundinsulation of walls. I don't know if that is accurate for that light-weight plates but if yes I would need 0,05 kg/m^2...that sound more like a foil than a plate.
Long story short, I don't have any real experience on that. That's all just numbers out of books.
So my main question I guess is...what would you try to do? Any thoughts on that?
Thanks
Andy
It's maybe not necessary to know for a general answer on how to preserve high frequencies but maybe it's good to know the context. This pic shows what I want to achieve. It's the equivalent absorption area. The gray lines mark my target area, the blue one is the actual state (with those edge absorbers) and the red lines mark the difference.
An "Equivalent absorption area" of 10m2 means I need 10m2 of an absorber with absorption factor 1. But I can also use 20m2 of absorbers with absorption factor 0.5. It's simply the product of the area and the alpha of the absorber.