Sonovo wrote on Tue, 06 November 2007 16:09 |
What, if any special requirements/treatments are neccessary when dealing with a surround room as opposed to a stereo room? |
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What about the sub? Are there special problems to deal with there? |
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Would a good stereo room be a good candidate for a suuround room, or are the requirements different? |
Sonovo wrote on Wed, 07 November 2007 17:44 |
Let me at least try it first and see, ok? |
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Ethan, does that mean I need to tune the subs response to both the speakers and to the room/nodes? Through both placement as well as crossover and level adjustment? Phase as well this far down? |
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Would a surround room without subs (but with response say down to around 34Hz or so) be usable, or is 20-35Hz an important part of the equation (for music, I expect film requirements will be different). |
franman wrote on Wed, 07 November 2007 19:39 |
In our best sounding rooms (LF wise) we have areas in the ceilings (large areas) with 3-4' deep broadband porous absorption! I know it sounds like a lot and it is.. It takes some serious real estate... that's why it's nice to have the ceiling height to put it up there!! |
StudioRhythm wrote on Sat, 10 November 2007 01:25 |
4' of 703 (or equivalent), a certain thickness (8"-12"?) of 703 plus an airgap stuffed with regular fluffy insulation, or...? |
Ethan Winer wrote on Sat, 10 November 2007 11:20 | ||
As fiberglass is made thicker / deeper, you can get away with lower density. So if you have four feet available you can get similar results for less money with the fluffy type. But I'll be interested to hear what Fran has to say about what he uses when four feet is available. --Ethan |
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From our experience, broadband traps have an effective low frequency cutoff equal to 1/7th of the wavelength absorbed. To compute this, measure the total panel length (i.e. L+S where L is the panel size and S is the hanger size per the diagram in the book) and use the formula on p.5 for wavelength. For example, a ceiling hung trap constructed of free-hanging fiberglass-covered panels, each measuring 24" long, on 6" hangers (total length = 30" or 2.5 feet) would be effective above 2.5' x 7 = 17.5' Effective cutoff wavelength 2.5 x 7 = 17.5 ft. (per formula on p.5) Frequency = V/l = 1130/17.5 = 64.5 Hz As long as the entire cavity was lined with absorptive materials and the trap blankets were at least 12" on center, this trap would be effective at all frequencies above 64.5 Hz. To decrease the cutoff frequency and make a basstrap effective down to 40 Hz, the panel length would need to be approximately 4 ft. long (including hangers). Very often, this type of broadband trap is constructed with varying sized panels, giving a more contoured absorption curve. An exact formula, relating panel spacing, size, hanger depth, etc. is empirical and not readily available. |
gullfo wrote on Mon, 19 November 2007 15:35 |
if you can make the room larger, then i second that approach. |