First in regard to the ceiling. Our idea was to build control room walls up to the existing ceiling, and not have to build a new ceiling on the walls. Honestly, I'm basing this idea mostly on pictures of audio post rooms in NYC. (If they can do it, why can't we?)
The basic answer there is another question: "Do they need as much isolation as you do?" Or "do they already have it from some other factor?"
All studios are different: it is seldom possible to look at what was done in one place and copy it exactly to a different place. Some things can be copied, but at best most things need to be adapted or modified, to work in other places.
You write that a good slab will prevent flanking of up to 60 or 70 dB. So why won't the ceiling act in the same way as the floor?
It most likely will act the same way, but that's part of the problem, not part of the solution!
OK, think of it this way: the basic concept of a room in a room is that the inner-room is as independent, separate, etc. as possible. You want it to be free to move, shake, rattle, roll and vibrate as much as it feels lie WITHOUT transmitting that to the building itself: that's the final goal.
But in reality, the room has to sit on something (unless you know some magic spells to make it levitate!
). You have no choice there. And since floating is out of the question, you sit it on the concrete slab floor, which all of a sudden imposes a flanking limit. Call it 60 db, to be reasonable. If you could levitate, then there would be no flanking limit at all: your isolation would be limited only by air transmission. But the simple fact of having to put it down, imposes that limit. Now it is firmly anchored on one of its six sides, but all the others can still move, shake, rattle, roll, etc. freely, without transmitting any of that to the rest of the building. A bit goes into the floor, which is why you now have a limit, but not much.
Now consider what happens if you attach the top end of everything to the ceiling: You just jammed the entire room solidly in place!
NONE of the sides can now shake, rattle etc. freely, since
ALL sides are firmly anchored at both ends! Sure, the wall panels themselves can still flex a bit in the middle, but even that is transmitted to both ends, since both ends are firmly anchored.
Not sure if that helps to understand the concept here: It's not exactly like that, of course (nothing is ever simple or intuitive in acoustics!), but it gives you an idea of what is going in if you anchor both ends of your walls to the structure.
Second, regarding natural light in the control room. How do you keep the light, but retain isolation?
That's a bit easier to explain: You have one pane of glass in the outer leaf probably the existing windows in the existing walls, then you put a second pane of glass in the right place in the inner leaf.
In the attached image, the windows appear to be double windows, which makes sense for isolation from the street & from other studios in the facility. Can you tell me - Is the inside window wall part of the room-within-a-room? From looking at the picture it's hard to tell if that wall is just built off the original wall a bit, rather than being one wall of the room-within-a-room.
It's really hard to say from those photos, since much of what goes in in acoustic isolation and acoustic treatment is frequently hidden from view (it is pretty ugly to look at!). So it could be that you are seeing acoustic fabric covering the gap between the inner and outer leaves, assuming that the room actually is built like that. Or you might just be seeing he surface of a very thick brick or concrete wall, if that's what it is made of. And if that is a 12" concrete wall, then that would explain why they didn't need a room-in-a-room: 12" of high density reinforced concrete will get you upwards of 50 dB of isolation all by itself, so if that room is built like that, and the don't have flanking issues in the building (elevators, HVAC equipment, footsteps, water pipes, traffic, subway, etc.), then they are fine like that, except for the windows. Since glass 6" thick is really expensive
(to match 12" walls), it just makes much more sense to put two panes of 3/4" glass on opposite sides of the wall, with a big air gap between them, to still get good isolation like that.
But like I said: it's impossible to tell from a photo what the situation of that room is. You'd have to directly ask the designer that built it, to see what he did and why he did it. And even if you did know that, you'd still get back to the same original issue I mentioned: all rooms are different, so what they did in their room to get it the way they need it, is most likely not applicable to you doing in your room to get it the way you need it. There are just too many variables involved in studio design to be able to take the "cookie cutter" approach. If you scan the forum for some of the threads where people tried to do that "Because my buddy saw that in a studio where he went once", you'll notice that 9 times out of 10, copying "what the buddy saw" leads to disaster, and the tenth time it did nothing useful, but at least didn't do anything bad either! Maybe that's a bit of exaggeration, but you get the point: What works for me in my studio is most likely not going to work the same for you in yours.
So basically, is it possible to get about 60 dB isolation by building four solid walls that go from existing floor to existing ceiling? That is what we'd like to achieve.
Physics to the rescue! Acoustics is a science, based on physics. There are equations that describe how it all works. The most basic is called "mass law", and this is what it says:
TL(dB)= 20log(W) + 20log(f) -47.2
"W" is the surface density of the wall, and "f" is the frequency. In real-world terms, that describes how much isolation you get at each frequency for a wall that has a given mass. So if you do that equation 20,000 times, for each of the 20,000 possible frequencies in the audio spectrum, you can draw a graph of how well a wall isolates!
Or for lazy folks like me, you can use the empirical version of the equation, which approximates all those 20,000 others, into one:
TL = 14.5 log (Ms * 0.205) + 23 dB
Where: Ms = Surface Mass in kg/m2. So all you need to know is the surface mass of your single wall! Let's talk about that really massive 12" thick concrete wall. Concrete density is roughly 2300 kg/m3, so a wall 12" thick has a surface density of about 782 kg per square meter. Plug that into the "mass law" equation, and you get 54.97 decibels. So if you find a building with 12" reinforced concrete walls, floor and ceiling, you can use it-as, since you are already getting pretty darn good isolation. Not the 60 dB you wanted (you'd need 30" of concrete for that...), but not far short.
That's the absolute truth: The laws of physics do not lie. to get high levels of isolation from a single leaf wall, you need amazing amounts of mass. Which is why studios are not usually built that way, and why I'm assuming that if the photo does show a studio with 8" or 10" concrete walls, then all they needed was 40-soemthing dB of isolation, not the 60 that you need. It is quite literally one hundred times harder to isolate to 60 dB than it is to 40 dB, simply because the dB scale is based on log10. So stepping up from 40 dB to 50 dB is ten TIMES the energy, and ten TIMES harder, but then going from 50 to 60 is ANOTHER ten times the energy, and ten times harder. So 10 x 10 = 100. Log math is a bitch! You think "I just need a few more dB, that's all!". But the logs catch you, as they apply to bank balances too...
OK, so what to do then, if there are no handy buildings with 30" thick concrete walls? That's when you resort to TWO-leaf walls. A 2-leaf wall uses a totally different principle of physics. Not mass, but resonance. That's why it can isolate to much higher levels with much less mass. In essence, it uses sound to kill sound. It uses the power of resonance to fight back. A two-leaf wall is a tuned resonant system, so all you need to do is tune it accordingly, and you get the isolation you need. You "tune" your wall by varying either the mass per unit area of the wall (surface density), or the size of the gap between the two leaves (or both). There are equations for figuring all this out too, but the they are a bit more complex than mass law. So in order to determine what you need to do with your SECOND leaf (the inner one that you plan to build, to create the "room", you first need to know what you already have with the outer leaf. The equation needs to know the surface density of BOTH leaves. So until you know what that is for your outer leaf, there's no real way of knowing what you need for the inner leaf or the air gap. If your outer leaf is paper-thin and has very little useful mass (eg, sheet metal), then you need a major big air gap and stacks of mass on your inner leaf. On the other hand, if the outer leaf is already pretty substantial (say 4" brick), then you can get by with a smaller air gap and less mass on the inner leaf. However, if your outer leaf turns out to already be two leaves, then you have yet another problem: Your inner leaf will make that into a three-leaf system, which very counter-intuitively REDUCES your isolation in low frequencies, so you need even MORE mass and/or even larger air gaps than you would have for a two-leaf...
I'm not trying to be evasive and skirt around answering your question: it's just that, without knowing some more details about what your outer leaf might be, I simply can't give you a valid answer! You are basically asking me "how long does my piece of string have to be", but neither of us knows yet what the string will be tied to, so it simply isn't possible to say.
So getting back to the actual question: "is it possible to get about 60 dB isolation by building four solid walls that go from existing floor to existing ceiling?", I would have to say "Yes", it is
possible, but it might not be
feasible in YOUR building, depending on how it is built. And seeing that tying the walls at both ends greatly reduces isolation due to the multiplied flanking paths (the walls are not decoupled), I would have to add that I wouldn't like to be the one trying to do that. Or rather, I'd love the challenge as a
designer, but I sure wouldn't want to be the guy paying the bills for the materials and labor!
On the other hand, achieving 60 dB with a fully-decoupled MSM "room-in-a-room" that ONLY sits on the floor, and is not attached to anything else, sure is possible, and also feasible, at reasonable cost.
Not sure if that was much help! But really, without knowing the basic parameters of the actual physical building that we are dealing with, that's the best I can do. Sorry!
- Stuart -
