Adding iso booth to small room with varied wall construction

[Beekman]

Overview - My objective for this project is to create additional isolation in an existing room in an office building. The room will be primarily used for voice over recordings (70% male, 30% female), and maybe 10% of the time used for vocal recording. No music recording. I have done some research at this point but am not sure how to proceed with regard to a two leaf vs three leaf issue and could use some advise.

The existing room is used as an office, located along a busy street. We are trying to isolate additional traffic noise from the road and talking in other portions of the building. The goal is to have an iso booth in the room, but also allow for two desk areas for computers.

One wall is an exterior of the building (left wall in picture). It is made of a 6" CMU, metal studs, R15 of fiberglass insulation, and 5/8" drywall. There is a long double pane, aluminum framed window on the bottom portion of the room that crosses into the adjacent room.

One wall separates the office section of the building from the warehouse of the building. (top wall). Coming from the warehouse, the wall is 5/8" drywall, 5/8" plywood for a portion of the wall, metal studs, r15 fiberglass insulation, and 5'8" drywall. The plywood extends roughly one half the length of the wall. There is a single pane aluminum framed window along part of the wall, and in the area of the window there is no plywood.

The other two walls are 5/8 drywall, metal stud, 5/8 drywall with no insulation. There is a 8' door and a 8' tall x 2' single pane side window that leads to a hallway on the right of the drawing.

Ceilings are 11' drop ceilings, with the actual roof at ~20ft and building mechanical (ducts, pipes, etc) overhead. Existing HVAC supply and return is overhead and is tied to several adjacent rooms. Flooring is concrete slab with industrial carpet over the slab. 24" OC for all studs. There is fire sprinkler in the ceiling as well.

Budget is approximately $3500 for materials. I am capable of framing, drywall, electrical work, etc. My goal is to maximize bang for the buck in terms of isolation for the investment.

My idea was to build a booth in the upper left corner of the room, between the exterior wall and the warehouse wall. I was thinking approximately 6' to 7' along these walls, which would avoid overlapping with the windows, and potentially making it a diamond shape to help with the door location and minimizing space taken up in the room. Thinking 8ft high on the exterior, so I could use a single sheet of drywall per layer with no seams. My thought was the challenges of the drop ceiling and windows would mean that a booth would provide bigger bang-for-the buck than attempting to isolate the entire room.

I see a few possible options for how to construct the booth walls -

Option 1 - Frame a single-leaf stand alone booth, with perhaps a 12" gap between the existing walls, and 7' for the more distant walls. Thinking two layers of drywall with green glue. Not sure if better to go with insulation inside to help with acoustics, or insulation out to maximize the air gap. (And then have to do more treatment inside, which would further reduce interior volume).

Option 2 - Use the two existing walls to act as the second leaf, perhaps beefing them up with an additional layer of drywall and green glue. Attach other walls to existing studs. Add a single leaf ceiling. So that would give me two leaves for two of the walls, and potentially three leaves for the others... with roughly a 7 ft airgap on the three leaf side walls, and a 3ft gap to the ceiling.

Option 3 - (not sure this is a good one) Frame a double wall stand alone booth like option 1. I suspect diminishing returns in terms of more work and materials and adding a fourth leaf, but wasn't sure if the 12" airgap changed that equation. Putting in the middle of the room is not a good option as I will lose the ability to use it for work areas/computers.

Option 4 - Spend the time trying to improve the isolation of the existing room. Perhaps another layer of drywall and green glue everywhere, soundproof seals on the door. Maybe replace windows with drywall. Build a small cabinet for computers to contain that noise. Not sure what if anything to do with ceiling.

Option 5 - Rip out the interior leaf of the existing walls, then build a booth for the entire volume of the room. I may have issues with the existing windows, fire sprinklers, etc. It will also mean I need to put computers insure the interior leaf, and build enclosures for them, etc. It also means I have to reverse the process when I move out. Not sure if my budget will allow this level of work, unless this is a so much superior option that I'd be foolish to not go this route.

Option 6 - Option 5, but I fabric coat the walls and leave the windows/doors, and build a smaller single leaf booth like option 1.

Questions -

A) Which of the above options seems to make the most sense?

B) If we go with a booth option, how do I best attach the walls to the floor? Rip out carpet where the wall is, caulk and anchor wall to slab?

[Soundman2020]

Hi "Beekman", and Welcome!

One wall is an exterior of the building (left wall in picture). It is made of a 6" CMU, metal studs, R15 of fiberglass insulation, and 5/8" drywall.

Are the metal studs attached to, or in contact with, the CMU? Or is there a gap between those studs and the CMU?

There is a long double pane, aluminum framed window on the bottom portion of the room that crosses into the adjacent room.

That's a problem. A big problem! Your booth needs to stay away from that area. What type/thickness of glass is that?

My goal is to maximize bang for the buck in terms of isolation for the investment.

First point in the process: define how much isolation you need, in decibels (not STC!).

My idea was to build a booth in the upper left corner of the room, between the exterior wall and the warehouse wall. I was thinking approximately 6' to 7' along these walls, which would avoid overlapping with the windows, and potentially making it a diamond shape to help with the door location and minimizing space taken up in the room.

That's fine, but you also have to be thinking about the acoustics inside the room. Small booths sound disgusting, and the smaller they are, the worse they sound. Try to optimize the interior volume of the room. You want it to be as big inside as you can possibly make it, and you should also try to avoid having direct relationship between the dimensions: So of the room is 7 feet long, it should NOT be 7 feet wide. Try to have at least 5% difference between the three sets of dimensions, and 10% if possible.

Thinking 8ft high on the exterior, so I could use a single sheet of drywall per layer with no seams.

Once again, maximize room volume. It's nice to optimize materials usage too, where possible, to avoid waste and extra work, but this is an ACOUSTIC project, so acoustics should be the first goal.

A) Which of the above options seems to make the most sense?

We don't know what your isolation goals are yet, so it's impossible to say. Once we know how much isolation you need, in decibels, and what frequency range you need it for, then we can help you decide on methods and materials.


- Stuart -

[Beekman]

Thanks for the help Stuart!

The metal studs are in contact with the CMU. I can't tell if the studs are screwed to the CMU at any regular interval, but they are pressing right against eachother.

The exterior glass is 1/2" tempered, with a 3/4" airgap in between the two panes. The interior glass is 1/4" tempered, single pane.

In terms of the isolation amount, I'd be thrilled if I could pick up an additional 30dB of reduction in the frequencies of traffic noise and human speech. I could probably live with 20dB.

Point taken on the room sizing. So I'm thinking maybe 6ft x 7ft for the walls. If I go 8ft for the ceiling, which would give me about 3ft distance from the drop ceiling. Better to go more like 9 or 10ft to increase the internal volume with the tradeoff of a shorter airgap with the ceiling?

[Beekman]

Anyone able to help?

My fundamental question is when building a iso booth inside an existing two-leaf structure, assuming no flanking paths other than a thick concrete floor, and assuming something like a 12" airgap.... will I have better low frequency performance with a single leaf structure, or a dual leaf structure? How could I quantify this difference?

[Soundman2020]

My fundamental question is when building a iso booth inside an existing two-leaf structure, assuming no flanking paths other than a thick concrete floor, and assuming something like a 12" airgap.... will I have better low frequency performance with a single leaf structure, or a dual leaf structure? How could I quantify this difference?

Assuming that you tune it correctly, you will get better low frequency isolation from a two-leaf structure, always. A 12" gap with good mass on both sides (eg, 2 layers of 5/8" drywall) will give you good isolation at low frequencies.

How could I quantify this difference?

Mass Law gives you the isolation for a single leaf:

TL(dB)= 20log(M) + 20log(f) -47.2

M is the surface density of the panel (mass per unit area (kg/m²) ), and
F is the center frequency of the third-octave measurement band


For a two-leaf wall, it's a bit more complex:


The equations for calculating total isolation of a two-leaf wall are simple:

First, for a single-leaf barrier you need the Empirical Mass Law equation (similar to the above, but for the entire spectrum, not a specific frequency):

TL = 14.5 log (M * 0.205) + 23 dB

Where: M = Surface density in kg/m2

For a two-leaf wall, you need to calculate that for EACH leaf separately (call the results "R1" and "R2").

Then you need to know the resonant frequency of the system, using the MSM resonance equation:

f0 = C [ (m1 + m2) / (m1 x m2 x d)]^0.5

Where:
C=constant (60 if the cavity is empty, 43 if you fill it with suitable insulation)
m1=mass of first leaf (kg/m^2)
m2 mass of second leaf (kg/m^2)
d=depth of cavity (m)

Then you use the following three equations to determine the isolation that your wall will provide for each of the three frequency ranges:

R = 20log(f (m1 + m2)) - 47 ...[for the region where f < f0]
R = R1 + R2 + 20log(f x d) - 29 ...[for the region where f0 < f < f1]
R = R1 + R2 + 6 ...[for the region where f > f1]

Where:
f0 is the resonant frequency from the MSM resonant equation,
f1 is 55/d Hz
R1 and R2 are the transmission loss numbers you calculated first, using the mass law equation

And that's it! Nothing complex. Any high school student can do that. It's just simple addition, subtraction, multiplication, division, square roots, and logarithms.


- Stuart -