Calculating transmission loss of a triple leaf system

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Waka
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Calculating transmission loss of a triple leaf system

Post by Waka »

Hi again,

To start with I'll show you my double leaf wall calculations and then I'll move on to the triple leaf questions.

(I'm going to show my working, so you can double check the method is correct)

For my double leaf walls:
First calculate MSM wall resonant frequency:

Fx = c[(m1+m2)/(m1m2d)]^0.5

18mm OSB3 and 15mm Plasterboard on each decoupled wall frame. Insulated cavity is 225mm panel to panel.
18mm OSB3 at 11.8kg/m2.
15mm Plasterboard at 10.25kg/m2.
Total: 22.05kg/m2

Giving:
Fx = 43 * [(22.05 + 22.05) / (22.05 * 22.05 * 0.225)] ^ 0.5
Fx = 43 * [ 44.1 / 109.4 ] ^ 0.5
Fx = 43 * 0.4 ^ 0.5
Fx = 43 * 0.63
Fx = 27.09Hz

Single leaf transmission loss for each leaf:
TL = 14.5 log(M * 0.205) + 23 dB

TL = 14.5 log(22.05 * 0.205) + 23
TL = 14.5 log(4.52) + 23
TL = 14.5 * 0.66 + 23
TL = 9.57 + 23
TL = 32.57dB

Now move on to transmission loss formulas:
R1 and R2 are using my single leaf transmission loss calculated above (32.57dB).
f0 is my wall resonant frequency.
f1 is 55/d Hz = 55 / 0.225 = 244.45Hz.

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

The lowest frequency I'm realistically aiming to isolate is the low E of a 4 string Bass guitar so I'm looking at 40Hz upwards.
So as 40Hz is above my resonant wall frequency and less than 244.45Hz I use formula two above.

R = R1 + R2 + 20log(f * d) - 29
R = 32.57 + 32.57 + 20 log( 40 * 0.225) - 29
R = 65.14 + 20 log( 9 ) - 29
R = 65.14 + 19.08 - 29
R = 55.22dB

I am happy with this level of isolation if I can achieve it.


Now moving onto the the triple leaf, this is my roof. I need a vented roof space so will have to seal the underside of the roof joists, this forces me to have a triple leaf ceiling.

How do you calculate the resonant frequencies is this system? As I have read this is not just a case of calculating them independently?

I've seen these equations listed on another thread here:
double-triple f0.jpg
The author of the thread listed the variables as:
m1,m2,m3 - mass per unit area (kg/m2)
c0 - speed of sound 343 m/s
ro0 - air density 1,18 kg/m3

These are more complicated formulas and I'm struggling to understand the triple leaf one especially. This is my working for the double leaf partition:

m0 = (2 * m1 * m2) / (m1 + m2)
m0 = (2 * 22.05 * 22.05) / (22.05 + 22.05)
m0 = 972.4 / 44.1
m0 = 22.05

f0 = 1 / (2π) * [(3.6 * ro0 * c0^2) / (m0 * d)] ^ 0.5
f0 = 0.16 * [(3.6 * 1.18 * 343^2) / ( 22.05 * 0.225)] ^ 0.5
f0 = 0.16 * [499772.95 / 21.07] ^ 0.5
f0 = 0.16 * 23719.65 ^ 0.5
f0 = 0.16 * 154.01
f0 = 24.64Hz

This is reasonably close to the simplified formula I've used before (although I'm not sure if insulation is figured in this formula at all??). Have I calculated this correctly?
Then can someone help me with the second formula for triple leaf partitions please?

This is my working so far on the triple leaf calcs:
m1 = 22.05
m2 = 22.05
m3 = 11.8 (single OSB3 18mm)
d1 and d2 = 0.2 (200mm cavities)

a = 1 / (2 * m2) * ( [(m1 + m2) / (m1 * d1)] + [(m2 + m3) / (m3 * d2)] )
a = 1 / ( 2 * 22.05) * ( [(22.05 + 22.05) / (22.05 * 0.2)] + [(22.05 + 11.8 ) / (11.8 * 0.2)] )
a = 0.023 * ( [44.1 / 4.41] + [33.85 / 2.36] )
a = 0.023 * ( 10 + 14.34)
a = 0.56

b = M / (m1 * m2 * m3 * d1 * d2)
I'm assuming capitalised M means the combined mass of the all of the layers on the leaves (as in the single leaf equation). So:
b = (22.05 + 22.05 + 11.8 ) / (22.05 * 22.05 * 11.8 * 0.2 * 0.2)
b = 55.9 / 229.49
b = 0.24

Now the actual resonant frequency formula:

For the first frequency:
fx = 1 / (2π) * [(3.6 * ro0 * c0 ^ 2) ^ 0.5] * [ ( a + [ (a^2 - b) ^ 0.5 ] ) ^ 0.5 ]
fx = 1 / (2π) * [(3.6 * 1.18 * 343^2) ^ 0.5] * [ ( 0.56 + [ (0.56^2 - 0.24) ^ 0.5 ] ) ^ 0.5 ]
fx = 0.16 * [499772.95 ^ 0.5] * [(0.56 + 0.27) ^ 0.5]
fx = 0.16 * 706.94 * 0.91
fx = 102.93Hz

For the second frequency:
fy = 1 / (2π) * [(3.6 * ro0 * c0 ^ 2) ^ 0.5] * [ ( a - [ (a^2 - b) ^ 0.5 ] ) ^ 0.5 ]
fy = 1 / (2π) * [(3.6 * 1.18 * 343^2) ^ 0.5] * [ ( 0.56 - [ (0.56^2 - 0.24) ^ 0.5 ] ) ^ 0.5 ]
fy = 0.16 * [499772.95 ^ 0.5] * [(0.56 - 0.27) ^ 0.5]
fy = 0.16 * 706.94 * 0.54
fy = 61Hz

These don't quite look right as the lowest frequency is almost 3 times what I calculated for the double leaf.

Thanks (I feel like I'm back at school) :oops:
Dan
Stay up at night reading books on acoustics and studio design, learn Sketchup, bang your head against a wall, redesign your studio 15 times, curse the gods of HVAC silencers and door seals .... or hire a studio designer.
TomSleebus
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Re: Calculating transmission loss of a triple leaf system

Post by TomSleebus »

Old thread, hope someone can pick it up

I found only a reference to green glue's pdf on https://www.greengluecompany.com/sites/ ... effect.pdf

Which seems to indicate indeed higher fx and fy figures.

Have you been able to confirm the formula's? I'm curious to see as I have a similar situation to solve...

As I cannot add mass to the shell (which is a double leaf) I currently have the inner shell (third leaf) to be built as far away from the existing wall as doable (and yes that is a compromise), not yet sure how to calculate correctly.
gullfo
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Re: Calculating transmission loss of a triple leaf system

Post by gullfo »

you have to take into account several factors - each layer attenuation before calculating the next leaf. then if the final (third leaf) is far away (e.g. 3') and it's mass is more substantial than the inner leafs, the 3rd leaf effect is much less significant. otherwise to mitigate - you need to change the layering in each leaf to avoid (as much as possible) coincident resonance frequencies. (2x 5/8, 2x 5/8, 1x 5/8) on each subsequent leaf. i did this for a client who had a 2 leaf construct from someone else but not enough isolation and adding the light 3rd leaf as a membrane absorber effect did the trick. neighbor 5' away couldn't hear anything when band inside was cranking 110 or more db... fusion jazz - 5 string bass, 7 string electrics
Glenn
TomSleebus
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Re: Calculating transmission loss of a triple leaf system

Post by TomSleebus »

Thanx Glenn,

I'll continue the triple leaf question in my thread on my studio project if you agree...

To return to the OP's initial question, I think he made a mistake in the formula

f0 = 1 / (2π) * [(3.6 * ro0 * c0^2) / (m0 * d)] ^ 0.5
f0 = 0.16 * [(3.6 * 1.18 * 343^2) / ( 22.05 * 0.225)] ^ 0.5
f0 = 0.16 * [499772.95 / 21.07] ^ 0.5
f0 = 0.16 * 23719.65 ^ 0.5
f0 = 0.16 * 154.01
f0 = 24.64Hz

In the third line, 21,07 is not the result of 22.05*0.225, when calculating the f0 in that case would be closer to 51Hz, which on its turn is much closer to the fy found for the triple leaf formula.

Still puzzles me why this' thread double leaf resonant frequency formula gets another result from the formula that I'm used to

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

This gives for the OP's wall 29Hz as resonant frequency with insulation, 40Hz without insulation.
gullfo
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Re: Calculating transmission loss of a triple leaf system

Post by gullfo »

keeping in mind, these formulas are very simplistic using simple mass calculations and changing the framing (narrower to raise resonances, staggered width 16" + 24", and plywood sheathing) can change the actual functioning quite a lot. the old studiotips MSM spreadsheets while useful are simply rough estimates and its really required to empirical tests to assess what the construction, layers, and materials actually do.
Glenn
Elusive Sounds
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Re: Calculating transmission loss of a triple leaf system

Post by Elusive Sounds »

I was just asking about this in my design thread. Will look at this closely. Thanks.
Elusive Sounds
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Re: Calculating transmission loss of a triple leaf system

Post by Elusive Sounds »

TomSleebus wrote: Still puzzles me why this' thread double leaf resonant frequency formula gets another result from the formula that I'm used to

f0 = C [ (m1 + m2) / (m1 x m2 x d)]^0.5
Tom, which value do you use for the constant C for your calculations? This is giving me a lot of confusion because I keep seeing the value 43 (representing a filled cavity for imperial units calculations) used while metric values for mass (KG/m^2) are used in the equation for fo.
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