Calculator finds out sound reduction in decibels (dB) for double layer wall with given density, thickness and air gap width using so-called mass law equation.

Beta version#

BETA TEST VERSION OF THIS ITEM
This online calculator is currently under heavy development. It may or it may NOT work correctly.
You CAN try to use it. You CAN even get the proper results.
However, please VERIFY all results on your own, as the level of completion of this item is NOT CONFIRMED.
Feel free to send any ideas and comments !

Calculations data: walls and air gap parameters#

The first wall
The material from which the wall is made
The density of the wall material
Wall thickness
The second wall
The material from which the wall is made
The density of the wall material
Wall thickness
Other
Cavity width

Transmission loss at various frequencies#

Frequency [Hz]	Transmission loss [dB]
31.25	44.54
62.5	50.56
125	56.58
250	62.6
500	68.62
1000	74.64
2000	80.66
4000	86.68
8000	92.7
16000	98.72

Sound insulation#

When a sound wave moves through the air meets a barrier in the form of a wall, part of the acoustic energy is reflected, part is absorbed inside the wall (converted to heat), and the another part is transmited out (on the other side of the wall). We can write it mathematically as follows:

$\alpha + \beta + \tau = 1$
where:
- $\alpha$ - absorption coefficient (determines the part of the energy that was absorbed inside the wall),
- $\beta$ - reflection coefficient (defines the part of the energy remaining in the first room),
- $\tau$ - transmission coefficient (defines the part of the energy that was emitted to the second room).
The transmission coefficient can be used as a measure of the acoustic insulation, because it determines the sound intensity ratio on both sides of the wall:

$\tau = \frac{I_t}{I_0}$
where:
- $I_t$ - intensity of the wave on the other side of the wall (sound intensity level audible in the second room),
- $I_0$ - incident wave intensity (sound intensity level audible in the first room).
In practice, the transmission factor is most often given in the logarithmic scale. In this way, we obtain a decrease in sound intensity given in decibels, so-called transmission loss:

$\Delta TL = -10 ~ log (\tau) = 10 ~ log \left(\frac{1}{\tau}\right)$

Some facts#

The acoustic insulation of a single wall is limited by its thickness and density of the material used. We can overcome these limits using two walls separated by air gap. The resulting system is called depending on the source:
- wall-air-wall,
- mass-air-mass,
- mass-spring-mass (more often found in theoretical papers where such a system is modeled by two masses connected with a spring).
Sound reduction index for double wall can be estimated using formulas introduced by London in 1950, updated later by Sharp in 1973:

$R(f) = \begin{cases} 20 ~ log\left[f \cdot (h_1 ~ \rho_1 + h_2 ~ \rho_2) \right] - 47, & \text{when } f <f_0 \\ R_1 + R_2 + 20 ~ log(f ~ d) - 29, & \text{when } f_0 <f <f_l \\ R_1 + R_2 + 6 & \text{when } f > f_l \end{cases}$
where:
- $R$ - decrease of the sound intensity level of a partition consisting of two walls in decibels,
- $R_1, R_2$ - sound level decrease calculated for the first and second walls separately,
- $h_1, h_2$ - thickness of the first and second walls,
- $\rho_1, \rho_2$ - material density of which the first and second walls are made,
- $d$ - distance between walls (cavity width),
- $f$ - frequency of the acoustic wave,
- $f_0$ - resonant frequency $f_0 = \sqrt{\frac{\rho_0 \cdot c_0^2}{d} \cdot \frac{h_1 ~ \rho_1 + h_2 ~ \rho_2} {h_1 ~ \rho_1 ~ h_2 ~ \rho_2}}$ ,
- $f_l$ - limit frequency $f_l \approx \sqrt{\frac{55}{d}}$ ,
- $c_0$ - speed of sound in the air,
- $\rho_0$ - air density

Room within the room#

A room whose all walls and ceiling are surrounded by an empty air gap (the room has no common walls with others except a common floor) is often called a room-within-the-room.
The disadvantage of this solution is high cost and permanent modification of the building.
For example, in order to isolate a medium-sized live room with a usable area of approx. 26 m² we need an additional approx. 24 m² of empty space for air-gap insulation (assuming a distance between the walls of 1 m). Additionally, it is necessary to raise the entire building or to give up part of the height of the separated room (to leave an empty space under the outer room ceiling).
For this reason, classic room-within-the-room solutions are used only in specialized buildings, which of purpose is permanently related to the need of high acoustic insulation such as recording studios or sound laboratories.

Tags and links to this website#

Tags:

mass_law_for_double_wall · mass_law_for_double_layer · wall_air_wall_calculator · mass_air_mass_calculator

Tags to Polish version:

prawo_masy_dla_sciany_podwojnej · prawo_masy_podwojna_przegroda · sciana_powietrze_sciana_kalkulator · masa_powietrze_masa_kalkulator

What tags this calculator has#

CALCULATOR · PHYSICS · AUDIO/SOUND · A -> Z (all)

Permalink#

This is permalink. Permalink is the link containing your input data. Just copy it and share your work with friends:

https://calculla.com/mass_law_for_double_wall?inSubstanceId1=-1&inDensity1=2500&inDensityUnitId1=kg_per_m3&inThickness1=50&inThicknessUnitId1=cm&inSubstanceId2=-1&inDensity2=2500&inDensityUnitId2=kg_per_m3&inThickness2=50&inThicknessUnitId2=cm&inCavityWidth=100&inCavityWidthUnitId=cm

Links to external sites (leaving Calculla?)#