Amount of Noise extinction in Acrylic glass per thickness in mm

[Tech_Rizzle]

Hey guys,

I need your help with the following topic on which I couldn't find sufficient information on the internet to solve.

In the need to further insulate my velux roof swing window against against airborne motor and tire sound from the nearby street, I came across the idea to install an additional pane out of acrylic glass into the existing wooden frame.

My question right now is whether this really absorbs sufficient sound for my use, and how much.

From my point of understanding, there has to be a determinable extinction capability of an acrylic pane measured in decibels, which is, as the logarithmic scale and measurement of the unit decibel implies, independent from the actual loudness of the noise, but is rather dependent on the thickness of the pane as well as the frequency of the sound being dealt with.

In this case, I assume we have motor/engine or more generally traffic noise with frequencies ranging from ca. 500 to 1000 hertz.
The acrylic glass pane, like the window to be attached to, measures 97 x 117 centimetres. Its thickness will be in the dimension of a few mm, I'm considering either 8 mm or 10 (which I consider important because it must influence its natural resonance frequency). The density of the acrylic glass it's made from is 1.19 g/ccm.

Now, my question: Given the frequency of sound, the dimensions of the glass pane, its density (which I consider to be important based on the assumption that the weight of a resonating body influences how strongly it resonates e.g. how much sonic energy it absorbs),

how many decibels of sound will the additional pane be able to absorb for a thickness of, say, 8 or 10 millimeters?

is it possible to calculate this in advance? and how?

I thank you very much for your help!

 

[Paul Colby]

Most of the sound dampening is due to sound energy being reflected by the impedance mismatch at the air-acrylic boundary. To get a number one would need the speed of sound in acrylic. For glass, the difference between 8 an 10 mm is likely small.

 

[Tech_Rizzle]

okay, thanks so far. But I still wouldn't know how to calculate this. At least, for the speed of sound in arylic I found 1430 m/s. This is, similar to the density difference, about half as much as for mineral glass - am I guessing right that a higher difference of sound speed at media boundary means a higher impedance mismatch? And what, by your judgement, would be the number?

Thanks a lot!

 

[Baluncore]

This is a complex case with many modes. For low frequencies, the sheet will act like the membrane on a drum with resonances. For higher frequencies it will act like a reflector due to the acoustic impedance mismatch at the air-polymer surfaces. There will also be a higher frequency at which it will appear to be transparent, as a quarter-wave transformer, or as a half-wave resonator.
https://en.wikipedia.org/wiki/Acoustic_impedance

 

[Baluncore]

Google gives us; https://soundproofingideas.com/acrylic-sheet-for-soundproofing/

 

[Paul Colby]

But I still wouldn't know how to calculate this.

Well, I could fit my acoustics knowledge in a very small thimble. Fortunately, as far as transmission line theory goes, acoustics should be the same as EM. Let's take the case of half the world is air and the other half is acrylic with a flat planer boundary between them. If the acoustic impedance of air is $z_1$ and the acoustic impedance of acrylic is $z_2$ the reflection coefficient is,

$R = \frac{z_2-z_1}{z_2+z_1}.$

This reflection coefficient is the ratio of the reflected sound amplitude divided by the incident amplitude. The reflected power ratio is $|R|^2$. Okay, what's the acoustic impedance? Well, if $\rho$ is the mass density of the medium and $v$ is the speed of sound in the medium then,

$z = \rho v.$

To compute the power making it into the house, we can take the approximation,

$P = 1-|R|^2.$

This assumes that the loss in the acrylic is zero and the exit boundary may be ignored. Ignoring the exit boundary is ignoring your question, "how thick should I make this window." It also ignores the mechanism suggested by @Baluncore.

Justifying ignoring @Baluncore mechanism is not a simple task. I've never noticed a marked decrease in sound transmission through a window even in all my years of banging on glass windows. I would think dampening the window vibration with my hand would have a marked effect if window vibrational modes were the dominant phenomena. That said, he could be right.

 

[sophiecentaur]

Could this link be useful? It would give a ball park figure for improvement. But spacing makes a difference and, of course, once you start to eliminate the more obvious noise, the leaks of sound through small cracks can dominate - so the added layer of plexiglass needs to have well sealed edges.

 

[sophiecentaur]

At least, for the speed of sound in arylic I found 1430 m/s. This is, similar to the density difference, about half as much as for mineral glass -

It tends to be more complicated than this because you are not just considering the reflection and transmission at the interface but the actual flexing of a fairly thin membrane. The dimensions of a window will affect what modes of oscillation it can have and it is the amplitudes of the surface displacement of the sheet that determines how much Energy is coupled to and from the air - and so on to the next pane. The information in my previous post shows how true this is; same material but different thickness, different result.

 

[Baluncore]

Drum membranes explain the low frequency transmissions through rectangular windows. Use smaller window panes to raise the frequency.

http://spiff.rit.edu/classes/phys283/lectures/two_d/two_d.html#2d

https://www.acs.psu.edu/drussell/Demos/rect-membrane/rect-mem.html

 

[Paul Colby]

Okay, just to complete the thought I've added window thickness to the transmission line theory. The wavelength in acrylic is,

$\lambda = \frac{1430}{1000} = 1.43 m$

which is much bigger than 10mm. For a lossless plastic, the impedance at the outer boundary is,

$Z(w) = z_2\frac{z_1+iz_2\tan(\beta w)}{z_2+iz_1\tan(\beta w)}$

where $w$ is the window thickness and $\beta=2\pi/\lambda .$ Since the window is real thin this can be written,

$Z(w) \simeq z_1 + i\frac{z_2^2-z_1^2}{z_2}\beta w$

One may just replace $Z(w)$ for $z_2$ in post #6 and compute the difference between an 8 and 10 mm window. Well, at least in this model. I've been informed that I need to go shopping so pushing the number is being left as an exercise.

 

[Paul Colby]

Shopping's done. If I did my arithmetic correctly,

$\frac{z_2}{z_1} = 4050.0$

The plate modal analysis is a physically more complete analysis for a finite area window. For normal incidence on an infinite area window, the analysis I've given is complete enough. Reflection at the window outer surface really dominates this problem.

Anyway, crunching the numbers I get,

8mm Window: -37.1 dB

10mm Window: -39.0 dB

I wouldn't sweat a difference of 2dB as it's likely dominated by other things.

 

[sophiecentaur]

. If I did my arithmetic correctly,

Are you convinced that the "arithmetic" is appropriate for the problem, though? Your sums should include the flexing of the pane, which is how the majority of the energy would pass through the window. (Your calculations are basically showing the transmission through a small diameter piston). The overall answer has to be that a thicker pane will work better; it will raise the frequency of any resonances / oscillatory modes, which may help and damp the low frequency transmission, which is affected by the overall mass of the pane, treated as a piston.
"The fatter the better" is the answer (as your arithmetic also predicts) and the fixings / supports should be a rigid as possible

 

[Paul Colby]

Are you convinced that the "arithmetic" is appropriate for the problem,

The more I think about it, yes. The surface modes of the window provide a complete basis to solve the complete boundary value problem. For an infinite area window, the transmission line equations are exact for the normal incidence of plane waves. For a finite-size window, one needs to expand the pressure wave in terms of the surface modes. Each mode will be driven by a tiny amount of the full spectral power so I'm not concerned with resonance phenomena. There is little doubt the calculation is an estimate. I think for the OP it's more than adequate.

The overall answer has to be that a thicker pane will work better;

Yes, it's about 2 dB better for 10mm versus 8mm in the limit of an infinite area.

"The fatter the better" is the answer (as your arithmetic also predicts) and the fixings / supports should be a rigid as possible

The impedance of acrylic is 4050 times that of air. The impedance difference drives the energy transfer between materials. My recommendation would be two 4mm thick windows with an air gap in between.

 

[sophiecentaur]

Each mode will be driven by a tiny amount of the full spectral power so I'm not concerned with resonance phenomena

I see where you're going with this but the modes are only natural frequencies of oscillation. In a situation where there's significant loss (damping) involved, the nuisance sounds don't need spectral components that coincide with those modes - as with simple damped harmonic motion, you can excite the oscillator with frequencies over a significant bandwidth. In many 'noisy rooms', the intrusive sounds can often be highly 'coloured', according to the response of the transmission mechanism and general noisy sounds are not monochromatic (or whatever the audio equivalent word is).

 

[Paul Colby]

modes are only natural frequencies of oscillation.

Acrylic is likely not anywhere near as high-Q as glass. The modes are all likely heavily damped.

Using a canonical solution, like the reflection from a plane, locally is a time-honored approximation. Take diffraction through an aperture. For a very small aperture size, using the illuminating field as the aperture field is a poor approximation. However, by the time one gets a significant fraction of a square wavelength, this becomes a reasonable estimate. You get the field near the edge wrong but the majority is spot on. Basically, a small aperture is all edge.

For the window problem specified, the window area is a significant fraction of a square wavelength at the wavelengths of interest.

I think people view the transmission line calculation as a different mechanism than the modal expansion of the finite window, they aren't. They are two approximations of a complicated boundary value problem.

I also find it interesting that the link given in #4 says,

"
The noise reduction mechanism of an acrylic sheet panel is to reflect sound instead of absorbing it. even 15 to 20mm acrylic sheet is capable of reducing about 32 to 35dB of noise.
"

which I agree with. It's the number of high impedance mismatches that gives good sound isolation. There are, of course, all the competing sources like small holes and the walls of the structure itself to consider.

 

[sophiecentaur]

My recommendation would be two 4mm thick windows with an air gap in between.

If you look at the observation windows between gallery and recording studio, you find the air gap is very large (at least the thickness of a standard (as massive as possible) brick wall. So your air gap would best be as wide as your ceiling / window frame will allow. Another case of the thicker the better for sound insulation.

 

[sophiecentaur]

I need a bit of help with this problem. I haven't the Maths to do it but I would think that one way to deal with the problem would be a 'lumped components' approach, treating the pane as a piston with a mass and a Young's modulus. Obviously, the same sort of factors are at work (density and bulk modulus) but I have a problem visualising (intuition alert !) the sound wave traveling through as a bulk wave (extremely high impedance), compared with a relatively low impedance diaphragm / piston. The wavelength of sounds, from higher up, is only around 1/5 of the wavelength in air - that means the pane is still a small fraction of a wavelength thick for virtually all the intrusive sounds.
So doesn't that imply the better way to approach the problem would be as a piston with restoring force?
I'll be quite happy if you can resolve this for me.

 

[Paul Colby]

I need a bit of help with this problem.

This paper looks like it works just this type of problem.

https://arxiv.org/abs/1310.8026

Okay, the model presented could be modified (simplified) to solve the problem where the window is treated as a membrain.

 

[Tech_Rizzle]

Okay guys, I find it very interesting what you all have said to this topic! The practicable way to go for me now would be to take a 10 mm pane and build it into the window in such a way that there is a space between the original and the new window of at least 10 mm. I'd go for the 2x 4mm option, but I think this is too difficult from the viewpoint of craftsmanship. From my intuition, I don't really believe that it will bring more than 30 decibels of noise reduction, but guessing that for this use, around 15 would already be sufficient, I hope that it will insulate thus far.

anyway, it was a pleasure reading all your contributions, although I didn't understand half of them.

I will, of course, keep you updated on any empirical findings.