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3rd September 2019
#1
Lives for gear

What frequency is reflected using 1cm thik timber slats?

Hi I was wondering if some knowledgeable acoustics person could answer a very straightforward question. If I have a completely dead room bass trapped to hell with low density fiberglass on all walls, and ceiling, and want to use timber slats with a thickness of 1cm over the 4 walls, what frequencies can I expect to be reflected back into the room? I am sure there is a threshold where frequencies below a certain point are absorbed into the treatment behind the slats. Can anyone help with this?

thanks
3rd September 2019
#2

Thickness?
Or do you mean they are 1 cm wide?
1 cm wide means 17.000 Hz and prolly some diffraction.
But of course they have also a length.
And there are gaps between the slats.

Questions, questions....
3rd September 2019
#3
Lives for gear

Quote:
what frequencies can I expect to be reflected back into the room?
There is a simple principle in acoustics: sound waves are only reflected by solid objects that are similar in size to the wavelength of the sound, or larger. In other words, if the object is 5 cm wide/high, it will only affect waves that are 5cm long, or shorter (shorter wave=higher frequency). That would mean waves with a frequency of about 7,000 Hertz, or higher. If the object is 10cm wide/high, then it would reflect frequencies above 3400 Hz. For an object 1 meter wide/high, that would reflect down to much lower frequencies: around 340 Hertz and higher.

There's also the issue that very thin, flexible objects might act like membranes or foils, in which case they could absorb some sound at the frequencies related to that, and there's the issue that the mass of the object itself will attenuate sound above certain frequencies (prevent i from passing through).

And finally, if you place wood slats over some type of cavity, such as the interior of a wall, then you have created a Helmholtz resonator, which will resonate at some frequency governed by the Helmholtz equations, and will therefore absorb that frequency very well, while reflecting higher frequencies, and perhaps somewhat absorbing lower frequencies, depending on mass, etc.

So there's lots of stuff to take into account. Its not really a simply as just nailing up some slats to get the high end back into the room. That's the general concept, yes, but there's a lot to take into account when you decide how to implement that in practice.

- Stuart -
3rd September 2019
#4

They also have length, Stuart :-).
3rd September 2019
#5

I wouldn't say I am "some knowledgeable acoustics person", more a person who over the years has aquired some bits and pieces knowledge about acoustics. Addition to above, in case you use larger sheet material instead of slats.

For reflection much would depend on the material's weight / m², for absorbtion on the "layer combination" (wood + air gap + possible insulation) plus how porous or not the outer layer is and íts stiffness which depends on distance between studs / framework (the smaller the distance the less effective as (panel) absorber as stiffness increases).

What isn't reflected is either absorbed or passing / transmitted through. In the attachements there are some clues which give some ball park figures for necessay wood thickness. The table is from 3 rd edition of Cox' & D'Antonios "Acoustic Absorbers and Diffusers." Depending on type of plywood the density varies a bit, around 500 kg/m³ could be "typical". For 1 cm thick plywood it is quite obvious it is very reflective for 1 kHz and above, 500 to 1000 Hz can be seen as a "transit region" both reflective and starting to act as a panel absorber if you compare with the figures for 9,5-12,7 mm with an air gap. For even lower frequencies the plywood gets less reflective and acts more and more like a panel absorber. The text attachement is from Kleiner's and Tichys' "Acoustics of Small Rooms". For high reflection you want a low transmission coefficient figure versus frequncies of interest.

If you use MDF as covering, paint it as MDF is a actually a bit porous.
Attached Thumbnails

3rd September 2019
#6
Lives for gear

Quote:
Originally Posted by bert stoltenborg
They also have length, Stuart :-).
Yup! That's why I specified the sizes as " xx cm wide/high", trying to clarify that I was talking about objects that have roughly the same width and height. I guess I should have been a bit more clear about, and specifically mentioned slat length too...

- Stuart -
4th September 2019
#7
Lives for gear

Quote:
Originally Posted by bert stoltenborg
Thickness?
Or do you mean they are 1 cm wide?
1 cm wide means 17.000 Hz and prolly some diffraction.
But of course they have also a length.
And there are gaps between the slats.

Questions, questions....
Hi Bert. I meant 1cm thick, not 1cm wide. The width of the slats are 9cm. The slats are about 2.5m tall and go from floor to ceiling. The gaps are 1cm between the slats. Now I hope can we get to the answer.... If not please ask away.
4th September 2019
#8
Lives for gear

Quote:
Originally Posted by Soundman2020
There is a simple principle in acoustics: sound waves are only reflected by solid objects that are similar in size to the wavelength of the sound, or larger. In other words, if the object is 5 cm wide/high, it will only affect waves that are 5cm long, or shorter (shorter wave=higher frequency). That would mean waves with a frequency of about 7,000 Hertz, or higher. If the object is 10cm wide/high, then it would reflect frequencies above 3400 Hz. For an object 1 meter wide/high, that would reflect down to much lower frequencies: around 340 Hertz and higher.
Hi Stuart, So I really was talking about the thickness of the slat being 1cm. The width of the slats are 9cm and the space between the slats are 1cm too. The slats go all the way to the ceiling. i basically have a room that is bass trapped with a lot of absorption and am looking at fixing slats over the walls so the room isn't dead. I found slats with these dimensions and wanted to use them and wanted to know above what frequency will be reflected back into the room. Obviously a 100hz sound wave will pass right through that. Hope that helps.

Quote:
There's also the issue that very thin, flexible objects might act like membranes or foils, in which case they could absorb some sound at the frequencies related to that, and there's the issue that the mass of the object itself will attenuate sound above certain frequencies (prevent i from passing through).
Ok I didn't know that.

Quote:
And finally, if you place wood slats over some type of cavity, such as the interior of a wall, then you have created a Helmholtz resonator, which will resonate at some frequency governed by the Helmholtz equations, and will therefore absorb that frequency very well, while reflecting higher frequencies, and perhaps somewhat absorbing lower frequencies, depending on mass, etc.
I am aware of this but in this case my main concern was low end control so I used thick superchunk treatment on all walls and ceiling to give me a dead environment down to 60hz then was looking at how to bring back the high end. I was recommended slats and now am looking at material. Can you make a suggestion here?
Quote:
So there's lots of stuff to take into account. Its not really a simply as just nailing up some slats to get the high end back into the room. That's the general concept, yes, but there's a lot to take into account when you decide how to implement that in practice.
great could you lay it down for me please? I was just told to use random slat sequence so if it is more complex than that please tell me so I can do it right.
4th September 2019
#9
Lives for gear

Quote:
Originally Posted by camomiletea
Hi I was wondering if some knowledgeable acoustics person could answer a very straightforward question. If I have a completely dead room bass trapped to hell with low density fiberglass on all walls, and ceiling, and want to use timber slats with a thickness of 1cm over the 4 walls, what frequencies can I expect to be reflected back into the room? I am sure there is a threshold where frequencies below a certain point are absorbed into the treatment behind the slats. Can anyone help with this?

thanks
Download AFMG demo version and type in your data using slotted panel. You will find that the simple diffraction rule does not apply ( "any wavelength smaller than the slat width is going to be reflected" ).

For the start, another tool:

http://www.acousticmodelling.com/mli...4=400&v24=6000

For 9 cm slats, the wavelength would be 3.7 khz.

But apparently, the absorber will not strictly behave like that, there will be a large transistion phase and this has to be taken into account when designing the room.

cheers

Last edited by Synthpark; 4th September 2019 at 11:54 AM..
4th September 2019
#10
Lives for gear

this is great thanks, some ACTUAL data. What could be better.

Quote:
Addition to above, in case you use larger sheet material instead of slats.
ok and how do things change if using slats instead of a solid piece of timber?

Quote:
For reflection much would depend on the material's weight / m², for absorbtion on the "layer combination" (wood + air gap + possible insulation) plus how porous or not the outer layer is and íts stiffness
in this case there would be no air gap. Just the slats laid on top of the giant absorbers.

Quote:
What isn't reflected is either absorbed or passing / transmitted through. In the attachements there are some clues which give some ball park figures for necessay wood thickness. The table is from 3 rd edition of Cox' & D'Antonios "Acoustic Absorbers and Diffusers." Depending on type of plywood the density varies a bit, around 500 kg/m³ could be "typical". For 1 cm thick plywood it is quite obvious it is very reflective for 1 kHz and above, 500 to 1000 Hz can be seen as a "transit region" both reflective and starting to act as a panel absorber
This is great but how do these figures change when the inner layer is not a solid piece of timber as is this case?
16th September 2019
#11
Lives for gear

Quote:
This is great but how do these figures change when the inner layer is not a solid piece of timber as is this case?
If you place wood slats over a porous absorber, then you get one of two possible results. Either you have made a "slot wall", which is a tuned Helmholtz device that has high absorption for some very specific frequencies, or you just have a bunch of absorption with reflective slats over it. In the second case, each slat will reflect the frequencies that I mentioned before: those whose wavelength is smaller than the size of the slats, or similar to the size of the slats, and the rest of the frequencies will get through to the absorption, where they will be absorbed to one degree or another. In the first case (tuned Helmholtz resonator slot wall with slats over a sealed cavity partly filled with insulation), you can tune the peak absorption frequency(ies) as needed to deal with specific issues in the room that are relate to that specific wall. You tune the slats by changing the dimensions: the height of each slat, the thickness of each slat, the size of the gap between the slats, and the depth of the cavity behind the slats, up to the solid wall.

- Stuart -

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