Soundproofing and room size

T

TMe

Senior Member
Is it true that a smaller room is easier to soundproof than a larger room?

If I build a room-within-a-room, and make it just large enough to hold my drum sound, will that small room leak less sound than if I make it twice the size, using the exact same wall structure, door, etc.? Bigger would be nicer, but not if more sound is going to leak out.

Or does room size make no difference, everything else being equal?
 
dboomer said:
The sound quality will suffer in a small room compared to a big room.
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I get that. I'm wondering if room size makes any difference to the amount of noise leaking out of the room.
 
TMe said:
I'm wondering if room size makes any difference to the amount of noise leaking out of the room.
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If I understand what you're after (sound containment only) in a nutshell, the answer is "not really." Containing your sound will mean not only a room in a room construction, but also a floating floor, acoustic grade sheetrock treatment, a whole lot of wall and ceiling dampening, a soundproof door and if you have any windows they'll also need to be soundproof grade. If you have A/C ducts going your room that's something else that needs attention. Think of it as a full on professional studio shrunk down to your room specs, but you still need to build the same quality construction in every detail of your room. If you plan to record in this room I think you need far more space as dboomer has noted above plus professional acoustic treatment on top of all the soundproofing build costs.
 
Also, don't forget that you have to breathe. I've built a soundproof room in the basement, but it's so small that I can only practice for 45 minutes at most before I get dizzy because all the oxygen is used up.
 
I suspect the benefits of a larger room will quickly outweigh any soundproofing advantages a small room may provide.
 
dboomer said:
The sound quality will suffer in a small room compared to a big room. You need at least some 16’ dimensions to develop 40Hz in a room

Audition your bass drum in the shower and see what you think ;)
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Our ears hear long Hz waves at their correct pitch regardless of the length of the airspace. It is false to say you need a full cycle of the wave. Consider headphones.

It can certainly be argued that the instrument sounds “better” in one space or another, but that’s due to how the whole range of harmonics gets reflected or dispersed—not whether the room is as long as a full wave cycle.
 
So far, everything I've heard (pardon the pun) suggests that making a sound booth smaller doesn't make it more soundproof - everything else being equal - so I should use all of what little space I have.
When I was younger I worked on rehearsal rooms in a few schools. The designer for that project believe that smaller rooms were easier to soundproof than large rooms - all else being equal. But I've never heard that from any other source, so I'll assume that was an erroneous belief of his.
 
bongoman said:
Our ears hear long Hz waves at their correct pitch regardless of the length of the airspace. It is false to say you need a full cycle of the wave. Consider headphones.

It can certainly be argued that the instrument sounds “better” in one space or another, but that’s due to how the whole range of harmonics gets reflected or dispersed—not whether the room is as long as a full wave cycle.
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I can tell you because I got to test this, the same drums but in a small church and in a room about the same size but carpeted floor, walls and ceiling, Eloy Casagrande could have been bashing the drums and if you were on the other side of the door you could barely hear it. The drums needed no muffling in that room, they were perfect for recording, but for the experience of playing them there live, it felt like you were playing cardboard boxes, the drums sounded very dead. In contrast because of the high ceilings and the stone walls of the church, the drums sounded huge, also you could not hear much of them right outside the door because they had 4 inch thick wood doors and very thick stone walls, no sound treatment at all. For me church sound all day.
 
bongoman said:
Our ears hear long Hz waves at their correct pitch regardless of the length of the airspace. It is false to say you need a full cycle of the wave. Consider headphones.
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The reason you can hear long wavelengths in headphones is because they create pressure. In normal atmosphere you need length.

Anyone that has mixed down a recording in their bedroom and then played it in the living room knows the low end gets louder. In your bedroom you keep adding low end EQ but it doesn’t really do anything if the dimensions of your room are too small. They when you play it back in a big room all the low end you added comes to life
 
dboomer said:
The reason you can hear long wavelengths in headphones is because they create pressure. In normal atmosphere you need length.
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Citation please. For another example, ask your bassist if they can hear low notes while standing in front of their speaker cab.
 
dboomer said:
Anyone that has mixed down a recording in their bedroom and then played it in the living room knows the low end gets louder. In your bedroom you keep adding low end EQ but it doesn’t really do anything if the dimensions of your room are too small. They when you play it back in a big room all the low end you added comes to life
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Wild fantasy.
 
bongoman said:
...ask your bassist if they can hear low notes while standing in front of their speaker cab.
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I don't know anything about this but... in my experience, when a bassist stands right next to his cab in a small practice room, he can't hear himself so he keeps turning up the volume until it's insufferable for someone on the opposite side of the room. They think they need an 8x10 cab in a room that's about 200 square feet. (Not that this is a pet peeve of mine or anything.) A different bassist puts a small combo amp in a corner opposite from where he's standing, and he can hear himself just fine. So there might be something to what dboomer is saying.
 
bongoman said:
Citation please. For another example, ask your bassist if they can hear low notes while standing in front of their speaker cab.
Click to expand...
You want the actual math or just quotes from the internet?

The general wave equation is described by a second order differential equation

y'' = 1/c^2 dy^2/dt^2

The prime denotes differentiation in respect with space. y is a state variable and usually denotes displacement. The constant c is the propagation speed of the wave in a medium.

The propagation speed depends on the material properties of the medium, and appears naturally in the wave equation when you calculate dynamic equilibrium for a system (force or energy equations).

To solve the wave equation you assume an harmonic solution of a propagating perturbarion. The most general solution is given by the D'Alambert solution that only states that a function of a coupled variable omega*t-kx is a solution. Omega is the angular frequency given by 2pi*freq and k is the wavenumber, which is the space frequency, inverse to the wavelength. This two quantities relates to the propagation velocity through

c = omega/k

Since omega = 2pi*freq and k=2pi/lambda , in which lambda = 2pi/wavelength, then c = lambda*frequency.

To obtain the wave speed in terms of the material properties for a specific wave, you have to look into its wave equation to see who forms the constant c.

For acoutic waves it depends if the medium is air or liquid. If its air, the wave speed is proportional to the medium pressure and mass density. If it is a incompressible liquid, it is given by the bulk modulus divided by the mass density. In general, the wave speed is proportional to a stiffness measure of the medium divided by its mass.





https://gearspace.com/board/so-much...orrect answers,for that information to travel.



https://www.quora.com/How-can-headphones-produce-low-notes-when-speakers-have-to-be-large



https://www.quora.com/Why-is-it-easier-to-hear-low-frequencies-with-earphones-rather-than-speakers
 
TMe said:
I don't know anything about this but... in my experience, when a bassist stands right next to his cab in a small practice room, he can't hear himself so he keeps turning up the volume until it's insufferable for someone on the opposite side of the room. They think they need an 8x10 cab in a room that's about 200 square feet. (Not that this is a pet peeve of mine or anything.) A different bassist puts a small combo amp in a corner opposite from where he's standing, and he can hear himself just fine. So there might be something to what dboomer is saying.
Click to expand...
There’s truth in there. The issue is not soundwaves “developing”, but the cancellation zones that can occur with many reflections. In a smaller space there will be more reflections and higher chances of null zones. So in fairness to dboomer, he might have experienced the things he described because he moved himself out of a null zone, and there are fewer of those in a bigger space. But even in the small space, simply changing where you stand relative to the speaker can make all the difference.

It’s also the case that it requires much more energy to amplify lows than highs, AND our ears hear mids and highs better: https://en.wikipedia.org/wiki/Equal-loudness_contour
That’s why a small speaker is easier to hear than a large one in the given example—the small one is amplifying all the upper harmonics that tell our ears the information needed to create the low note in our hearing, without so much power going to actual low frequencies—meaning the amp doesn’t need to be turned as high, and it’s less overall volume in the room.
 
dboomer said:
You want the actual math or just quotes from the internet?

The general wave equation is described by a second order differential equation

y'' = 1/c^2 dy^2/dt^2

The prime denotes differentiation in respect with space. y is a state variable and usually denotes displacement. The constant c is the propagation speed of the wave in a medium.

The propagation speed depends on the material properties of the medium, and appears naturally in the wave equation when you calculate dynamic equilibrium for a system (force or energy equations).

To solve the wave equation you assume an harmonic solution of a propagating perturbarion. The most general solution is given by the D'Alambert solution that only states that a function of a coupled variable omega*t-kx is a solution. Omega is the angular frequency given by 2pi*freq and k is the wavenumber, which is the space frequency, inverse to the wavelength. This two quantities relates to the propagation velocity through

c = omega/k

Since omega = 2pi*freq and k=2pi/lambda , in which lambda = 2pi/wavelength, then c = lambda*frequency.

To obtain the wave speed in terms of the material properties for a specific wave, you have to look into its wave equation to see who forms the constant c.

For acoutic waves it depends if the medium is air or liquid. If its air, the wave speed is proportional to the medium pressure and mass density. If it is a incompressible liquid, it is given by the bulk modulus divided by the mass density. In general, the wave speed is proportional to a stiffness measure of the medium divided by its mass.





https://gearspace.com/board/so-much-gear-so-little-time/1290621-how-come-headphones-produce-so-low-frequencies.html#:~:text=You have the correct answers,for that information to travel.



https://www.quora.com/How-can-headphones-produce-low-notes-when-speakers-have-to-be-large



https://www.quora.com/Why-is-it-easier-to-hear-low-frequencies-with-earphones-rather-than-speakers
Click to expand...
The first and third links only address power usage, not wave development in space. The second link is about pressure, which —fair enough— is a factor in power usage and dispersal of energy. But again, not about distance.

The math provided does describe how the waves travel, but does not address hearing in any way.
 
TMe said:
I don't know anything about this but... in my experience, when a bassist stands right next to his cab in a small practice room, he can't hear himself so he keeps turning up the volume until it's insufferable for someone on the opposite side of the room. They think they need an 8x10 cab in a room that's about 200 square feet. (Not that this is a pet peeve of mine or anything.) A different bassist puts a small combo amp in a corner opposite from where he's standing, and he can hear himself just fine. So there might be something to what dboomer is saying.
Click to expand...

In his case it is more about the actual distance he is from the cabinet and the frequency he is trying to hear. Heks probably standing in a null and not a peak. For any single frequency there are always a large number of nulls (dead spots). If you were to play a single 40 Hz tone and measure different spots you can clearly see this. Unfortunately you really need to do it in an free space (anechoic chamber), but outside in the middle of a field work pretty good too.

Play a 40 Hz tone and measure the level with an SPL meter. Now backup slowly and you will be able to see the level drop drastically at 4 feet. At 8 feet it will get much louder, at 12 feet it will dip, at 16 feet back up in level. In a room with walls and a ceiling and floor (unfortunately you still have a floor in a field.

When I used to teach classes I would play a 40 Hz tone and then a 50 Hz tone (sine wave) and ask the members of the class whether the sound got louder of quieter. About half the class would say yes, louder and half softer. But the level of both frequencies was exactly the same. The reason this happens is because the dimensions of the peaks and dips changes with frequency.
 
bongoman said:
The first and third links only address power usage, not wave development in space. The second link is about pressure, which —fair enough— is a factor in power usage and dispersal of energy. But again, not about distance.

The math provided does describe how the waves travel, but does not address hearing in any way.
Click to expand...
Sorry you don’t understand the relationship. Wavelength changes with differences in pressure.
 
dboomer said:
Sorry you don’t understand the relationship. Wavelength changes with differences in pressure.
Click to expand...
Are you proposing that frequencies get higher as pressure increases? That the lows in headphones are actually highs?

Instead, the pressure in the air chamber affects the amount of power needed to propagate the wave, not the length of the wave.
 
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