A question that comes to mind: can an aperiodic vent be made by drilling small holes in the woodwork? I've placed a solid wooden plate over the former bass reflex port.
Then, I could easily make the aperiodic vent by drilling very small holes.. Any thoughts?
Then, I could easily make the aperiodic vent by drilling very small holes.. Any thoughts?
Aperiodic Loudspeaker Enclosure Design
Here's an article describing aperiodic venting. There are some links at the bottom of that page that you should read through as well to get a handle on what an aperiodic vent does and how to apply it to your project.
Here's an article describing aperiodic venting. There are some links at the bottom of that page that you should read through as well to get a handle on what an aperiodic vent does and how to apply it to your project.
Yes, you definitely can, many examples in the past.
Do a search here at DIYAudio for "aperiodic" and you'll find a lot of info, some from Planet10, some from others.
To be most effective an aperiodic vent's Sd should be as a minimum 1/3 to 1/2 the Sd of the driver.
Have a look at some of the files below and also the links suggested by puppet.
Attachments
Well, I'm first going to try isolating the bass cabinet with long hair sheep wool and maybe line it with bitumen. After that, we'll see where we stand.
tl;dr: To my understanding applying a layer of sound absorbing material directly onto the walls of the cabinet is not a good idea as it basically forms an anti-reflective coating similar to what is used for eyeglasses helping the waves to escape the cabinet while you actually want them to dissipate to heat inside the cabinet. This affects regular and standing waves alike.
Secondly, standing waves can only be reduced in areas where air moves. In general, the closer you get to the wall the lower the average particle velocity for standing waves will be. Additionally, you should consider the position of the driver in the cabinet as some standing waves will not be excited.
The acoustic impedance determines the ratio of sound which is reflected back at an interface and which share is transmitted into the other medium. The acoustic impedance is calculated based on the speed of sound in the medium and the density (Z = rho * c). Based on the impedance of two materials the reflection factor can be calculated (r=(Z1-Z2)/(Z1+Z2)) (This is only valid for perpendicular incidence and presumably infinite dimensions).
Now let’s consider two examples: A cabinet made of beech and a second cabinet made of beech with an additional layer of porous absorber on the inside. On the all-knowing the internet I found the following numbers for speed of sound and density and used them to calculate the impedance of the materials:
Air: c=330 m/s, rho=1,292 kg/m³ -> Z=428 Ns/m³
Absorber: c=280 m/s, rho=150 kg/m³ -> Z=42000 Ns/m³
Beech: c=5100 m/s, rho=780 kg/m³ -> Z=3978000 Ns/m³
With this the share of reflected sound at each interface can be calculated for each individual interaction:
Air - Beech: r= 0,9997
Air - Absorber: r= 0,9798
Absorber - Beech: r= 0,9791
At first glance those numbers don’t look too different but let see what happens if a sound wave bounces around 300-times allowing to switch between materials each time unless it reached the outside.
In the first example: Inside-the-cabinet (Air) - Beech - Outside (Air)
The starting conditions look like this:
Inside 1 - Beech 0 - Outside 0
After 300 bounces:
Inside 0,9394 - Beech 0,0587 - Outside 0,0019
In the second example: Inside-the-cabinet (Air) - Absorber - Beech - Outside (Air)
The starting conditions are again:
Inside 1 - Absorber 0 - Beech 0 - Outside 0
After 300 bounces:
Inside 0,3303 - Absorber 0,3281 - Beech 0,3249 - Outside 0,0166
In example 1 the sound wave bounces back and forth in the inside and hardly leaves the cabinet. If a piece of absorber is placed far away from the walls of the cabinet, the sound wave will pass through it multiple times turning it partially to heat each time ultimately getting rid of it efficiently.
In example 2 the absorber close to the walls helps the sound wave to enter the beech which will then slowly leak to the outside. Once in the beech, absorber material placed inside the cabinet has no more impact and, overall, more sound will leak to the outside.
So, imho, the way to go to reduce regular sound waves (in contrast to standing waves) in the cabinet is to not apply an absorber to the wall which would work as an anti-reflective layer but use absorber somewhere else in the cabinet. Just from a perspective of turning sound into heat: the more absorber the better.
Standing waves are a different topic.
In order for a standing wave to be problematic in your build it has to fulfil all of the following conditions:
- It has to exist (duh). The shape and dimensions of your cabinet allow only a specific (but infinite) number of standing waves to exist. All standing waves have in common that their average particle velocity is 0 at the walls of the cabinet. So, again, absorber doesn’t help there.
- It has to be excited. While the particle velocity is always 0 at a wall, the pressure change is at a maximum there. In general, wherever the particle velocity is 0 the pressure change is at a maximum and the other way round.
Now think of your driver as an array of point sources for pressure equally distributed as its membrane. For all standing waves oscillating between the front and the back wall of the cabinet all these point sources are in an ideal position to excite all standing waves, no exceptions. But for standing waves oscillating between top and bottom or left and right only some of the point sources may be positioned in a way to excite a specific standing wave while others may even excite them with a 180° turned phase effectively cancelling each other out immediately. To be more precise, standing waves which have a zero-crossing for pressure changes exactly in the middle of the driver are cancelled out completely. Standing wave with the zero-crossing shifted to either side are still weakened. This means overall there will be a minimum in particle velocity due to standing waves exactly in the middle of the driver as these waves are cancelled out during excitation.
So out of all the existing standing waves mainly those exclusively oscillating between front and back have to be addressed. Waves with a component in another direction get much less excited. Especially high frequency waves with components in other directions get much less excited since several peaks with different algebraic sign are excited simultaneously.
- It has to find a way out of the cabinet. Overall, high frequencies have a much harder time getting out of the cabinet than lower frequencies. The cabinet in general is ideally quite sound proof. The weak points are mainly a bass reflex port and the drivers. The driver will always be in a position where the particle velocity is at a minimum and the pressure change at a maximum since it is causing the standing waves in the first place.
Standing waves inside the bass reflex port will be excited by high particle velocity since it is open-ended. So, ideally you put the inside opening in a spot with low particle velocity. That is either very close to the wall or at 50% of either dimension or at the exact position of the driver. I already mentioned multiple times that the particle velocity will be 0 at the wall but beyond that every other wave will have a zero-crossing at 50% of either dimension making this the second best choice for the average particle velocity but on the other hand this leaves you with all the remaining particle velocity packed into a few strong resonances. Finally, the exact position of the driver may already be taken by... the driver but since all waves are symmetrical you will encounter the same lower particle velocity at several spots of the cabinet. Say your driver is mounted 5 cm from top and left, respectively, on the front plate you will have the same conditions 5cm from top and right, 5cm from bottom and left, 5 cm from bottom and right both at the front plate as well as at the backplate.
- It must not be absorbed. With all other design decisions made, it is finally time to get to the point and answer the question where to put the absorber. The job of the absorber is only to take care of what’s left of standing waves considering all the other effects mentioned. There are two approaches and imho both should be followed:
1. Reduce the overall amount of standing waves.
2. Put special focus on waves which may escape through either weak point in the cabinet.
Regarding 1.:
First of all, absorber is very efficient for high frequency waves and considering all the other points you can pretty much ignore them and focus only on the low frequency waves.
Then, waves left-right and top-bottom are much weaker than front-back, so while you should address all of them the focus has to be front-back. To address them specifically you would need to put an absorber plate parallel to the front wall inside the cabinet. The most efficient spots are between 10-40% and 60-90% of the axis since only high frequencies show significant particle velocity closer to the walls. Since all waves are symmetrically to the centre, putting something at e.g. 10-20% and 80-90% would address the same waves to a higher degree but would not address a broader range of waves, whereas if you would put something at e.g. 10-20% and 70-80% you would instead address a broader range of waves. For symmetry reasons these plates wouldn’t even need to be as large as the whole front plate but putting them into either quarter (from 0-50% along the other two axis) would address still the same range of standing waves.
Regarding 2.:
You also want to reduce specifically the waves which may couple with standing waves inside the bass reflex port. As I mentioned the waves are symmetrically to the centre of each axis. So, if you put absorber mirrored to the position of the opening of the bass reflex port along either axis you can reliably reduce the particle velocity due to standing waves at the opening of the bass reflex port.
Secondly, standing waves can only be reduced in areas where air moves. In general, the closer you get to the wall the lower the average particle velocity for standing waves will be. Additionally, you should consider the position of the driver in the cabinet as some standing waves will not be excited.
The acoustic impedance determines the ratio of sound which is reflected back at an interface and which share is transmitted into the other medium. The acoustic impedance is calculated based on the speed of sound in the medium and the density (Z = rho * c). Based on the impedance of two materials the reflection factor can be calculated (r=(Z1-Z2)/(Z1+Z2)) (This is only valid for perpendicular incidence and presumably infinite dimensions).
Now let’s consider two examples: A cabinet made of beech and a second cabinet made of beech with an additional layer of porous absorber on the inside. On the all-knowing the internet I found the following numbers for speed of sound and density and used them to calculate the impedance of the materials:
Air: c=330 m/s, rho=1,292 kg/m³ -> Z=428 Ns/m³
Absorber: c=280 m/s, rho=150 kg/m³ -> Z=42000 Ns/m³
Beech: c=5100 m/s, rho=780 kg/m³ -> Z=3978000 Ns/m³
With this the share of reflected sound at each interface can be calculated for each individual interaction:
Air - Beech: r= 0,9997
Air - Absorber: r= 0,9798
Absorber - Beech: r= 0,9791
At first glance those numbers don’t look too different but let see what happens if a sound wave bounces around 300-times allowing to switch between materials each time unless it reached the outside.
In the first example: Inside-the-cabinet (Air) - Beech - Outside (Air)
The starting conditions look like this:
Inside 1 - Beech 0 - Outside 0
After 300 bounces:
Inside 0,9394 - Beech 0,0587 - Outside 0,0019
In the second example: Inside-the-cabinet (Air) - Absorber - Beech - Outside (Air)
The starting conditions are again:
Inside 1 - Absorber 0 - Beech 0 - Outside 0
After 300 bounces:
Inside 0,3303 - Absorber 0,3281 - Beech 0,3249 - Outside 0,0166
In example 1 the sound wave bounces back and forth in the inside and hardly leaves the cabinet. If a piece of absorber is placed far away from the walls of the cabinet, the sound wave will pass through it multiple times turning it partially to heat each time ultimately getting rid of it efficiently.
In example 2 the absorber close to the walls helps the sound wave to enter the beech which will then slowly leak to the outside. Once in the beech, absorber material placed inside the cabinet has no more impact and, overall, more sound will leak to the outside.
So, imho, the way to go to reduce regular sound waves (in contrast to standing waves) in the cabinet is to not apply an absorber to the wall which would work as an anti-reflective layer but use absorber somewhere else in the cabinet. Just from a perspective of turning sound into heat: the more absorber the better.
Standing waves are a different topic.
In order for a standing wave to be problematic in your build it has to fulfil all of the following conditions:
- It has to exist (duh). The shape and dimensions of your cabinet allow only a specific (but infinite) number of standing waves to exist. All standing waves have in common that their average particle velocity is 0 at the walls of the cabinet. So, again, absorber doesn’t help there.
- It has to be excited. While the particle velocity is always 0 at a wall, the pressure change is at a maximum there. In general, wherever the particle velocity is 0 the pressure change is at a maximum and the other way round.
Now think of your driver as an array of point sources for pressure equally distributed as its membrane. For all standing waves oscillating between the front and the back wall of the cabinet all these point sources are in an ideal position to excite all standing waves, no exceptions. But for standing waves oscillating between top and bottom or left and right only some of the point sources may be positioned in a way to excite a specific standing wave while others may even excite them with a 180° turned phase effectively cancelling each other out immediately. To be more precise, standing waves which have a zero-crossing for pressure changes exactly in the middle of the driver are cancelled out completely. Standing wave with the zero-crossing shifted to either side are still weakened. This means overall there will be a minimum in particle velocity due to standing waves exactly in the middle of the driver as these waves are cancelled out during excitation.
So out of all the existing standing waves mainly those exclusively oscillating between front and back have to be addressed. Waves with a component in another direction get much less excited. Especially high frequency waves with components in other directions get much less excited since several peaks with different algebraic sign are excited simultaneously.
- It has to find a way out of the cabinet. Overall, high frequencies have a much harder time getting out of the cabinet than lower frequencies. The cabinet in general is ideally quite sound proof. The weak points are mainly a bass reflex port and the drivers. The driver will always be in a position where the particle velocity is at a minimum and the pressure change at a maximum since it is causing the standing waves in the first place.
Standing waves inside the bass reflex port will be excited by high particle velocity since it is open-ended. So, ideally you put the inside opening in a spot with low particle velocity. That is either very close to the wall or at 50% of either dimension or at the exact position of the driver. I already mentioned multiple times that the particle velocity will be 0 at the wall but beyond that every other wave will have a zero-crossing at 50% of either dimension making this the second best choice for the average particle velocity but on the other hand this leaves you with all the remaining particle velocity packed into a few strong resonances. Finally, the exact position of the driver may already be taken by... the driver but since all waves are symmetrical you will encounter the same lower particle velocity at several spots of the cabinet. Say your driver is mounted 5 cm from top and left, respectively, on the front plate you will have the same conditions 5cm from top and right, 5cm from bottom and left, 5 cm from bottom and right both at the front plate as well as at the backplate.
- It must not be absorbed. With all other design decisions made, it is finally time to get to the point and answer the question where to put the absorber. The job of the absorber is only to take care of what’s left of standing waves considering all the other effects mentioned. There are two approaches and imho both should be followed:
1. Reduce the overall amount of standing waves.
2. Put special focus on waves which may escape through either weak point in the cabinet.
Regarding 1.:
First of all, absorber is very efficient for high frequency waves and considering all the other points you can pretty much ignore them and focus only on the low frequency waves.
Then, waves left-right and top-bottom are much weaker than front-back, so while you should address all of them the focus has to be front-back. To address them specifically you would need to put an absorber plate parallel to the front wall inside the cabinet. The most efficient spots are between 10-40% and 60-90% of the axis since only high frequencies show significant particle velocity closer to the walls. Since all waves are symmetrically to the centre, putting something at e.g. 10-20% and 80-90% would address the same waves to a higher degree but would not address a broader range of waves, whereas if you would put something at e.g. 10-20% and 70-80% you would instead address a broader range of waves. For symmetry reasons these plates wouldn’t even need to be as large as the whole front plate but putting them into either quarter (from 0-50% along the other two axis) would address still the same range of standing waves.
Regarding 2.:
You also want to reduce specifically the waves which may couple with standing waves inside the bass reflex port. As I mentioned the waves are symmetrically to the centre of each axis. So, if you put absorber mirrored to the position of the opening of the bass reflex port along either axis you can reliably reduce the particle velocity due to standing waves at the opening of the bass reflex port.
Thank you for this very thorough explanation.
I've just spend the day lining the walls with bitumen and 5 mm of felt.
Tomorrow I expect a shipment with long hair sheep wool to fill the cabinets (mid and bass) and with.
Afterwards I'll spend some time listening to it all and then decide what to do.
Since the bass reflex port is sealed shut the bass drivers will be the primary exit point of the lower frequencies (the mid and tweet are in a separate cabinet.
Hoping I can do some mearuments to see what's going on, but so far I've learned a lot. I'm curious if all these inner cabinet stuff actually will matter in my room with a wooden floor and stone walls. But we'll see.
After the the bitumen lining, the bass cabinet vibrates less though.
I've just spend the day lining the walls with bitumen and 5 mm of felt.
Tomorrow I expect a shipment with long hair sheep wool to fill the cabinets (mid and bass) and with.
Afterwards I'll spend some time listening to it all and then decide what to do.
Since the bass reflex port is sealed shut the bass drivers will be the primary exit point of the lower frequencies (the mid and tweet are in a separate cabinet.
Hoping I can do some mearuments to see what's going on, but so far I've learned a lot. I'm curious if all these inner cabinet stuff actually will matter in my room with a wooden floor and stone walls. But we'll see.
After the the bitumen lining, the bass cabinet vibrates less though.
@TroyMC
The transmission of sound energy through a solid is equal to the total sound energy on impact minus the sound energy reflected minus the sound energy transformed into heat. It seems to me you ignore this basic understanding of acoustics in the first part of your plead.
Furthermore I’d say you are simplifying things in an attractive way, but also ignoring other more or less important details. It sadly isn’t this elegant. The conclusion however could stand in certain situations.
The transmission of sound energy through a solid is equal to the total sound energy on impact minus the sound energy reflected minus the sound energy transformed into heat. It seems to me you ignore this basic understanding of acoustics in the first part of your plead.
Furthermore I’d say you are simplifying things in an attractive way, but also ignoring other more or less important details. It sadly isn’t this elegant. The conclusion however could stand in certain situations.
I guess I don't see any conflict betwen these two statements, since the bitumen and felt are an attempt to convert any sound energy transmitted to the walls into heat, above and beyond attempting to absorb the energy bouncing around the interior of the enclosure...
Damping vibrations on the interior surface with layers of absorptive material may be less effective than stuffing the center of the enclosure, bracing or adding mass to the walls. Until empirically proven to be ineffective, will continue to be a long-standing practice for reducing unwanted sound transmission through the walls of a sealed speaker enclosure.
Empirically speaking, it is hard to argue against the knuckle rap test... an more inert wall should be less amenable to vibrating as a result of the modes excited within the cabinet by one or more drivers. Could that be due to the added mass? or could it be due to some of the energy being converted to heat? perhaps.
It is difficult to measure heat conversion, so we have to rely on the measured SPL graph as an indicator that cabinet resonances are mitigated to the builder's satisfaction, within the construction and financial means available to them.
I love how this thread is combinig theory and practice, and helping others evolve their thinking about stuffing, bracing, and damping strategies.
Thanks! - Six - Minneapolis.
Exactly, after all this reading on different views I decided to just try something affordable and doable.
It is of course an idealized description and nothing I personally came up with (e.g. Acoustic Impedance - an overview | ScienceDirect Topics). It does consider all the effect you mention though except for sound energy being transformed into heat at the interface. I am not aware of a particularly high amount of energy being dissipated into heat at the interface but certainly sound slowly turns into heat while travelling which is true independetly if it travels back and forth inside of the box or leaks outside. So imho the general point remains true.
If you can point out specific errors or can point out more or less important details missing in the second consideration I am very happy to alter my view as this only leads to better cabinets.
From a different perspective, bitumen and maybe also felt helps reducing structure-borne vibrations by increasing the loss modulus of the plate meaning that once the vibrations are transferred into the cabinet plates, they are dissipated faster into heat giving them less time to radiate to the outside. A viscoelastic material is thereby especially effective when confined by two layers.
You certainly address right and true issues. So keep on examining. But when you state
you are missing the quintessence of damping materials. A wall with damping wedges like those in an anechoic chamber absorbs virtually all acoustic energy thrown at it. There mainly is a relation between the thickness of the layer and the cutoff frequency. That’s absorption for you and it helps sound isolation of cabinet walls, not worsen it.
Maybe just a communication issue. I can see a way that Troy is correct, but I would rather ask to explain that paragraph again..
My comments may land somewhat blunt, apologies where needed. It’s just that stacking assumptions only brings you anywhere if all of them are right. And I noticed quite some long shot assumptions not backed up by observations (measurements).
Well in any case, hopefully the bitumen, felt and wool will make a positive difference.
Can't wait to measure again.
Can't wait to measure again.
markbakk, please clarify :
So the contention is about the sound energy after the absorption material. You say significant portion gets absorbed by the insulation and do not make it through the cabinet.
TroyMC says a significant amount of sound energy gets though the insulation and reaches the cabinet, then gets transferred to the outside through the cabinet.
Is is that for all group of frequencies?
SMathews (absolute novice here)
So the contention is about the sound energy after the absorption material. You say significant portion gets absorbed by the insulation and do not make it through the cabinet.
TroyMC says a significant amount of sound energy gets though the insulation and reaches the cabinet, then gets transferred to the outside through the cabinet.
Is is that for all group of frequencies?
SMathews (absolute novice here)
I do not have measurements to back up anything I wrote. I am just using fundamental relations and their consequences in, admittedly, an idealized and simplified scenario. I dont claim to know everything about this as I am a hobbyist myself but I am happy to participate in delving deeper into that topic. Although, I am unsure if this is the right thread for this as it focuses on Geertidow's approach to this problem.
However, since you brought up anechoic chambers I will comment on that. I did some reasearch in Room Acoustics, fourth edition (184-187). It is written there that the wedges are used to gradually increase the flow resistance and subsequently the impedance starting from air to that of a continious layer of absorber intentionally forming an anti-reflective layer so that there is no sharp air-absorber interface which would reflect too much sound for a proper anechoic chamber. It is also written there that the "anti-reflective character" of the wedges are only effective if they are at least 1/3 of the wavelength. As such one can argue if the "anti-reflective character" matters for a thin bitumen - felt treatment. Still one would assume that the overall impedance of the cabinet perceived by longer wavelengths is lower and therefore less advantageous than when using e.g. steel instead of felt. In the example shown there the wedges also arent directly mounted to the wall but in a 10 cm distance and with some spacing between the wedges to form an Helmholtz resonator for the bass region where the wedges alone arent enough.
Back on topic, I am fully convinced that your bitumen - felt arrangement changes things to the better and I am looking forward to your assessment.
However, since you brought up anechoic chambers I will comment on that. I did some reasearch in Room Acoustics, fourth edition (184-187). It is written there that the wedges are used to gradually increase the flow resistance and subsequently the impedance starting from air to that of a continious layer of absorber intentionally forming an anti-reflective layer so that there is no sharp air-absorber interface which would reflect too much sound for a proper anechoic chamber. It is also written there that the "anti-reflective character" of the wedges are only effective if they are at least 1/3 of the wavelength. As such one can argue if the "anti-reflective character" matters for a thin bitumen - felt treatment. Still one would assume that the overall impedance of the cabinet perceived by longer wavelengths is lower and therefore less advantageous than when using e.g. steel instead of felt. In the example shown there the wedges also arent directly mounted to the wall but in a 10 cm distance and with some spacing between the wedges to form an Helmholtz resonator for the bass region where the wedges alone arent enough.
Back on topic, I am fully convinced that your bitumen - felt arrangement changes things to the better and I am looking forward to your assessment.
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I mainly stated the impedance calculations of TroyMcC did not take absorption into account. Now absorption of porous materials isn’t very simple to describe, but every porous material has a transmission loss factor. This factor is frequency dependent.
Calculations of absorption in porous materials are only relatively simple for perpendicular angles of incidence. The aforementioned wedges combined with resonators aren’t calculated simple. Absorption also is altered by nearby boundaries (read enclosure walls with porous materials on them).
Calculating transmission of enclosure walls also isn’t basic math. While mass law helps us to begin with, clamps and corners really mess simplicity up. In short for us relative novices it is often more simple to measure than to calculate. Even renowned speaker brands rather use BEM techniques.
So to answer your question more or less, in sound transmission of enclosure walls with porous lining, there is frequency dependency.
