Helmholtz absorber size question

[Ethan Winer]

The type of bass traps I'm most familiar with are either large slabs
of fiberglass or foam (broadband), or large vibrating panels over a
sealed cavity that are tuned by varying the panel mass / thickness and
cavity depth. Those are easy to understand because sound waves strike
a large surface area.

But I don't understand the absorbing mechanism for Helmholtz
resonators like this:

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/cavity.html

How can a small round opening or slot absorb a meaningful amount? I
understand the cavity is filled with absorbing "fuzz" to damp the
resonance. But the exposed "surface area" of the hole or slot seems
way too small. Obviously I'm missing something important, but what? Is
there a large amount of turbulence at the opening?

Thanks!

--Ethan

[GregS]

Post by Ethan Winer
The type of bass traps I'm most familiar with are either large slabs
of fiberglass or foam (broadband), or large vibrating panels over a
sealed cavity that are tuned by varying the panel mass / thickness and
cavity depth. Those are easy to understand because sound waves strike
a large surface area.
But I don't understand the absorbing mechanism for Helmholtz
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/cavity.html
How can a small round opening or slot absorb a meaningful amount? I
understand the cavity is filled with absorbing "fuzz" to damp the
resonance. But the exposed "surface area" of the hole or slot seems
way too small. Obviously I'm missing something important, but what? Is
there a large amount of turbulence at the opening?
Thanks!
--Ethan

There is more action at the opening. I guess the bigger
the opening the less resonance you have, but there
must some best sized opening. The opening shown is small.
Seems to me the types I have seen before are some sort of
tube with the entire end open.

greg

[Szczepan Bialek]

Post by Ethan Winer
The type of bass traps I'm most familiar with are either large slabs
of fiberglass or foam (broadband), or large vibrating panels over a
sealed cavity that are tuned by varying the panel mass / thickness and
cavity depth. Those are easy to understand because sound waves strike
a large surface area.

Look at this: http://hyperphysics.phy-astr.gsu.edu/hbase/waves/cavh.html#c1

Sound waves strike the large surface area of the brass Helmholtz resonators
tuned to a scale of frequencies.

Post by Ethan Winer
But I don't understand the absorbing mechanism for Helmholtz
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/cavity.html
How can a small round opening or slot absorb a meaningful amount? I
understand the cavity is filled with absorbing "fuzz" to damp the
resonance. But the exposed "surface area" of the hole or slot seems
way too small. Obviously I'm missing something important, but what?

"If extra air is pushed into the volume and then released, the pressure will
drive it out. But, acting somewhat like a mass on a spring which is pulled
down and then released, it will overshoot and produce a slight vacuum in the
cavity. The air will oscillate into and out of the container for a few
cycles at a natural frequency."

The above is true but the resonator as the aid works in the opposite way.
Brass absorbs and air is pushed into the ear.

Post by Ethan Winer
Is there a large amount of turbulence at the opening?

Rather no.
S*

[Answerman]

Post by Ethan Winer
The type of bass traps I'm most familiar with are either large slabs
of fiberglass or foam (broadband), or large vibrating panels over a
sealed cavity that are tuned by varying the panel mass / thickness and
cavity depth. Those are easy to understand because sound waves strike
a large surface area.
But I don't understand the absorbing mechanism for Helmholtz
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/cavity.html
How can a small round opening or slot absorb a meaningful amount? I
understand the cavity is filled with absorbing "fuzz" to damp the
resonance. But the exposed "surface area" of the hole or slot seems
way too small. Obviously I'm missing something important, but what? Is
there a large amount of turbulence at the opening?
Thanks!
--Ethan

It's more about what you're assuming than what you're missing. In fact,
you are assuming two things, both of which are incorrect. One is that the
description in the link that you give pertains to sound absorbtion. The
other is that, in general, a single Helmholtz resonator is an effective
sound absorber.

Broadly speaking, Helmholtz resonators have two applications. One is as a
lumped element in an acoustic circuit. Examples of this application would
be a notch filter in a duct or a resonant element as part of a complex
filter to shape the response of a hearing aid. In these applications, the
function of the Helmholtz resonator is to provide a desired frequency-
dependent impedance within a acoustic network involving other acoustic
elements, which may or may not involve the absorbtion of sound.

The application is a two-dimensional array of a multiplicity of individual
Helmholtz resonators such as a perforated panel sound absorber. In this
application the panel ideally provides, over a limited frequency range,
the same specific acoustic impedance as that of the incident traveling
wave. As a result, ideally, there is no wave reflection, which essentially
amounts complete absorption.

[Ethan Winer]

Thanks very much to all who replied. I have a few follow-up questions,

Post by Szczepan Bialek
Sound waves strike the large surface area of the brass Helmholtz resonators

Are you saying the sound waves excite air inside the brass bowls by
going through the outer brass surface? I've seen Helmholtz absorbers
built into construction bricks, where it seems unlikely the sound is
going through the brick walls. RPG sells a product like that:

http://www.rpginc.com/products/diffusorblox/index.htm

Also, page 318 of the 4th edition of Everest's Master Handbook of
Acoustics shows a solid brick wall with small slots sparsely placed
between the bricks. It looks like maybe 1 percent total surface is
slots and the rest is solid brick.

Post by Szczepan Bialek
The application is a two-dimensional array of a multiplicity of individual
Helmholtz resonators such as a perforated panel sound absorber.

So in this case a series of many holes effectively becomes a large
absorbing surface?

What I don't understand are the designs that have a large cavity with
a small opening or narrow slot. Where the total opening seems way too
small to present enough absorbing surface to do anything useful.

I'll explain why I'm asking. Bag End sells an active bass trap called
the E-Trap:

http://www.bagend.com/

(There's no way to post a direct link. Click Products on the left
side, then Professional Acoustics on the far right top.)

This bass trap is basically a small (10-inch) subwoofer that plays a
countering signal into the room. I always figured it was like a panel
absorber, where the speaker diaphragm acts as a "shock absorber" for
the waves that strike its surface. Since a 10-inch speaker has a small
surface area, it's either set up to provide more than 100 percent
diaphragmatic absorption, or mimic what happens around the opening of
a Helmholtz absorber. So my point here is to learn what's possible and
what is not with surfaces that intuitively seem too small to absorb.

--Ethan

[GregS]

Post by Ethan Winer
Thanks very much to all who replied. I have a few follow-up questions,

Post by Szczepan Bialek
Sound waves strike the large surface area of the brass Helmholtz resonators

Are you saying the sound waves excite air inside the brass bowls by
going through the outer brass surface? I've seen Helmholtz absorbers
built into construction bricks, where it seems unlikely the sound is

A round object has very good strength to non-movement.

Post by Ethan Winer
http://www.rpginc.com/products/diffusorblox/index.htm
Also, page 318 of the 4th edition of Everest's Master Handbook of
Acoustics shows a solid brick wall with small slots sparsely placed
between the bricks. It looks like maybe 1 percent total surface is
slots and the rest is solid brick.

I have seen this used in a motor room.
There is also fiberglass inside the space in the block.

Post by Ethan Winer

Post by Szczepan Bialek
The application is a two-dimensional array of a multiplicity of individual
Helmholtz resonators such as a perforated panel sound absorber.

So in this case a series of many holes effectively becomes a large
absorbing surface?
What I don't understand are the designs that have a large cavity with
a small opening or narrow slot. Where the total opening seems way too
small to present enough absorbing surface to do anything useful.
I'll explain why I'm asking. Bag End sells an active bass trap called
http://www.bagend.com/
(There's no way to post a direct link. Click Products on the left
side, then Professional Acoustics on the far right top.)
This bass trap is basically a small (10-inch) subwoofer that plays a
countering signal into the room. I always figured it was like a panel
absorber, where the speaker diaphragm acts as a "shock absorber" for
the waves that strike its surface. Since a 10-inch speaker has a small
surface area, it's either set up to provide more than 100 percent
diaphragmatic absorption, or mimic what happens around the opening of
a Helmholtz absorber. So my point here is to learn what's possible and
what is not with surfaces that intuitively seem too small to absorb.

If its active, its just going to produce an anti sound in reverse polarity.
Be it closed box, vented, or bandpass. Closed box would be easier to
deal with phase. Some of the active subwoofers use feedback
of the diaphram to get even response. You would need a microphone
feedback system to produce the opposite polarity cancelling component.
It will also have a limitation on SPL an be tricky to maintain
polarity over the passband.

Any inactive speakers in a room will have a small effect on that
rooms response. Most mid to large sized speakers have woofer free air
resonance below 100 Hz acting lik an absorber.

I wish I could remember what I came across in Speaker Builder magazine, but looking through
25 years issues is time consuming. I don't think I have ever came across an inactive
absorber with a real small hole. Maybe a hole 8 inches or more perhaps.

greg

[Szczepan Bialek]

Post by Ethan Winer
Thanks very much to all who replied. I have a few follow-up questions,

Post by Szczepan Bialek
Sound waves strike the large surface area of the brass Helmholtz resonators

Are you saying the sound waves excite air inside the brass bowls by
going through the outer brass surface?

Yes. But it apply to the hearing aid. (Answerman wrote: " In fact,
you are assuming two things, both of which are incorrect. One is that the
description in the link that you give pertains to sound absorbtion.")

Post by Ethan Winer
I've seen Helmholtz absorbers
built into construction bricks, where it seems unlikely the sound is
http://www.rpginc.com/products/diffusorblox/index.htm

Helmholtz absorbers are like the optic absorbers (chamber with the small
openings). It is not effective ( The
other is that, in general, a single Helmholtz resonator is an effective
sound absorber.)

Post by Ethan Winer
Also, page 318 of the 4th edition of Everest's Master Handbook of
Acoustics shows a solid brick wall with small slots sparsely placed
between the bricks. It looks like maybe 1 percent total surface is
slots and the rest is solid brick.

Post by Szczepan Bialek
The application is a two-dimensional array of a multiplicity of individual
Helmholtz resonators such as a perforated panel sound absorber.

So in this case a series of many holes effectively becomes a large
absorbing surface?

Absorbing is a combination the all waves phenomenons: reflection,
interference, scattering and so on. Experts know why to use them.

Post by Ethan Winer
What I don't understand are the designs that have a large cavity with
a small opening or narrow slot. Where the total opening seems way too
small to present enough absorbing surface to do anything useful.
I'll explain why I'm asking. Bag End sells an active bass trap called
http://www.bagend.com/
(There's no way to post a direct link. Click Products on the left
side, then Professional Acoustics on the far right top.)
This bass trap is basically a small (10-inch) subwoofer that plays a
countering signal into the room. I always figured it was like a panel
absorber, where the speaker diaphragm acts as a "shock absorber" for
the waves that strike its surface. Since a 10-inch speaker has a small
surface area, it's either set up to provide more than 100 percent
diaphragmatic absorption, or mimic what happens around the opening of
a Helmholtz absorber. So my point here is to learn what's possible and
what is not with surfaces that intuitively seem too small to absorb.

S*

[jerry]

Post by Ethan Winer
Thanks very much to all who replied. I have a few follow-up questions,

Post by Szczepan Bialek
Sound waves strike the large surface area of the brass Helmholtz resonators

Are you saying the sound waves excite air inside the brass bowls by
going through the outer brass surface? I've seen Helmholtz absorbers
built into construction bricks, where it seems unlikely the sound is
http://www.rpginc.com/products/diffusorblox/index.htm
Also, page 318 of the 4th edition of Everest's Master Handbook of
Acoustics shows a solid brick wall with small slots sparsely placed
between the bricks. It looks like maybe 1 percent total surface is
slots and the rest is solid brick.

Post by Szczepan Bialek
The application is a two-dimensional array of a multiplicity of individual
Helmholtz resonators such as a perforated panel sound absorber.

So in this case a series of many holes effectively becomes a large
absorbing surface?
What I don't understand are the designs that have a large cavity with
a small opening or narrow slot. Where the total opening seems way too
small to present enough absorbing surface to do anything useful.
I'll explain why I'm asking. Bag End sells an active bass trap called
http://www.bagend.com/
(There's no way to post a direct link. Click Products on the left
side, then Professional Acoustics on the far right top.)
This bass trap is basically a small (10-inch) subwoofer that plays a
countering signal into the room. I always figured it was like a panel
absorber, where the speaker diaphragm acts as a "shock absorber" for
the waves that strike its surface. Since a 10-inch speaker has a small
surface area, it's either set up to provide more than 100 percent
diaphragmatic absorption, or mimic what happens around the opening of
a Helmholtz absorber. So my point here is to learn what's possible and
what is not with surfaces that intuitively seem too small to absorb.
--Ethan

Ethan,
What you are looking for is an "effectiveness factor" for a given
absorber, and in general there isn't a simple answer--these things are
determined by calculating overlap integrals of the boundary conditions
at the absorber mouth (area) times the pressure (or velocity) for a
given room mode. Room modes are previously determined by the shape of
the room and solving Helmholtz' equation for the pressure
distribution--in a non-rectangular room this can require finite
element analysis. Thus, the placement and number of such absorbers is
reduced to a trial and error approach in many instances, i.e. an
"art", as the audiophiles would like to say. It's not an art, but
just a complex phenomenon with several factors, that responds much
better to empirical test than to calculation. Overall, you would
expect the surface area of a suitable absorber to be about the same as
that of a suitable woofer for the room. Keep in mind that the
effective area of the mouth of a Helmholtz resonator is larger than
the throat diameter...see "Acoustics" by Beranek for some formulas for
this effect...it results in the effective diameter being about 1.5
times the real diameter, but the factor depends on frequency.

[Anserman]

Post by Ethan Winer
Thanks very much to all who replied. I have a few follow-up questions,

Post by Szczepan Bialek
Sound waves strike the large surface area of the brass Helmholtz resonators

Are you saying the sound waves excite air inside the brass bowls by
going through the outer brass surface? I've seen Helmholtz absorbers
built into construction bricks, where it seems unlikely the sound is
http://www.rpginc.com/products/diffusorblox/index.htm
Also, page 318 of the 4th edition of Everest's Master Handbook of
Acoustics shows a solid brick wall with small slots sparsely placed
between the bricks. It looks like maybe 1 percent total surface is
slots and the rest is solid brick.

Post by Szczepan Bialek
The application is a two-dimensional array of a multiplicity of
individual Helmholtz resonators such as a perforated panel sound
absorber.

So in this case a series of many holes effectively becomes a large
absorbing surface?

It's not the holes. It's the large two-dimensional array consisting of a
multiplicity of identical Helmholtz resonators that absorbs sound at the
frequency to which the resonators are tuned.

Post by Ethan Winer
What I don't understand are the designs that have a large cavity with
a small opening or narrow slot. Where the total opening seems way too
small to present enough absorbing surface to do anything useful.

You are incorrectly assuming that surface area of the openings in a
Helmholtz panel absorber is the main factor that determines its
efffectiveness. It's the impedance of absorber at its resonant frequency
relative to the the impedance of air that determines its absorption
effectiveness, not the area of the holes.

Post by Ethan Winer
I'll explain why I'm asking. Bag End sells an active bass trap called
http://www.bagend.com/
(There's no way to post a direct link. Click Products on the left
side, then Professional Acoustics on the far right top.)
This bass trap is basically a small (10-inch) subwoofer that plays a
countering signal into the room. I always figured it was like a panel
absorber, where the speaker diaphragm acts as a "shock absorber" for
the waves that strike its surface. Since a 10-inch speaker has a small
surface area, it's either set up to provide more than 100 percent
diaphragmatic absorption, or mimic what happens around the opening of
a Helmholtz absorber. So my point here is to learn what's possible and
what is not with surfaces that intuitively seem too small to absorb.

If by "bass trap" you mean "absorber" then this device is not a bass trap.
When you add absorption to a room, reverberation time is reduced. This
does not reduce the remveberation time of the room because it does not
absorb sound. All it does is to partially cancel the sound in a very
limited frequency range. According to the description, this device
operates on the same principle as an active noise canceler. The room
will ring just as much when the device is on as when it is off. The only
difference is that signal level is a bit lower in amplitude than the
original signal, but the decay time will be the same.

[Kari Pesonen]

Post by Ethan Winer
The type of bass traps I'm most familiar with are either large slabs
of fiberglass or foam (broadband), or large vibrating panels over a
sealed cavity that are tuned by varying the panel mass / thickness and
cavity depth. Those are easy to understand because sound waves strike
a large surface area.
But I don't understand the absorbing mechanism for Helmholtz
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/cavity.html

When the wave length of sound is much larger than the neck and cavity
dimensions
"effective" absorption occurs only at, or within, a narrow band around
the natural frequency,
when the device is resonating. The cavity acts as a compressible spring
and
the air in the neck as an incompressible mass plug.
Absorption means less reflected sound energy from the area of neck
opening than from an identical area without the opening.
One of the mechanisms to wipe out sound energy is air flow friction
losses
inside the neck. The air plug end in the neck opening acts as a piston
radiating sound. Absorption power or efficiency depends on the area and
vibration velocity of the piston, as well as on the phase in relation
to incident wave.

Putting absorption material inside the cavity changes the effective
volume
("spring constant") of the cavity. This is seen as a change in the
natural frequency
and usually as widening the band of absorption, but usually also as
worsening
absorption at the peak frequency of the absorption band.

One method to broaden the band of absorption is to use several
resonators fine toned to different natural frequencies to cover a wider
band.

Kari Pesonen

Post by Ethan Winer
How can a small round opening or slot absorb a meaningful amount? I
understand the cavity is filled with absorbing "fuzz" to damp the
resonance. But the exposed "surface area" of the hole or slot seems
way too small. Obviously I'm missing something important, but what? Is
there a large amount of turbulence at the opening?
Thanks!
--Ethan

[Ethan Winer]

Thanks again to all who replied. I'm still not quite getting the
answer I hoped for. Let me rephrase:

One sabin of absorption is what you get from a one square foot opening
to the outdoors. This is the largest amount of absorption possible, at
least for the porous absorbers I'm familiar with such as fiberglass or
foam. But one sabin is nowhere near enough to do much to reduce peaks
and nulls and modal ringing, even in a small room.

Let's say a large cement box is built as a Helmholtz absorber, but the
slot opening at one end is only two square feet total. How is it
possible to get a useful amount of absorption from such a small
absorbing surface? I'm not concerned with the bandwidth. It could be
0.001 Hz for all I care. What I still don't see is how a Helmholtz
absorber can have more sabins than the total slot surface area could
give even if it were an acoustic "black hole," so to speak.

As for the Bag End active E-Trap, it seems they could get more than
100 percent absorption by playing a larger countering signal than what
arrives at the trap's sensing microphone. But in that case the
improvement in the room would be localized. Sort of like trying to use
EQ to flatten the response in a small room. EQ might help at one
place, but even four inches away the response can be made worse. This
is ultimately what I'm trying to determine. My contact at Bag End says
the E-Trap's improvement is not localized. But I don't see how this is
possible, unless there's an absorbing mechanism I don't understand.
This is what led me to wonder how a small slot in a Helmholtz absorber
can absorb more sabins than the slot size would imply.

--Ethan

[GregS]

Post by Ethan Winer
Thanks again to all who replied. I'm still not quite getting the
One sabin of absorption is what you get from a one square foot opening
to the outdoors. This is the largest amount of absorption possible, at
least for the porous absorbers I'm familiar with such as fiberglass or
foam. But one sabin is nowhere near enough to do much to reduce peaks
and nulls and modal ringing, even in a small room.
Let's say a large cement box is built as a Helmholtz absorber, but the
slot opening at one end is only two square feet total. How is it
possible to get a useful amount of absorption from such a small
absorbing surface? I'm not concerned with the bandwidth. It could be
0.001 Hz for all I care. What I still don't see is how a Helmholtz
absorber can have more sabins than the total slot surface area could
give even if it were an acoustic "black hole," so to speak.
As for the Bag End active E-Trap, it seems they could get more than
100 percent absorption by playing a larger countering signal than what
arrives at the trap's sensing microphone. But in that case the
improvement in the room would be localized. Sort of like trying to use
EQ to flatten the response in a small room. EQ might help at one
place, but even four inches away the response can be made worse. This
is ultimately what I'm trying to determine. My contact at Bag End says
the E-Trap's improvement is not localized. But I don't see how this is
possible, unless there's an absorbing mechanism I don't understand.
This is what led me to wonder how a small slot in a Helmholtz absorber
can absorb more sabins than the slot size would imply.

Its acting just like any other bass absorber. If padding the corners does not
help, why do it ? It does help. Its just more efficient, but thats not
good if it removes the sound alltogether.

greg

[Szczepan Bialek]

Post by Ethan Winer
As for the Bag End active E-Trap, it seems they could get more than
100 percent absorption by playing a larger countering signal than what
arrives at the trap's sensing microphone. But in that case the
improvement in the room would be localized. Sort of like trying to use
EQ to flatten the response in a small room. EQ might help at one
place, but even four inches away the response can be made worse. This
is ultimately what I'm trying to determine. My contact at Bag End says
the E-Trap's improvement is not localized.

They probably use the directinal pattern of the dipole:
http://paws.kettering.edu/~drussell/Demos/rad2/mdq.html
"The regions where sound is cancelled shows up along the diagonals (where
the grid motion is almost zero). Furthermore, there is a 180o phase
difference between the horizontal and vertical wavefronts."

Post by Ethan Winer
But I don't see how this is
possible, unless there's an absorbing mechanism I don't understand.

It is interference the waves from many sources. They should be properly
oriented to the room.

Post by Ethan Winer
This is what led me to wonder how a small slot in a Helmholtz absorber
can absorb more sabins than the slot size would imply.

It acts as the source?
S*

[Answerman]

Post by Ethan Winer
Thanks again to all who replied. I'm still not quite getting the
One sabin of absorption is what you get from a one square foot opening
to the outdoors. This is the largest amount of absorption possible, at
least for the porous absorbers I'm familiar with such as fiberglass or
foam. But one sabin is nowhere near enough to do much to reduce peaks
and nulls and modal ringing, even in a small room.
Let's say a large cement box is built as a Helmholtz absorber, but the
slot opening at one end is only two square feet total. How is it
possible to get a useful amount of absorption from such a small
absorbing surface? I'm not concerned with the bandwidth. It could be
0.001 Hz for all I care. What I still don't see is how a Helmholtz
absorber can have more sabins than the total slot surface area could
give even if it were an acoustic "black hole," so to speak.
As for the Bag End active E-Trap, it seems they could get more than
100 percent absorption by playing a larger countering signal than what
arrives at the trap's sensing microphone. But in that case the
improvement in the room would be localized. Sort of like trying to use
EQ to flatten the response in a small room. EQ might help at one
place, but even four inches away the response can be made worse. This
is ultimately what I'm trying to determine. My contact at Bag End says
the E-Trap's improvement is not localized. But I don't see how this is
possible, unless there's an absorbing mechanism I don't understand.
This is what led me to wonder how a small slot in a Helmholtz absorber
can absorb more sabins than the slot size would imply.
--Ethan

I think that part of the confusion is that you are not looking at this
prblem from the proper perspective. For whatever reason you seem to be
putting the cart before the horse and are looking for an intuitive
explanation based on the physical area of the opening of a single
Helmholtz resonator. A different approach would be to consider the
problem from a reverberation time perspective. For example, you have a
room mode with a reveberation time t1 and you want t2. The desired
reduction in reverberation time for that mode determines the amount of
absorption in sabins that is requrired. If you want to translate that
into panel area, you need to know how many sabins of absorbption a
perforated Helmholtz panel absorber produces per unit area. I don't know
the latter relationship, but it appears that the problem has been worked
out.

http://www.springerlink.com/content/yu371741164861k3/

[Tony]

Post by Ethan Winer
How can a small round opening or slot absorb a meaningful amount?

It's not easy to give an answer to this that's intuitively convincing, but
it does work. Various text books have mathematical derivations of the
"absorption cross-section" i.e. the effective area of 100% absorption. For
a Helmholtz absorber optimised for maximum absorption, this works out as the
square of the wavelength divided by (2 x PI). As you see this is
independent of the neck area! It looks as if this can be very large for low
frequencies, but in this optimised condition the bandwidth is very narrow.
So for a useful absorber the absorption cross-section would normally be less
unless you are only trying to absorb a very narrow frequency band.

[Szczepan Bia³ek]

Post by Tony

Post by Ethan Winer
How can a small round opening or slot absorb a meaningful amount?

It's not easy to give an answer to this that's intuitively convincing, but
it does work. Various text books have of the mathematical derivations
"absorption cross-section" i.e. the effective area of 100% absorption.
For a Helmholtz absorber optimised for maximum absorption, this works out
as the square of the wavelength divided by (2 x PI). As you see this is
independent of the neck area! It looks as if this can be very large for
low frequencies, but in this optimised condition the bandwidth is very
narrow. So for a useful absorber the absorption cross-section would
normally be less unless you are only trying to absorb a very narrow
frequency band.

Here no mathematical derivations:
http://www.audioholics.com/education/acoustics-principles/helmholtz-resonant-absorber/helmholtz-resonant-absorber-page-2

We can see that the front cover is large and only the round opening is
small.
It is simply the sound source activated by the primary source. The front
cover radiate. The violin has also the small opening and the front cover.

The active E-Traps are also acivated by the primary source.

The rest is the interference.
S*

[Ethan Winer]

Post by Szczepan Bia³ek
http://www.audioholics.com/education/acoustics-principles/helmholtz-resonant-absorber/helmholtz-resonant-absorber-page-2
We can see that the front cover is large and only the round opening is small.

Yes, exactly - the opening seems way too small to absorb a useful
amount for any room, even if the hole area absorbs 100 percent of the
sound.

Since it's clear the larger body of the trap is not absorbing, there
are only two explanations I can imagine:

1) There's a high amount of turbulence at the opening, making the
opening equivalent to one much larger.

2) These traps don't really work very well, which is why they're not
that common. (Ignoring that tuned absorbers are not usually a good
choice anyway for most rooms.)

--Ethan

[Don Pearce]

On Sat, 14 Nov 2009 10:10:09 -0800 (PST), Ethan Winer

Post by Ethan Winer

Post by Szczepan Bia³ek
http://www.audioholics.com/education/acoustics-principles/helmholtz-resonant-absorber/helmholtz-resonant-absorber-page-2
We can see that the front cover is large and only the round opening is small.

Yes, exactly - the opening seems way too small to absorb a useful
amount for any room, even if the hole area absorbs 100 percent of the
sound.
Since it's clear the larger body of the trap is not absorbing, there
1) There's a high amount of turbulence at the opening, making the
opening equivalent to one much larger.
2) These traps don't really work very well, which is why they're not
that common. (Ignoring that tuned absorbers are not usually a good
choice anyway for most rooms.)
--Ethan

I really don't see much value in tuned traps. Broadband ones generally
do a better job anyway at the problem frequency. At other frequencies
where the room damping is already adequate they won't make a
significant difference.

Most room problems are in any case modal, which although in any given
location they appear narrow band, that frequency changes as you move
about the room, so a broadband solution is in any case needed for
these.

Broadband traps have the added advantage that it is easy to define the
absorption factor to set the resulting T60.

d

[Tony]

Post by Szczepan Bia³ek
http://www.audioholics.com/education/acoustics-principles/helmholtz-resonant-absorber/helmholtz-resonant-absorber-page-2
We can see that the front cover is large and only the round opening is
small.
It is simply the sound source activated by the primary source. The front
cover radiate. The violin has also the small opening and the front cover.

The front cover of a Helmholtz resonator does not radiate or absorb. The
absorption is done by the hole. The absorber could have a spherical body
buried in the wall and it would work just the same.
I had a look through my files and found a report of a BBC studio treated
mainly with Helmholtz resonators in 1952. The BBC did some lab experiments
and found that the maximum absorption cross-section was considerably less
than the theoretical figure that I gave before, which is the one given in
textbooks. But it was still a great deal more than the area of the hole.
They found that in their test room they got only about one-third of the
absorption that I quoted i.e. the square of the wavelength divided by (2 x
PI). But it was still dependent on the wavelength - not the area of the
hole.
The BBC used this experimental data to treat a large orchestral music studio
that was being newly built. The interpretation of the finished results is
rather complicated, as often happens in large room because of structural
absorption and other factors. But their main problem was not too little
absorption but too much, so they had to block up some of the absorbers. (An
advantage of Helmholtz absorbers is that it is very easy to turn them off if
you want to.)
Reference: BBC Research Department Report no. B.052 (1952/20).

[Ethan Winer]

The absorber could have a spherical body buried in the wall and it would work just the same.

That's what I figured too.

The BBC did some lab experiments and found that the maximum absorption cross-section was considerably less than the theoretical figure that I gave before, which is the one given in textbooks. But it was still a great deal more than the area of the hole.

This is the crux of my question. How is it possible to have more than
100 percent absorption? The only explanation I can imagine is there's
increased turbulence at the hole, and the "negative waves" coming back
out of the hole are more than 100 percent.

Now, the next question is, does "more than 100 percent" mean the
improvement in peaks and nulls and ringing in the room are localized?
Where room EQ can make things worse in some places, broadband
absorption improves the response and decay time everywhere in the
room. So are the improvements from a Helmholtz absorber also
localized?

--Ethan

[Szczepan Bialek]

Post by Ethan Winer

The absorber could have a spherical body buried in the wall and it would
work just the same.

That's what I figured too.

The BBC did some lab experiments and found that the maximum absorption
cross-section was considerably less than the theoretical figure that I
gave before, which is the one given in textbooks. But it was still a
great deal more than the area of the hole.

This is the crux of my question. How is it possible to have more than
100 percent absorption? The only explanation I can imagine is there's
increased turbulence at the hole, and the "negative waves" coming back
out of the hole are more than 100 percent.

When a sonnd wave reflects from a hard surface the pressure is doubled. See
the Kundt's tube.
So the "negative waves" coming back out of the hole are close to 200 %.

Post by Ethan Winer
Now, the next question is, does "more than 100 percent" mean the
improvement in peaks and nulls and ringing in the room are localized?
Where room EQ can make things worse in some places, broadband
absorption improves the response and decay time everywhere in the
room. So are the improvements from a Helmholtz absorber also
localized?

Probably simmilar like the E-traps. For the both the interference works.
S*

[Answerman]

Post by Ethan Winer

The absorber could have a spherical body buried in the wall and it
would work just the same.

That's what I figured too.

The BBC did some lab experiments and found that the maximum
absorption cross-section was considerably less than the theoretical
figure that I gave before, which is the one given in textbooks. But
it was still a great deal more than the area of the hole.

This is the crux of my question. How is it possible to have more than
100 percent absorption? The only explanation I can imagine is there's
increased turbulence at the hole, and the "negative waves" coming back
out of the hole are more than 100 percent.
Now, the next question is, does "more than 100 percent" mean the
improvement in peaks and nulls and ringing in the room are localized?
Where room EQ can make things worse in some places, broadband
absorption improves the response and decay time everywhere in the
room. So are the improvements from a Helmholtz absorber also
localized?
--Ethan

The reason that you are confused is because you are thinking in terms of
percent absorption without either defining or thinking about what you
really mean by 100% absorption.

The proper path toward underatnding this this apparent paradox is to
consider absorbed power which depends on "absorption area" which is
different than the physical area. The absorbed power equals the incident
intensity (incident power per unit physical area) times the absorption
area. When one calculates the power that is absorbed by a reasonator
having an area A at the opening, the result is that the absorption area can
easily be two orders of magnitude (100x) greater than the physical area A
of the opening.

http://books.google.com/books?id=ysznrLBo5OMC&pg=PA109&lpg=PA109
&dq=sound++absorption+cross-section+helmholtz&source=bl&ots=cPbujMkwV3
&sig=mIerdI4QZk892gjJLxmmQAy90xs&hl=en&ei=gWcAS9vSM4ncsgPe_ISICw&sa=X&oi=bo
ok_result&ct=result&resnum=10&ved=0CC8Q6AEwCQ#v=onepage&q=sound%20%
20absorption%20cross-section%20helmholtz&f=false

[Ethan Winer]

Post by Answerman
The reason that you are confused is because you are thinking in terms of
percent absorption without either defining or thinking about what you
really mean by 100% absorption.

That's why earlier I reiterated the definition of one sabin as a one
square foot opening to the outdoors.

Post by Answerman
http://books.google.com/books?id=ysznrLBo5OMC&pg=PA109&lpg=PA109
&dq=sound++absorption+cross-section+helmholtz&source=bl&ots=cPbujMkwV3
&sig=mIerdI4QZk892gjJLxmmQAy90xs&hl=en&ei=gWcAS9vSM4ncsgPe_ISICw&sa=X&oi=bo
ok_result&ct=result&resnum=10&ved=0CC8Q6AEwCQ#v=onepage&q=sound%20%
20absorption%20cross-section%20helmholtz&f=false

Thanks, that looks useful, even if the math is a bit over my head. :-

Post by Answerman
)

--Ethan

[Answerman]

Post by Ethan Winer

Post by Answerman
The reason that you are confused is because you are thinking in terms
of percent absorption without either defining or thinking about what
you really mean by 100% absorption.

That's why earlier I reiterated the definition of one sabin as a one
square foot opening to the outdoors.

I understand that, but as the reference illustrates, the definition of one
sabin has nothing to do with the calculation of the absorption that is
provided by a Helmholtz resonator having an opening with physical area A.

Post by Ethan Winer

Post by Answerman
http://books.google.com/books?id=ysznrLBo5OMC&pg=PA109&lpg=PA109
&dq=sound++absorption+cross-section+helmholtz&source=bl&ots=cPbujMkwV3
&sig=mIerdI4QZk892gjJLxmmQAy90xs&hl=en&ei=gWcAS9vSM4ncsgPe_ISICw&sa=X&
oi=bo
ok_result&ct=result&resnum=10&ved=0CC8Q6AEwCQ#v=onepage&q=sound%20%
20absorption%20cross-section%20helmholtz&f=false

Thanks, that looks useful, even if the math is a bit over my head. :-

Post by Answerman
)

Based on the table of contents, this looks like an excellent reference
book. I just ordered a new copy from Barnes & Noble for $54(US) with free
shipping within the USA.

[Don Pearce]

Post by Answerman
I understand that, but as the reference illustrates, the definition of one
sabin has nothing to do with the calculation of the absorption that is
provided by a Helmholtz resonator having an opening with physical area A.

Why should a Helmholz resonator absorb anything? It is a resonant
structure which, if perfect, give back every scrap of energy it
receives. There must be some kind of dissipating mechanism associated
with it in order for it to absorb. If that dissipation is large, the
structure will cease to be particularly resonant. Resonator Q is
defined as the energy stored in one cycle divided by the energy
dissipated in one cycle. Can anyone offer clues regarding this aspect?

d

[Answerman]

Post by Don Pearce

Post by Answerman
I understand that, but as the reference illustrates, the definition of
one sabin has nothing to do with the calculation of the absorption
that is provided by a Helmholtz resonator having an opening with
physical area A.

Why should a Helmholz resonator absorb anything? It is a resonant
structure which, if perfect, give back every scrap of energy it
receives. There must be some kind of dissipating mechanism associated
with it in order for it to absorb. If that dissipation is large, the
structure will cease to be particularly resonant. Resonator Q is
defined as the energy stored in one cycle divided by the energy
dissipated in one cycle. Can anyone offer clues regarding this aspect?

The real part of the input impedance is determined primarily by viscous
losses in the tube. At resonance, it's the real part that provides the
dissipation. By trading off cross-sectional area and length, it is
possible manipulate the real part while keeping the acoustic mass and
resonant frequency the same. Further manipulation of the real part is also
possible by adding resistance in the tube as well as the volume.

[Don Pearce]

Post by Answerman

Post by Don Pearce

Post by Answerman
I understand that, but as the reference illustrates, the definition of
one sabin has nothing to do with the calculation of the absorption
that is provided by a Helmholtz resonator having an opening with
physical area A.

Why should a Helmholz resonator absorb anything? It is a resonant
structure which, if perfect, give back every scrap of energy it
receives. There must be some kind of dissipating mechanism associated
with it in order for it to absorb. If that dissipation is large, the
structure will cease to be particularly resonant. Resonator Q is
defined as the energy stored in one cycle divided by the energy
dissipated in one cycle. Can anyone offer clues regarding this aspect?

The real part of the input impedance is determined primarily by viscous
losses in the tube. At resonance, it's the real part that provides the
dissipation. By trading off cross-sectional area and length, it is
possible manipulate the real part while keeping the acoustic mass and
resonant frequency the same. Further manipulation of the real part is also
possible by adding resistance in the tube as well as the volume.

So to increase the absorption, you must make the port narrower. But
surely that also diminishes the effective area of absorption in the
room? So the net increase in absorption remains largely unchanged.

d

[Answerman]

Post by Don Pearce

Post by Answerman

Post by Don Pearce

Post by Answerman
I understand that, but as the reference illustrates, the definition
of one sabin has nothing to do with the calculation of the
absorption that is provided by a Helmholtz resonator having an
opening with physical area A.

Why should a Helmholz resonator absorb anything? It is a resonant
structure which, if perfect, give back every scrap of energy it
receives. There must be some kind of dissipating mechanism
associated with it in order for it to absorb. If that dissipation is
large, the structure will cease to be particularly resonant.
Resonator Q is defined as the energy stored in one cycle divided by
the energy dissipated in one cycle. Can anyone offer clues regarding
this aspect?

The real part of the input impedance is determined primarily by
viscous losses in the tube. At resonance, it's the real part that
provides the dissipation. By trading off cross-sectional area and
length, it is possible manipulate the real part while keeping the
acoustic mass and resonant frequency the same. Further manipulation
of the real part is also possible by adding resistance in the tube as
well as the volume.

So to increase the absorption, you must make the port narrower. But
surely that also diminishes the effective area of absorption in the
room? So the net increase in absorption remains largely unchanged.

That's probably why ports are large and the required real part is provide
by added resistance in either the tube or the volume.

[Don Pearce]

Post by Answerman

Post by Don Pearce

Post by Answerman

Post by Don Pearce

Post by Answerman
I understand that, but as the reference illustrates, the definition
of one sabin has nothing to do with the calculation of the
absorption that is provided by a Helmholtz resonator having an
opening with physical area A.

Why should a Helmholz resonator absorb anything? It is a resonant
structure which, if perfect, give back every scrap of energy it
receives. There must be some kind of dissipating mechanism
associated with it in order for it to absorb. If that dissipation is
large, the structure will cease to be particularly resonant.
Resonator Q is defined as the energy stored in one cycle divided by
the energy dissipated in one cycle. Can anyone offer clues regarding
this aspect?

The real part of the input impedance is determined primarily by
viscous losses in the tube. At resonance, it's the real part that
provides the dissipation. By trading off cross-sectional area and
length, it is possible manipulate the real part while keeping the
acoustic mass and resonant frequency the same. Further manipulation
of the real part is also possible by adding resistance in the tube as
well as the volume.

So to increase the absorption, you must make the port narrower. But
surely that also diminishes the effective area of absorption in the
room? So the net increase in absorption remains largely unchanged.

That's probably why ports are large and the required real part is provide
by added resistance in either the tube or the volume.

That makes sense. Looking in that book, I couldn't find a formula for
the effective area of absorption, though.

d

[Answerman]

Post by Don Pearce

Post by Answerman

Post by Don Pearce

Post by Answerman

Post by Don Pearce

Post by Answerman
I understand that, but as the reference illustrates, the
definition of one sabin has nothing to do with the calculation of
the absorption that is provided by a Helmholtz resonator having an
opening with physical area A.

Why should a Helmholz resonator absorb anything? It is a resonant
structure which, if perfect, give back every scrap of energy it
receives. There must be some kind of dissipating mechanism
associated with it in order for it to absorb. If that dissipation
is large, the structure will cease to be particularly resonant.
Resonator Q is defined as the energy stored in one cycle divided
by the energy dissipated in one cycle. Can anyone offer clues
regarding this aspect?

The real part of the input impedance is determined primarily by
viscous losses in the tube. At resonance, it's the real part that
provides the dissipation. By trading off cross-sectional area and
length, it is possible manipulate the real part while keeping the
acoustic mass and resonant frequency the same. Further manipulation
of the real part is also possible by adding resistance in the tube
as well as the volume.

So to increase the absorption, you must make the port narrower. But
surely that also diminishes the effective area of absorption in the
room? So the net increase in absorption remains largely unchanged.

That's probably why ports are large and the required real part is
provide by added resistance in either the tube or the volume.

That makes sense. Looking in that book, I couldn't find a formula for
the effective area of absorption, though.
d

Page 108, equation (4.2).

[Answerman]

Post by Don Pearce

Post by Answerman

Post by Don Pearce

Post by Answerman
I understand that, but as the reference illustrates, the definition
of one sabin has nothing to do with the calculation of the
absorption that is provided by a Helmholtz resonator having an
opening with physical area A.

Why should a Helmholz resonator absorb anything? It is a resonant
structure which, if perfect, give back every scrap of energy it
receives. There must be some kind of dissipating mechanism
associated with it in order for it to absorb. If that dissipation is
large, the structure will cease to be particularly resonant.
Resonator Q is defined as the energy stored in one cycle divided by
the energy dissipated in one cycle. Can anyone offer clues regarding
this aspect?

The real part of the input impedance is determined primarily by
viscous losses in the tube. At resonance, it's the real part that
provides the dissipation. By trading off cross-sectional area and
length, it is possible manipulate the real part while keeping the
acoustic mass and resonant frequency the same. Further manipulation
of the real part is also possible by adding resistance in the tube as
well as the volume.

So to increase the absorption, you must make the port narrower.

Not necssarily. A real part that is too high is just as bad as a real
part that is too low, as both will result in an impedance mismatch and in
wave reflection rather than optimum wave (transmission) absorption.

[Don Pearce]

Post by Answerman

Post by Don Pearce

Post by Answerman

Post by Don Pearce

Post by Answerman
I understand that, but as the reference illustrates, the definition
of one sabin has nothing to do with the calculation of the
absorption that is provided by a Helmholtz resonator having an
opening with physical area A.

Why should a Helmholz resonator absorb anything? It is a resonant
structure which, if perfect, give back every scrap of energy it
receives. There must be some kind of dissipating mechanism
associated with it in order for it to absorb. If that dissipation is
large, the structure will cease to be particularly resonant.
Resonator Q is defined as the energy stored in one cycle divided by
the energy dissipated in one cycle. Can anyone offer clues regarding
this aspect?

The real part of the input impedance is determined primarily by
viscous losses in the tube. At resonance, it's the real part that
provides the dissipation. By trading off cross-sectional area and
length, it is possible manipulate the real part while keeping the
acoustic mass and resonant frequency the same. Further manipulation
of the real part is also possible by adding resistance in the tube as
well as the volume.

So to increase the absorption, you must make the port narrower.

Not necssarily. A real part that is too high is just as bad as a real
part that is too low, as both will result in an impedance mismatch and in
wave reflection rather than optimum wave (transmission) absorption.

I wasn't really thinking of that kind of extreme case.

d

[Answerman]

Post by Don Pearce

Post by Answerman

Post by Don Pearce

Post by Answerman

Post by Don Pearce

Post by Answerman
I understand that, but as the reference illustrates, the
definition of one sabin has nothing to do with the calculation of
the absorption that is provided by a Helmholtz resonator having an
opening with physical area A.

Why should a Helmholz resonator absorb anything? It is a resonant
structure which, if perfect, give back every scrap of energy it
receives. There must be some kind of dissipating mechanism
associated with it in order for it to absorb. If that dissipation
is large, the structure will cease to be particularly resonant.
Resonator Q is defined as the energy stored in one cycle divided
by the energy dissipated in one cycle. Can anyone offer clues
regarding this aspect?

The real part of the input impedance is determined primarily by
viscous losses in the tube. At resonance, it's the real part that
provides the dissipation. By trading off cross-sectional area and
length, it is possible manipulate the real part while keeping the
acoustic mass and resonant frequency the same. Further manipulation
of the real part is also possible by adding resistance in the tube
as well as the volume.

So to increase the absorption, you must make the port narrower.

Not necssarily. A real part that is too high is just as bad as a real
part that is too low, as both will result in an impedance mismatch and
in wave reflection rather than optimum wave (transmission) absorption.

I wasn't really thinking of that kind of extreme case.

You had several exchanges with Bob Cain in rec.audio.pro. He seems that
he's dropped off the face of the earth. What if anything do you know?

[Don Pearce]

Post by Answerman

Post by Don Pearce

Post by Answerman

Post by Don Pearce

Post by Answerman

Post by Don Pearce

Post by Answerman
I understand that, but as the reference illustrates, the
definition of one sabin has nothing to do with the calculation of
the absorption that is provided by a Helmholtz resonator having an
opening with physical area A.

Why should a Helmholz resonator absorb anything? It is a resonant
structure which, if perfect, give back every scrap of energy it
receives. There must be some kind of dissipating mechanism
associated with it in order for it to absorb. If that dissipation
is large, the structure will cease to be particularly resonant.
Resonator Q is defined as the energy stored in one cycle divided
by the energy dissipated in one cycle. Can anyone offer clues
regarding this aspect?

The real part of the input impedance is determined primarily by
viscous losses in the tube. At resonance, it's the real part that
provides the dissipation. By trading off cross-sectional area and
length, it is possible manipulate the real part while keeping the
acoustic mass and resonant frequency the same. Further manipulation
of the real part is also possible by adding resistance in the tube
as well as the volume.

So to increase the absorption, you must make the port narrower.

Not necssarily. A real part that is too high is just as bad as a real
part that is too low, as both will result in an impedance mismatch and
in wave reflection rather than optimum wave (transmission) absorption.

I wasn't really thinking of that kind of extreme case.

You had several exchanges with Bob Cain in rec.audio.pro. He seems that
he's dropped off the face of the earth. What if anything do you know?

Nothing, I'm afraid. He vanished.

d

[Szczepan Bia³ek]

Post by Tony

Post by Szczepan Bia³ek
http://www.audioholics.com/education/acoustics-principles/helmholtz-resonant-absorber/helmholtz-resonant-absorber-page-2
We can see that the front cover is large and only the round opening is
small.
It is simply the sound source activated by the primary source. The front
cover radiate. The violin has also the small opening and the front cover.

The front cover of a Helmholtz resonator does not radiate or absorb. The
absorption is done by the hole. The absorber could have a spherical body
buried in the wall and it would work just the same.

It seems to me that a Helmholtz resonator absorb and radiate like the all
instruments:
[edit]" Acoustic instruments
See also: Musical acoustics
It is sometimes said that the sounding board or soundbox "amplifies" the
sound of the strings. Technically speaking, no amplification occurs, because
all of the energy to produce sound comes from the vibrating string. What
really happens is that the sounding board of the instrument provides a
larger surface area to create sound waves than that of the string. A larger
vibrating surface moves more air, hence produces a louder sound."

It absorbs the primary sound and "amplifies" it. But the amplified sound
must be in such phase that the interference cancel the both. The hole
absorbs and radiate. In such approach the area of the hole does not matter
(of course the chamber must be tuned).

Post by Tony
I had a look through my files and found a report of a BBC studio treated
mainly with Helmholtz resonators in 1952. The BBC did some lab experiments
and found that the maximum absorption cross-section was considerably less
than the theoretical figure that I gave before, which is the one given in
textbooks. But it was still a great deal more than the area of the hole.
They found that in their test room they got only about one-third of the
absorption that I quoted i.e. the square of the wavelength divided by (2 x
PI). But it was still dependent on the wavelength - not the area of the
hole.
The BBC used this experimental data to treat a large orchestral music studio
that was being newly built. The interpretation of the finished results is
rather complicated, as often happens in large room because of structural
absorption and other factors. But their main problem was not too little
absorption but too much, so they had to block up some of the absorbers.
(An
advantage of Helmholtz absorbers is that it is very easy to turn them off if
you want to.)
Reference: BBC Research Department Report no. B.052 (1952/20).

It seems to me that Helmholtz resonators do not absorb but cancel by
interference.

S*

[Tony]

Post by Szczepan Bia³ek
It seems to me that Helmholtz resonators do not absorb but cancel by
interference.

I am beginning to wonder if this is a troll. But in case you really don't
know, there is a fundamental difference between absorption and cancellation.
With absorption, the energy in the sound wave is converted to heat. With
cancellation it is not, so there is no loss of energy. So generally, if a
sound is cancelled in one place, it will be increased in another place.
For an explanation of the "magic" absorption that seems more than is
possible from the hole area, look at Fahy, "Foundations of Engineering
Acoustics" page 175.
My last word to you is: diffraction.

[Answerman]

Post by Tony

Post by Szczepan Bia³ek
It seems to me that Helmholtz resonators do not absorb but cancel by
interference.

I am beginning to wonder if this is a troll. But in case you really
don't know, there is a fundamental difference between absorption and
cancellation. With absorption, the energy in the sound wave is
converted to heat. With cancellation it is not, so there is no loss
of energy. So generally, if a sound is cancelled in one place, it
will be increased in another place. For an explanation of the "magic"
absorption that seems more than is possible from the hole area, look
at Fahy, "Foundations of Engineering Acoustics" page 175.
My last word to you is: diffraction.

FWIW, I stoped wondering after his first nonsensical post.

[Szczepan Bia³ek]

Post by Answerman

Post by Tony

Post by Szczepan Bia³ek
It seems to me that Helmholtz resonators do not absorb but cancel by
interference.

I am beginning to wonder if this is a troll. But in case you really
don't know, there is a fundamental difference between absorption and
cancellation. With absorption, the energy in the sound wave is
converted to heat. With cancellation it is not, so there is no loss
of energy. So generally, if a sound is cancelled in one place, it
will be increased in another place. For an explanation of the "magic"
absorption that seems more than is possible from the hole area, look
at Fahy, "Foundations of Engineering Acoustics" page 175.
My last word to you is: diffraction.

FWIW, I stoped wondering after his first nonsensical post.

After Your first post I know that You are an expert.

I am not, but I am interesting in the Helmholtz Electrodynamics. In this the
electric waves are longitudinal like the sound. In the Helmholtz laboratory
Hertz did his famous experiments. Helmoholtz was the genius in acoustic,
fluids, whirls, electricity and many others. It is impossible that he was
wrong.

Up to now I see the full analogy of the electric waves and the soud waves.
Simple question: Who is right: Heaviside or Helmholtz?

S*

[Szczepan Bia³ek]

Post by Tony

Post by Szczepan Bia³ek
It seems to me that Helmholtz resonators do not absorb but cancel by
interference.

I am beginning to wonder if this is a troll. But in case you really don't
know, there is a fundamental difference between absorption and
cancellation.
With absorption, the energy in the sound wave is converted to heat. With
cancellation it is not, so there is no loss of energy.

I know this and and I am trying to find the analogy to radio waves and
radars.

Post by Tony
So generally, if a
sound is cancelled in one place, it will be increased in another place.

In radio and acoustics people use the directional pattern of the many
sources.
The goal is to cancel in some area (the room). Where it will be increased I
do not know.

Post by Tony
For an explanation of the "magic" absorption that seems more than is
possible from the hole area, look at Fahy, "Foundations of Engineering
Acoustics" page 175.
My last word to you is: diffraction.

The all necessary information to understand are in: http://www.bagend.com/

(There's no way to post a direct link. Click Products on the left
side, then Professional Acoustics on the far right top.)


"Typically passive bass traps, since they are large, can also affect the
midrange and upper frequencies and require careful integration into a room.
Reactive absorbers such as Helmholtz resonators (HR) or quarter wave tubes
are commonly used as the most effective solution for treating the low
frequency standing waves. "

Passive traps convert the sound into heat. The reactive (like the Helmholtz
resonators) no. The active also no. How they work than?
S*

[Kari Pesonen]

Post by Tony

Post by Szczepan Bia³ek
It seems to me that Helmholtz resonators do not absorb but cancel by
interference.

I am beginning to wonder if this is a troll. But in case you really don't
know, there is a fundamental difference between absorption and cancellation.
With absorption, the energy in the sound wave is converted to heat.
With
cancellation it is not, so there is no loss of energy.

Do you think/assume that we had noise cancellation systems/device that
do not
need any external power, or do you think that, for example, electrical
amplifiers
powering cancelling sound sources (loudspeakers, vibrators) feed the
power cancelled to the mains network (net energy consumption = zero)? ;)
- -

Think, for example, a pendulum driven by a vibrator and the movement
cancelled by another vibrator: although the cancelling device stops the
movement (energy = force x distance moved in time unit = force x
velocity
= zero), we need external energy to prevent movement. Both the original
vibrator and the cancelling vibrator need external energy.

A little more complicated imaginary exercise to solve is the case where
the
sound field at/close a sound source is modified by a set of cancelling
sound sources
with such a way that the original source does not radiate sound or
radiates
less than before we switched on the cancelling system.

Kari Pesonen

[Szczepan Bia³ek]

Post by Tony

Post by Szczepan Bia³ek
It seems to me that Helmholtz resonators do not absorb but cancel by
interference.

I am beginning to wonder if this is a troll. But in case you really don't
know, there is a fundamental difference between absorption and cancellation.
With absorption, the energy in the sound wave is converted to heat. With
cancellation it is not, so there is no loss of energy.

Do you think/assume that we had noise cancellation systems/device that do
not
need any external power,

Each resonator need the external power. The power is from the wave emmited
by the primary source. The energy is very small so it takes time to achive
the full amplitude.
S*

[Ethan Winer]

Post by Szczepan Bia³ek
It seems to me that Helmholtz resonators do not absorb but cancel by
interference.

I agree with the others that this doesn't make sense. A bass trap MUST
absorb in order to reduce modal ringing decay times. Otherwise it does
only half the job.

--Ethan

[Szczepan Bialek]

Post by Szczepan Bia³ek

Post by Szczepan Bia³ek
It seems to me that Helmholtz resonators do not absorb but cancel by

interference.
I agree with the others that this doesn't make sense. A bass trap MUST

absorb

But we have the three types of the traps: passive, reactive and active.
The passive convert the sound into the heat. The reactive and active no.

Post by Szczepan Bia³ek
in order to reduce modal ringing decay times. Otherwise it does

only half the job.

"The downside is that in some places in the room, for a given note,
compression meets compression, amplifying that frequency (constructive
interference), and in other parts of the room, compression meets
rarefaction, canceling at that frequency (destructive interference). From:
http://www.realtraps.com/art_waves.htm
S*

[Peter Davis]

Post by Ethan Winer
The type of bass traps I'm most familiar with are either large slabs
of fiberglass or foam (broadband), or large vibrating panels over a
sealed cavity that are tuned by varying the panel mass / thickness and
cavity depth. Those are easy to understand because sound waves strike
a large surface area.
But I don't understand the absorbing mechanism for Helmholtz
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/cavity.html
How can a small round opening or slot absorb a meaningful amount? I
understand the cavity is filled with absorbing "fuzz" to damp the
resonance. But the exposed "surface area" of the hole or slot seems
way too small. Obviously I'm missing something important, but what? Is
there a large amount of turbulence at the opening?
Thanks!
--Ethan

It's not the size, it's the frequency that matters...that's what she said!

[Ethan Winer]

Post by Peter Davis
It's not the size, it's the frequency that matters...that's what she said!

Groan. :->)

In the mean time, Angelo Campanella was nice enough to email me off
list with further information. As I understand his comments, the waves
travel into the slots, so the depth of the cavity behind increases the
apparent "surface area" exposed to the room. Angelo also described an
electrical equivalent - the slot opening is a series inductor and the
cavity behind is a shunt capacitor, which makes sense. Thanks Angelo.

--Ethan

[a***@home]

Post by Ethan Winer
In the mean time, Angelo Campanella was nice enough to email me off
list with further information. As I understand his comments, the waves
travel into the slots, so the depth of the cavity behind increases the
apparent "surface area" exposed to the room.

Sounds like gibberish to me. Perhaps something got lost in the
translation.

Post by Ethan Winer
Angelo also described an
electrical equivalent - the slot opening is a series inductor and the
cavity behind is a shunt capacitor, which makes sense.

That is correct, but that lumped-element electrical equivalent network
model only accounts for the approximate resonant frequency of the Helmholtz
absorber. It has nothing whatsoever to do with the effective absorption
surface area.

Post by Ethan Winer
Thanks Angelo.

Yea. Thanks Algelo for sharing your input with the rest of us.

[Don Pearce]

On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer

Post by Ethan Winer

Post by Peter Davis
It's not the size, it's the frequency that matters...that's what she said!

Groan. :->)
In the mean time, Angelo Campanella was nice enough to email me off
list with further information. As I understand his comments, the waves
travel into the slots, so the depth of the cavity behind increases the
apparent "surface area" exposed to the room. Angelo also described an
electrical equivalent - the slot opening is a series inductor and the
cavity behind is a shunt capacitor, which makes sense. Thanks Angelo.
--Ethan

Think also of the Yagi TV antenna. It has about 16dB of gain over a
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.

d

[a***@home]

Post by Don Pearce
On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer

Post by Ethan Winer

Post by Peter Davis
It's not the size, it's the frequency that matters...that's what she said!

Groan. :->)
In the mean time, Angelo Campanella was nice enough to email me off
list with further information. As I understand his comments, the waves
travel into the slots, so the depth of the cavity behind increases the
apparent "surface area" exposed to the room. Angelo also described an
electrical equivalent - the slot opening is a series inductor and the
cavity behind is a shunt capacitor, which makes sense. Thanks Angelo.
--Ethan

Think also of the Yagi TV antenna. It has about 16dB of gain over a
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.
d

If a shotgun mic has 16dB of gain over an omni, does that mean that it
intercepts power from an area 40 times as great as its physical dimensions?

[Don Pearce]

Post by a***@home

Post by Don Pearce
On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer

Post by Ethan Winer

Post by Peter Davis
It's not the size, it's the frequency that matters...that's what she said!

Groan. :->)
In the mean time, Angelo Campanella was nice enough to email me off
list with further information. As I understand his comments, the waves
travel into the slots, so the depth of the cavity behind increases the
apparent "surface area" exposed to the room. Angelo also described an
electrical equivalent - the slot opening is a series inductor and the
cavity behind is a shunt capacitor, which makes sense. Thanks Angelo.
--Ethan

Think also of the Yagi TV antenna. It has about 16dB of gain over a
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.
d

If a shotgun mic has 16dB of gain over an omni, does that mean that it
intercepts power from an area 40 times as great as its physical dimensions?

Yes, it does.

d

[a***@home]

Post by Don Pearce

Post by a***@home

Post by Don Pearce
On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer

Post by Ethan Winer

Post by Peter Davis
It's not the size, it's the frequency that matters...that's what she said!

Groan. :->)
In the mean time, Angelo Campanella was nice enough to email me off
list with further information. As I understand his comments, the
waves travel into the slots, so the depth of the cavity behind
increases the apparent "surface area" exposed to the room. Angelo
also described an electrical equivalent - the slot opening is a
series inductor and the cavity behind is a shunt capacitor, which
makes sense. Thanks Angelo.
--Ethan

Think also of the Yagi TV antenna. It has about 16dB of gain over a
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.
d

If a shotgun mic has 16dB of gain over an omni, does that mean that it
intercepts power from an area 40 times as great as its physical dimensions?

Yes, it does.
d

I disagree, but that's a matter for discussion on another day.

[Don Pearce]

Post by a***@home

Post by Don Pearce

Post by a***@home

Post by Don Pearce
On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer

Post by Ethan Winer

Post by Peter Davis
It's not the size, it's the frequency that matters...that's what she said!

Groan. :->)
In the mean time, Angelo Campanella was nice enough to email me off
list with further information. As I understand his comments, the
waves travel into the slots, so the depth of the cavity behind
increases the apparent "surface area" exposed to the room. Angelo
also described an electrical equivalent - the slot opening is a
series inductor and the cavity behind is a shunt capacitor, which
makes sense. Thanks Angelo.
--Ethan

Think also of the Yagi TV antenna. It has about 16dB of gain over a
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.
d

If a shotgun mic has 16dB of gain over an omni, does that mean that it
intercepts power from an area 40 times as great as its physical dimensions?

Yes, it does.
d

I disagree, but that's a matter for discussion on another day.

I think you may be disagreeing with something I didn't say. Most
directional mics don't have gain over an omni - they have loss in
other directions., which doesn't result in added effective area. A
shotgun mic is different in that it has real directional gain..

Also think of absorbers and their Sabin rating. They can be greater
than unity because the sides go on absorbing even if they are not in
the direct line of the sound. Wavefront diffraction is a wonderful
thing.

d

[a***@home]

Post by Don Pearce

Post by a***@home

Post by Don Pearce

Post by a***@home

Post by Don Pearce
On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer

Post by Ethan Winer

Post by Peter Davis
It's not the size, it's the frequency that matters...that's what she said!

Groan. :->)
In the mean time, Angelo Campanella was nice enough to email me off
list with further information. As I understand his comments, the
waves travel into the slots, so the depth of the cavity behind
increases the apparent "surface area" exposed to the room. Angelo
also described an electrical equivalent - the slot opening is a
series inductor and the cavity behind is a shunt capacitor, which
makes sense. Thanks Angelo.
--Ethan

Think also of the Yagi TV antenna. It has about 16dB of gain over a
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.
d

If a shotgun mic has 16dB of gain over an omni, does that mean that it
intercepts power from an area 40 times as great as its physical dimensions?

Yes, it does.
d

I disagree, but that's a matter for discussion on another day.

I think you may be disagreeing with something I didn't say. Most
directional mics don't have gain over an omni - they have loss in
other directions., which doesn't result in added effective area. A
shotgun mic is different in that it has real directional gain..
Also think of absorbers and their Sabin rating. They can be greater
than unity because the sides go on absorbing even if they are not in
the direct line of the sound. Wavefront diffraction is a wonderful
thing.
d

I know what you said, and I stipulated a shotgun mic for the very reason
you just stated. There are two issues here. One is whether or not your
interpretation of the on-axis gain increase is correct. The other is
whether the analogy between a Helmholtz absorber and a Yagi antenna is
valid.

[Don Pearce]

Post by a***@home

Post by Don Pearce

Post by a***@home

Post by Don Pearce

Post by a***@home

Post by Don Pearce
On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer

Post by Ethan Winer

Post by Peter Davis
It's not the size, it's the frequency that matters...that's what she said!

Groan. :->)
In the mean time, Angelo Campanella was nice enough to email me off
list with further information. As I understand his comments, the
waves travel into the slots, so the depth of the cavity behind
increases the apparent "surface area" exposed to the room. Angelo
also described an electrical equivalent - the slot opening is a
series inductor and the cavity behind is a shunt capacitor, which
makes sense. Thanks Angelo.
--Ethan

Think also of the Yagi TV antenna. It has about 16dB of gain over a
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.
d

If a shotgun mic has 16dB of gain over an omni, does that mean that it
intercepts power from an area 40 times as great as its physical dimensions?

Yes, it does.
d

I disagree, but that's a matter for discussion on another day.

I think you may be disagreeing with something I didn't say. Most
directional mics don't have gain over an omni - they have loss in
other directions., which doesn't result in added effective area. A
shotgun mic is different in that it has real directional gain..
Also think of absorbers and their Sabin rating. They can be greater
than unity because the sides go on absorbing even if they are not in
the direct line of the sound. Wavefront diffraction is a wonderful
thing.
d

I know what you said, and I stipulated a shotgun mic for the very reason
you just stated. There are two issues here. One is whether or not your
interpretation of the on-axis gain increase is correct. The other is
whether the analogy between a Helmholtz absorber and a Yagi antenna is
valid.

My analogies - my apologies - the Helmholz absorber is incorrect, but
the Yagi is quite correct. I used this not so much as an analogy as an
illustration that the effective dimensions can be much larger than the
physical dimensions.

d

[GregS]

Post by Don Pearce

Post by a***@home

Post by Don Pearce

Post by a***@home

Post by Don Pearce
On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer

Post by Ethan Winer

Post by Peter Davis
It's not the size, it's the frequency that matters...that's what she said!

Groan. :->)
In the mean time, Angelo Campanella was nice enough to email me off
list with further information. As I understand his comments, the
waves travel into the slots, so the depth of the cavity behind
increases the apparent "surface area" exposed to the room. Angelo
also described an electrical equivalent - the slot opening is a
series inductor and the cavity behind is a shunt capacitor, which
makes sense. Thanks Angelo.
--Ethan

Think also of the Yagi TV antenna. It has about 16dB of gain over aesonant shotgum mic has little
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.
d

If a shotgun mic has 16dB of gain over an omni, does that mean that it
intercepts power from an area 40 times as great as its physical dimensions?

Yes, it does.
d

I disagree, but that's a matter for discussion on another day.

I think you may be disagreeing with something I didn't say. Most
directional mics don't have gain over an omni - they have loss in
other directions., which doesn't result in added effective area. A
shotgun mic is different in that it has real directional gain..

A resonant shotgun mic has little directionality. Its like a bunch of omnis
connected in parallel, and the diameter is small, so there
is little phase cancellaon off to the side.
greg

Post by Don Pearce
Also think of absorbers and their Sabin rating. They can be greater
than unity because the sides go on absorbing even if they are not in
the direct line of the sound. Wavefront diffraction is a wonderful
thing.
d

[Angelo Campanella]

Post by a***@home

Post by Don Pearce
On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer

Post by Ethan Winer

Post by Peter Davis
It's not the size, it's the frequency that matters...that's what she said!

Groan. :->)
In the mean time, Angelo Campanella was nice enough to email me off
list with further information. As I understand his comments, the waves
travel into the slots, so the depth of the cavity behind increases the
apparent "surface area" exposed to the room. Angelo also described an
electrical equivalent - the slot opening is a series inductor and the
cavity behind is a shunt capacitor, which makes sense. Thanks Angelo.
--Ethan

Think also of the Yagi TV antenna. It has about 16dB of gain over a
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.
d

If a shotgun mic has 16dB of gain over an omni, does that mean that it
intercepts power from an area 40 times as great as its physical dimensions?

When the JASA CD's covering JASA from 1929 forward came available, I
searched for the "shotgun microphone" paper, found it and determined that it
is a bundle of small tubes, quarter-wave resonant in the 1,000 to 2,000 Hz
range, abutting the diaphragm of a condenser microphone. That frequency
range favors human speech intelligibility.

This phenomenon is similar to the "conch shell" effect (sea shell
held by ear amplifies environmental sounds). One of our participants here,
Greg, I believe, managed to measure that effect, noting that it, too was
around 1,500 Hz range, at a level also of about 15 dB.

It's a small world, after all!

What we re seeing here is what I might call "Impedance
amplification", or simply Impedance Matching. The energy if the sound waves
about us is quite enough to stimulate our hearing cells, but it can be in
such a form that it is not immediately detectable. Sound in water vs. sound
in air is a good example of that fact. Our ear system makes use of that
fact. The hair cells are in a liquid medium of density near 1.0, while sound
critical to our existence propagates through air whose density is only
1/1,000th of that. The ear drum, and he stapes bone amount to an impedance
matching transformer to purvey sound energy out of air (low impedance) into
the oval window of the stapes into the cochlear fluid (high impedance). Like
any transformer, it has lower and upper cut off frequencies, with 3 dB
points of 500 Hz and about 5,000 Hz (less with NIPTS).

Another story: When swimming underwater to listen to the performance
of the Lubell underwater loudspeaker (http://www.lubell.com/), I found
amazing low frequency response was occurring.. I then realized that the
impedance match (terrible for the air entrained in the ear canal) was really
GOOD for bone conduction;then entire skull sphere being driven as a monopole
vibrator. This bypasses the ear canal and stapes, conducting vibrations into
the stapes itself. The 500 Hz rollover frequency does not apply.

If you ever swim in the field of an underwater loudspeaker -
typically music in a swimming pool - take note that the acoustic pressure
release near the top water surface eliminates most low frequency sound
(sounds "tinny"), while swimming underwater down to the bottom and
especially a corner, full brilliant low frequency sound is heard. This
demonstrates the coupling of water acoustic pressure to the skull (my
interpretation of the facts).

Ange




--- news://freenews.netfront.net/ - complaints: ***@netfront.net ---

[Angelo Campanella]

Post by Angelo Campanella
Another story: When swimming underwater to listen to the
performance of the Lubell underwater loudspeaker (http://www.lubell.com/),
I found amazing low frequency response was occurring.. I then realized
that the impedance match (terrible for the air entrained in the ear canal)
was really GOOD for bone conduction;then entire skull sphere being driven
as a monopole vibrator. This bypasses the ear canal and stapes, conducting
vibrations into the "stapes"

Correction: "into the cochlear fluid itself"

Post by Angelo Campanella
--- itself. The 500 Hz rollover frequency does not apply.

Ange



--- news://freenews.netfront.net/ - complaints: ***@netfront.net ---

[Ethan Winer]

Post by Don Pearce
On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer
Think also of the Yagi TV antenna. It has about 16dB of gain over a
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.

Yes, though this comes back to my original concern: Usually, when the
"gain" is increased, it comes at the expense of coverage. This is
exactly what I'm trying to get to the bottom of. If a bass trap
absorbs more than 100 percent, does that mean the benefits are more
localized than having more coverage from traps that absorb "only" 100
percent?

--Ethan

[Don Pearce]

On Thu, 31 Dec 2009 08:38:35 -0800 (PST), Ethan Winer

Post by Ethan Winer

Post by Don Pearce
On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer
Think also of the Yagi TV antenna. It has about 16dB of gain over a
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.

Yes, though this comes back to my original concern: Usually, when the
"gain" is increased, it comes at the expense of coverage. This is
exactly what I'm trying to get to the bottom of. If a bass trap
absorbs more than 100 percent, does that mean the benefits are more
localized than having more coverage from traps that absorb "only" 100
percent?
--Ethan

I understand the problem. A trap that covers a large area can be seen
to work - it absorbs the acoustic energy that covers that area. But a
relatively small opening only "sees" a small amount of the energy. My
Yagi analogy fell rather flat, although it did serve to show that a
far greater area can be absorbed than the physical size would suggest.
I'm still trying to rationalize the Helmholz case in my head, to see
if it can exploit the same effect.

d

[a***@home]

Post by Don Pearce
On Thu, 31 Dec 2009 08:38:35 -0800 (PST), Ethan Winer

Post by Ethan Winer

Post by Don Pearce
On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer
Think also of the Yagi TV antenna. It has about 16dB of gain over a
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.

Yes, though this comes back to my original concern: Usually, when the
"gain" is increased, it comes at the expense of coverage. This is
exactly what I'm trying to get to the bottom of. If a bass trap
absorbs more than 100 percent, does that mean the benefits are more
localized than having more coverage from traps that absorb "only" 100
percent?
--Ethan

I understand the problem. A trap that covers a large area can be seen
to work - it absorbs the acoustic energy that covers that area. But a
relatively small opening only "sees" a small amount of the energy. My
Yagi analogy fell rather flat, although it did serve to show that a
far greater area can be absorbed than the physical size would suggest.
I'm still trying to rationalize the Helmholz case in my head, to see
if it can exploit the same effect.
d

After doing some reading on antenna theory, it looks likes your Yagi
analogy didn't fall as short as you might think. As near as I can tell,
the gain of the Yagi is irrelevant and only serves complicate appreciation
of the analogy in terms of fundamental principles between a receiving
antenna and a resonant (Helmholtz) absorber.

I believe that the correct analogy is between an isotropic antenna which by
definition is onidirectional. and an omnidirectional vibrating surface. In
antenna theory, one refers to effective area, whereas for the sound
absorber one refers to the absorption cross-secction. However, BOTH the
effective area and the absorption cross-section are defined as the
functionally equivalent area from which the device absorbs energy from an
incident wave. Also, in BOTH cases, the equivalent area is determied by
calculating the amount of power that is absorbed from the the incident
wave. Lastly, in BOTH cases, maximum effective area occurs at the resonant
frequency and when the absorbed power equals the re-radiated power, and is
equal to the wavelength squared divided by four pi.








According to what I have read, both the antenna and the Helmholtz absorber
behave in the following manner for an incident plane wave:

1) both absorb and re-radiate incident power
2) maximum absorbed power occurs

[a***@home]

Post by Ethan Winer

Post by Don Pearce
On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer
Think also of the Yagi TV antenna. It has about 16dB of gain over a
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.

Yes, though this comes back to my original concern: Usually, when the
"gain" is increased, it comes at the expense of coverage. This is
exactly what I'm trying to get to the bottom of.

Based on my understanding of what I have read on antenna theory and the
theory of resonant absorbers, gain is a separate issue that only
complicates one's understanding of the fundamental principles involved with
the absorption of a resonant (Helmholtz) absorber.

Post by Ethan Winer
If a bass trap
absorbs more than 100 percent, does that mean the benefits are more
localized than having more coverage from traps that absorb "only" 100
percent?
--Ethan

The maximum amount of power that a Helmholtz aborber can absorb is 50% of
the power of the incident wave. That occurs when it absorbs as much power
as it re-radiates, which occurs at resonance and when the input impedance
and its radiation impedance of the absorber are equal.

[GregS]

Post by Don Pearce
On Mon, 30 Nov 2009 10:22:29 -0800 (PST), Ethan Winer

Post by Ethan Winer

Post by Peter Davis
It's not the size, it's the frequency that matters...that's what she said!

Groan. :->)
In the mean time, Angelo Campanella was nice enough to email me off
list with further information. As I understand his comments, the waves
travel into the slots, so the depth of the cavity behind increases the
apparent "surface area" exposed to the room. Angelo also described an
electrical equivalent - the slot opening is a series inductor and the
cavity behind is a shunt capacitor, which makes sense. Thanks Angelo.
--Ethan

Think also of the Yagi TV antenna. It has about 16dB of gain over a
simple dipole. That means that it intercepts power from an area 40
times as great as its physical dimensions would suggest is possible.
This is a similar effect.

It makes power by zooming in on the target with phase additions.. You can better do this
with a dish. A two element Yagi has maybe 3 dB gain.

greg