Max sound force /power from a driver, dB, effect

This is what I believe (like a guess):

Ex. In a hifi-store I listened to a 2-way speaker (2 x 6½" + 1 tweeter)
with reasonable sensitivity ~91dB 1W/1m. The amp was a large branded amp,
something like 250W.

At 3m the sound power was perhaps 100dB and above that the sound
charachteristic was simply ****.

My guess was that the cone area MUST be bigger to "drive the amount of air".
1. Is the area important (if amp and sensitivity is enough) ???
2. Can the sound max strength be calculated for a mid at, lets say 1Khz?

Let's say we intend to build a 3-way with 2x 10"woofers.
# The question is if 1x 4" mid (0,8-5khz) will be sufficient to MATCH the
woofers? We don't want 120$ woofers to be limited by the 20$ mid that chokes
in its lower freq, do we!?


Morgan O.

On Sun, 25 Sep 2005 08:12:14 GMT, Morgan Ohlson wrote:

Max sound force /power from a driver, dB, effect

This is what I believe (like a guess):

Ex. In a hifi-store I listened to a 2-way speaker (2 x 6½" + 1 tweeter)
with reasonable sensitivity ~91dB 1W/1m. The amp was a large branded amp,
something like 250W.

At 3m the sound power was perhaps 100dB and above that the sound
charachteristic was simply ****.

My guess was that the cone area MUST be bigger to "drive the amount of air".
1. Is the area important (if amp and sensitivity is enough) ???
2. Can the sound max strength be calculated for a mid at, lets say 1Khz?

Let's say we intend to build a 3-way with 2x 10"woofers.
# The question is if 1x 4" mid (0,8-5khz) will be sufficient to MATCH the
woofers? We don't want 120$ woofers to be limited by the 20$ mid that chokes
in its lower freq, do we!?


Morgan O.

The maximum sound level is related to the volume of air that the speaker
can shift. To know this you need two things - the area of the cone and the
maximum linear excursion. Both of these are inculded in Thiel and Small's
set of parameters. With these and the basic sensitivity of the driver units
you can calculate the maximium clean sound level you should be able to get
from any given speaker design.

There are plenty of shareware programmes out there for designing boxes
according to a relatively simple set of equations, but there is no single
answer to the question, because a speaker system can have a variety of
response types (which you need to choose among), and the answers come out
differently for each.

If you need a detailed explanation of this, Dick Pierce has made plenty of
posts going through the theory. Search the archives, and I'm sure you will
find something that explains it all for you.

d

Don Pearce wrote:
On Sun, 25 Sep 2005 08:12:14 GMT, Morgan Ohlson wrote:
Max sound force /power from a driver, dB, effect

This is what I believe (like a guess):

Ex. In a hifi-store I listened to a 2-way speaker (2 x 6=BD"
+ 1 tweeter) with reasonable sensitivity ~91dB 1W/1m. The
amp was a large branded amp, something like 250W.

At 3m the sound power was perhaps 100dB and above that the
sound charachteristic was simply ****.

My guess was that the cone area MUST be bigger to "drive the
amount of air". 1. Is the area important (if amp and
sensitivity is enough) ???

2. Can the sound max strength be calculated for a mid at,
lets say 1Khz?

Let's say we intend to build a 3-way with 2x 10"woofers.
3. The question is if 1x 4" mid (0,8-5khz) will be sufficient
to MATCH the woofers? We don't want 120$ woofers to be limited
by the 20$ mid that chokes in its lower freq, do we!?

The maximum sound level is related to the volume of air that
the speaker can shift. To know this you need two things - the
area of the cone and the maximum linear excursion.

Almost, but not quite. The crucial missing parameter is
frequency.

The acoustic power output of a vibrating diaphragm is:

Pa =3D p c^2 Sd^2 x^2 w^4

whe

Pa is acoustic power in watts
p is the density of air, about 1.18 kg/m^2
c is the veloxity of sound, about 342 m/s
Sd is the emissive area of the diaphragm in m^2
x is the excursion of the diaphragm in m
and w is radian frequency, i.e., 2 pi f
Note: assumed is that the diaphragm is mounted on a baffle
substantially larger than the wavelength, and itself is
substantially smaller than a wavelength.

So, yes, it suggests that all other things being equal, i.e.,
same frequency and same excursion, power output is proportional
to the square of the diaphragm area.

To the specific questions the original poster asked:

1. Is the area important (if amp and sensitivity is enough)

Yes, but the maximum linear excursion (Xmax) and the frequency
being produced are equally important, indeed, frequency is
"more" important since power output goes as frequency to the
fourth power, while power goes only as the square of area and
excursion.

2. Can the sound max strength be calculated for a mid at,
lets say 1Khz?

Most assuredly, it can. Since you have already determined one
parameter (frequency) all you need to know are the other two.
Say your midrange is a nominal 4" unit, whose typical Sd or
emissive area is about 0.008m^2. And let's say that its linear
excursion is, oh, 4 mm or 0.004m. By the above equation, we
find that the excursion-limited power output of such a driver
at 1000 Hz is around 500 watts. That would generate a sound
pressure level at 1 meter of about 136 dB SPL. Pretty damned
loud. Now, assume said driver has an electroacoustic efficiency
of 1%: to get that 500 watts, you'd need 50,000 watts of electrical
power going into the driver.

But, notice what the limits are from the same driver at different
frequencies:

Freq (Hz) Pa (W) Max SPL
20 6x10^-4 W 68.5
50 0.0035 84.4
100 0.055 96.4
200 0.88 108.4
500 34 124.4
800 226 132.5
1000 550 136.4

Now, mind you, these are the theoretical outputs. limited only by
area and excursion. They ignore completely efficiency and thermal
power dissipation issues.

3. The question is if 1x 4" mid (0,8-5khz) will be sufficient
to MATCH the [2 x 10"woofers] woofers?

Let's say your hypothetical system's woofers are capable of linear
excursions of 10mm (0.01m). The system's power output will look
like (6 dB higher than the single case due to coupling and such):

Freq (Hz) Pa (W) Max SPL
50 0.4 100.3 dB
100 2.2 112.3
200 34 124.4
500 1349 140.3
800 8841 148.5

Well, from this, i looks like your lowly 4" midrange could never
hope to match the two 10" woofer. However, this conclusion isgnores
the fact that the broadband limit of the woofers is set not at 800
Hz, but low frequencies, where the limit is much lower. Assume the
peak of the musical spectrum is 200 Hz: the woofer limit is 124 dB,
and is easily handled by the low end of the midrange, whose
excursion limit at the low end (132 dB @800 Hz) exceeds that of the
woofer (124.4 @200Hz).

Again, these are the excursion-limited power outputs ONLY. If we
assume the woofer and midrange are of equal efficiency and both are
a reasonable 1%, that 124 dB limit require the driver each handle
several thosand watts: certainly not the most realistic and practical
scenario.

On 25 Sep 2005 06:48:28 -0700, wrote:

Don Pearce wrote:
On Sun, 25 Sep 2005 08:12:14 GMT, Morgan Ohlson wrote:
Max sound force /power from a driver, dB, effect

This is what I believe (like a guess):

Ex. In a hifi-store I listened to a 2-way speaker (2 x 6½"
+ 1 tweeter) with reasonable sensitivity ~91dB 1W/1m. The
amp was a large branded amp, something like 250W.

At 3m the sound power was perhaps 100dB and above that the
sound charachteristic was simply ****.

My guess was that the cone area MUST be bigger to "drive the
amount of air". 1. Is the area important (if amp and
sensitivity is enough) ???

2. Can the sound max strength be calculated for a mid at,
lets say 1Khz?

Let's say we intend to build a 3-way with 2x 10"woofers.
3. The question is if 1x 4" mid (0,8-5khz) will be sufficient
to MATCH the woofers? We don't want 120$ woofers to be limited
by the 20$ mid that chokes in its lower freq, do we!?

The maximum sound level is related to the volume of air that
the speaker can shift. To know this you need two things - the
area of the cone and the maximum linear excursion.

Almost, but not quite. The crucial missing parameter is
frequency.

The acoustic power output of a vibrating diaphragm is:

Pa = p c^2 Sd^2 x^2 w^4

whe

Pa is acoustic power in watts
p is the density of air, about 1.18 kg/m^2
c is the veloxity of sound, about 342 m/s
Sd is the emissive area of the diaphragm in m^2
x is the excursion of the diaphragm in m
and w is radian frequency, i.e., 2 pi f
Note: assumed is that the diaphragm is mounted on a baffle
substantially larger than the wavelength, and itself is
substantially smaller than a wavelength.

So, yes, it suggests that all other things being equal, i.e.,
same frequency and same excursion, power output is proportional
to the square of the diaphragm area.

To the specific questions the original poster asked:

1. Is the area important (if amp and sensitivity is enough)

Yes, but the maximum linear excursion (Xmax) and the frequency
being produced are equally important, indeed, frequency is
"more" important since power output goes as frequency to the
fourth power, while power goes only as the square of area and
excursion.

2. Can the sound max strength be calculated for a mid at,
lets say 1Khz?

Most assuredly, it can. Since you have already determined one
parameter (frequency) all you need to know are the other two.
Say your midrange is a nominal 4" unit, whose typical Sd or
emissive area is about 0.008m^2. And let's say that its linear
excursion is, oh, 4 mm or 0.004m. By the above equation, we
find that the excursion-limited power output of such a driver
at 1000 Hz is around 500 watts. That would generate a sound
pressure level at 1 meter of about 136 dB SPL. Pretty damned
loud. Now, assume said driver has an electroacoustic efficiency
of 1%: to get that 500 watts, you'd need 50,000 watts of electrical
power going into the driver.

But, notice what the limits are from the same driver at different
frequencies:

Freq (Hz) Pa (W) Max SPL
20 6x10^-4 W 68.5
50 0.0035 84.4
100 0.055 96.4
200 0.88 108.4
500 34 124.4
800 226 132.5
1000 550 136.4

Now, mind you, these are the theoretical outputs. limited only by
area and excursion. They ignore completely efficiency and thermal
power dissipation issues.

3. The question is if 1x 4" mid (0,8-5khz) will be sufficient
to MATCH the [2 x 10"woofers] woofers?

Let's say your hypothetical system's woofers are capable of linear
excursions of 10mm (0.01m). The system's power output will look
like (6 dB higher than the single case due to coupling and such):

Freq (Hz) Pa (W) Max SPL
50 0.4 100.3 dB
100 2.2 112.3
200 34 124.4
500 1349 140.3
800 8841 148.5

Well, from this, i looks like your lowly 4" midrange could never
hope to match the two 10" woofer. However, this conclusion isgnores
the fact that the broadband limit of the woofers is set not at 800
Hz, but low frequencies, where the limit is much lower. Assume the
peak of the musical spectrum is 200 Hz: the woofer limit is 124 dB,
and is easily handled by the low end of the midrange, whose
excursion limit at the low end (132 dB @800 Hz) exceeds that of the
woofer (124.4 @200Hz).

Again, these are the excursion-limited power outputs ONLY. If we
assume the woofer and midrange are of equal efficiency and both are
a reasonable 1%, that 124 dB limit require the driver each handle
several thosand watts: certainly not the most realistic and practical
scenario.

Yes, and the tougher to handle is the woofer... (???) so the mid is even
more likely to keep up.... if serious boxing etc. Isn't it???

It would then be quite obvious that most 3w systems with 2x mids are
oversized, sometimes up to ((132dB+6dB)-124)/3 = 4,6 (~450%)

Conclusions (Right or wrong?)
1) 1x 4" mid is in most systems some overkill
2) Woofers importance regarded, but heavily undereestemated.
3) Only argument for 2x mids is to bring sensitivity up.

2x 6½ mids would only be matched with somthing like
64x 10" woofers ...or
~14x 12" (slightly bigger movement) ;o)

If this is even close to reality one would ask: Why aren't there more high
sensitive mid drivers designed and manufactured to match multiple woofers?


Morgan O.

In article , Morgan Ohlson wrote:
On 25 Sep 2005 06:48:28 -0700, wrote:

Don Pearce wrote:
On Sun, 25 Sep 2005 08:12:14 GMT, Morgan Ohlson wrote:
Max sound force /power from a driver, dB, effect

This is what I believe (like a guess):

Ex. In a hifi-store I listened to a 2-way speaker (2 x 6½"
+ 1 tweeter) with reasonable sensitivity ~91dB 1W/1m. The
amp was a large branded amp, something like 250W.

At 3m the sound power was perhaps 100dB and above that the
sound charachteristic was simply ****.

My guess was that the cone area MUST be bigger to "drive the
amount of air". 1. Is the area important (if amp and
sensitivity is enough) ???

2. Can the sound max strength be calculated for a mid at,
lets say 1Khz?

Let's say we intend to build a 3-way with 2x 10"woofers.
3. The question is if 1x 4" mid (0,8-5khz) will be sufficient
to MATCH the woofers? We don't want 120$ woofers to be limited
by the 20$ mid that chokes in its lower freq, do we!?

The maximum sound level is related to the volume of air that
the speaker can shift. To know this you need two things - the
area of the cone and the maximum linear excursion.

Almost, but not quite. The crucial missing parameter is
frequency.

The acoustic power output of a vibrating diaphragm is:

Pa = p c^2 Sd^2 x^2 w^4

whe

Pa is acoustic power in watts
p is the density of air, about 1.18 kg/m^2
c is the veloxity of sound, about 342 m/s
Sd is the emissive area of the diaphragm in m^2
x is the excursion of the diaphragm in m
and w is radian frequency, i.e., 2 pi f
Note: assumed is that the diaphragm is mounted on a baffle
substantially larger than the wavelength, and itself is
substantially smaller than a wavelength.

So, yes, it suggests that all other things being equal, i.e.,
same frequency and same excursion, power output is proportional
to the square of the diaphragm area.

To the specific questions the original poster asked:

1. Is the area important (if amp and sensitivity is enough)

Yes, but the maximum linear excursion (Xmax) and the frequency
being produced are equally important, indeed, frequency is
"more" important since power output goes as frequency to the
fourth power, while power goes only as the square of area and
excursion.

2. Can the sound max strength be calculated for a mid at,
lets say 1Khz?

Most assuredly, it can. Since you have already determined one
parameter (frequency) all you need to know are the other two.
Say your midrange is a nominal 4" unit, whose typical Sd or
emissive area is about 0.008m^2. And let's say that its linear
excursion is, oh, 4 mm or 0.004m. By the above equation, we
find that the excursion-limited power output of such a driver
at 1000 Hz is around 500 watts. That would generate a sound
pressure level at 1 meter of about 136 dB SPL. Pretty damned
loud. Now, assume said driver has an electroacoustic efficiency
of 1%: to get that 500 watts, you'd need 50,000 watts of electrical
power going into the driver.

But, notice what the limits are from the same driver at different
frequencies:

Freq (Hz) Pa (W) Max SPL
20 6x10^-4 W 68.5
50 0.0035 84.4
100 0.055 96.4
200 0.88 108.4
500 34 124.4
800 226 132.5
1000 550 136.4

Now, mind you, these are the theoretical outputs. limited only by
area and excursion. They ignore completely efficiency and thermal
power dissipation issues.

3. The question is if 1x 4" mid (0,8-5khz) will be sufficient
to MATCH the [2 x 10"woofers] woofers?

Let's say your hypothetical system's woofers are capable of linear
excursions of 10mm (0.01m). The system's power output will look
like (6 dB higher than the single case due to coupling and such):

Freq (Hz) Pa (W) Max SPL
50 0.4 100.3 dB
100 2.2 112.3
200 34 124.4
500 1349 140.3
800 8841 148.5

Well, from this, i looks like your lowly 4" midrange could never
hope to match the two 10" woofer. However, this conclusion isgnores
the fact that the broadband limit of the woofers is set not at 800
Hz, but low frequencies, where the limit is much lower. Assume the
peak of the musical spectrum is 200 Hz: the woofer limit is 124 dB,
and is easily handled by the low end of the midrange, whose
excursion limit at the low end (132 dB @800 Hz) exceeds that of the
woofer (124.4 @200Hz).

Again, these are the excursion-limited power outputs ONLY. If we
assume the woofer and midrange are of equal efficiency and both are
a reasonable 1%, that 124 dB limit require the driver each handle
several thosand watts: certainly not the most realistic and practical
scenario.

Yes, and the tougher to handle is the woofer... (???) so the mid is even
more likely to keep up.... if serious boxing etc. Isn't it???

It would then be quite obvious that most 3w systems with 2x mids are
oversized, sometimes up to ((132dB+6dB)-124)/3 = 4,6 (~450%)

Conclusions (Right or wrong?)
1) 1x 4" mid is in most systems some overkill
2) Woofers importance regarded, but heavily undereestemated.
3) Only argument for 2x mids is to bring sensitivity up.

2x 6½ mids would only be matched with somthing like
64x 10" woofers ...or
~14x 12" (slightly bigger movement) ;o)

If this is even close to reality one would ask: Why aren't there more high
sensitive mid drivers designed and manufactured to match multiple woofers?


They are called horns.

I see in many PA type speakers, using too large of a driver, just
to increase sensitivity and handling, at the expense of beaming,
instead of making better use of vertically placed smaller drivers.
Other than using a horn, making a higher efficiency midrange is
tough.

greg

Morgan O.

Morgan Ohlson wrote:
On 25 Sep 2005 06:48:28 -0700, wrote:

Again, these are the excursion-limited power outputs ONLY. If we
assume the woofer and midrange are of equal efficiency and both are
a reasonable 1%, that 124 dB limit require the driver each handle
several thosand watts: certainly not the most realistic and practical
scenario.

Yes, and the tougher to handle is the woofer... (???)

I assume the ??? means you want to know why. There are a couple
of reasons:

1=2E To maintain a given sound pressure level, the excursion
of a driver goes as the inverse square of frequency. That
means to produce the same SPL that a driver is producing
at 200 Hz, that same driver must move 4 times farther at
100 Hz, 16 times farther at 50 Hz, and 64 times farther
at 25 Hz.

2=2E In general, the distribution of energy vs frequency of most
music is not even across the spectrum, far from it. Most
music tends to exhibit a peak in the spectrum around 200 Hz,
and dcreases above and below that. Now, there are certainly
exceptions to this general tendency, but the overall spectrum
remarkably, does not vary to much from this overall average
from genre to genre.

so the mid is even more likely to keep up....

Remeber that several times I quite explicitly stated that the
figure I was talking about were theoretical based on excursion
limitations alone and did NOT take into consideration ANY other
details, for example, thermal power limits. In most cases, the
thermal power handling capabilities of woofers exceeds that of
midranges, which exceed that of tweeters. The reason what now
might seem like very fragile tweeters don't just burn up in a
quick puff of smoke is that there is simply less overall energy
in their operating band.

And, do recall, a lot of tweeters DO just burn up in a quick
puff of smoke.

if serious boxing etc. Isn't it???

I have no idea what you mean by "serious boxing."

It would then be quite obvious that most 3w systems with 2x mids are
oversized, sometimes up to ((132dB+6dB)-124)/3 =3D 4,6 (~450%)

It does not seem obvious at all to me, perhaps you might
explain how you came to that conclusion.

Conclusions (Right or wrong?)
1) 1x 4" mid is in most systems some overkill

Wrong. Again, please recall that the figures I discussed were
ONLY based on THEORETICAL limitations based on EXCURSION only.
There's a LOT more to designing a speaker than that one issue.


2) Woofers importance regarded, but heavily undereestemated.

I am not sure how you arrived at that conclusion, either.

3) Only argument for 2x mids is to bring sensitivity up.

Wrong. It's a lot easire to make a more efficient midrange than
it is to make a more efficient woofer, as the midrawnges are
often not contrained to work to their fundamental resonances,
moving mass is much less, and the resulting Bl/mass ratio works
much more favorably towards higher midrange efficiency. I don't
know where you got the idea that midranges are less efficient
inherently than woofers, but it is a conclusion not supported by
actual fact.

Besides there are a LOT of reasons why two midranges may be desired
or required. One reasons, for example, is that using two midranges
with a tweeter in between enables one to implemenbt the so-called
D'Appolito configuration, which results in symmetrical dispersion
through the crossover region.

2x 6=BD mids would only be matched with somthing like
64x 10" woofers ...or
~14x 12" (slightly bigger movement) ;o)

I really hope that really is a smiley.

If this is even close to reality one would ask: Why aren't there
more high sensitive mid drivers designed and manufactured to
match multiple woofers?

I would ask: where are you getting your data, because in the several
thousand drivers of various size I have measured, I have found that,
in fact, the opposite is generally true: that drivers suitable for
midranges of suitable sensitivity are generally very easy to find
and often have higher sensitivity than woofers.

I suspect your investigation has not been wide enough to give you
a reasonably accurate sampling of what's out there and what's
possible.
=20
Morgan O.

On 27 Sep 2005 11:48:08 -0700, wrote:

Morgan Ohlson wrote:
On 25 Sep 2005 06:48:28 -0700, wrote:

Again, these are the excursion-limited power outputs ONLY. If we
assume the woofer and midrange are of equal efficiency and both are
a reasonable 1%, that 124 dB limit require the driver each handle
several thosand watts: certainly not the most realistic and practical
scenario.

Yes, and the tougher to handle is the woofer... (???)

I assume the ??? means you want to know why. There are a couple
of reasons:

1. To maintain a given sound pressure level, the excursion
of a driver goes as the inverse square of frequency. That
means to produce the same SPL that a driver is producing
at 200 Hz, that same driver must move 4 times farther at
100 Hz, 16 times farther at 50 Hz, and 64 times farther
at 25 Hz.

2. In general, the distribution of energy vs frequency of most
music is not even across the spectrum, far from it. Most
music tends to exhibit a peak in the spectrum around 200 Hz,
and dcreases above and below that. Now, there are certainly
exceptions to this general tendency, but the overall spectrum
remarkably, does not vary to much from this overall average
from genre to genre.

so the mid is even more likely to keep up....

Remeber that several times I quite explicitly stated that the
figure I was talking about were theoretical based on excursion
limitations alone and did NOT take into consideration ANY other
details, for example, thermal power limits. In most cases, the
thermal power handling capabilities of woofers exceeds that of
midranges, which exceed that of tweeters. The reason what now
might seem like very fragile tweeters don't just burn up in a
quick puff of smoke is that there is simply less overall energy
in their operating band.

And, do recall, a lot of tweeters DO just burn up in a quick
puff of smoke.

if serious boxing etc. Isn't it???

I have no idea what you mean by "serious boxing."

It would then be quite obvious that most 3w systems with 2x mids are
oversized, sometimes up to ((132dB+6dB)-124)/3 = 4,6 (~450%)

It does not seem obvious at all to me, perhaps you might
explain how you came to that conclusion.

Only based on moving air amount

....and not taking heat and power limitations into it... ;o)

But driving a speaker to its power limits I guess that heat is a matter...
which I didn't think of at first.

/.../
3) Only argument for 2x mids is to bring sensitivity up.

Wrong. It's a lot easire to make a more efficient midrange than
it is to make a more efficient woofer,

All the same, catalogues I have seen often have lots of highly sensitive
woofers, and only a limited, perhaps 1 or 2 Hifi mid drivers 91dB@1m1W


as the midrawnges are
often not contrained to work to their fundamental resonances,
moving mass is much less, and the resulting Bl/mass ratio works
much more favorably towards higher midrange efficiency. I don't
know where you got the idea that midranges are less efficient
inherently than woofers, but it is a conclusion not supported by
actual fact.

Must ask... The efficiency you talk about is basicly (exactly?) the same as
sensitivity?


Besides there are a LOT of reasons why two midranges may be desired
or required. One reasons, for example, is that using two midranges
with a tweeter in between enables one to implemenbt the so-called
D'Appolito configuration, which results in symmetrical dispersion
through the crossover region.

2x 6½ mids would only be matched with somthing like
64x 10" woofers ...or
~14x 12" (slightly bigger movement) ;o)

I really hope that really is a smiley.

If this is even close to reality one would ask: Why aren't there
more high sensitive mid drivers designed and manufactured to
match multiple woofers?

I would ask: where are you getting your data,

Only trying to put things together... and I understand I'm a bad pupil at
the moment. /

But perhaps its also about that I was driving the theoritical discussion
into an oversimplyfied non-existing reality.

...because in the several
thousand drivers of various size I have measured, I have found that,
in fact, the opposite is generally true: that drivers suitable for
midranges of suitable sensitivity are generally very easy to find
and often have higher sensitivity than woofers.

At the moment I have at least to examples of companies. A have 1 5 1/4
polypropylene cone 93dB. B has none at all above 91dB. That is mid drivers
on a budget 40$.


I suspect your investigation has not been wide enough to give you
a reasonably accurate sampling of what's out there and what's
possible.

No, my research has no intention to be scientific... I was just referring to
what I had actually seen, no more, no less.


Morgan O.

On Tue, 27 Sep 2005 18:15:22 GMT, GregS wrote:

In article , Morgan Ohlson wrote:
On 25 Sep 2005 06:48:28 -0700, wrote:

Don Pearce wrote:
On Sun, 25 Sep 2005 08:12:14 GMT, Morgan Ohlson wrote:
Max sound force /power from a driver, dB, effect

This is what I believe (like a guess):

Ex. In a hifi-store I listened to a 2-way speaker (2 x 6½"
+ 1 tweeter) with reasonable sensitivity ~91dB 1W/1m. The
amp was a large branded amp, something like 250W.

At 3m the sound power was perhaps 100dB and above that the
sound charachteristic was simply ****.

My guess was that the cone area MUST be bigger to "drive the
amount of air". 1. Is the area important (if amp and
sensitivity is enough) ???

2. Can the sound max strength be calculated for a mid at,
lets say 1Khz?

Let's say we intend to build a 3-way with 2x 10"woofers.
3. The question is if 1x 4" mid (0,8-5khz) will be sufficient
to MATCH the woofers? We don't want 120$ woofers to be limited
by the 20$ mid that chokes in its lower freq, do we!?

The maximum sound level is related to the volume of air that
the speaker can shift. To know this you need two things - the
area of the cone and the maximum linear excursion.

Almost, but not quite. The crucial missing parameter is
frequency.

The acoustic power output of a vibrating diaphragm is:

Pa = p c^2 Sd^2 x^2 w^4

whe

Pa is acoustic power in watts
p is the density of air, about 1.18 kg/m^2
c is the veloxity of sound, about 342 m/s
Sd is the emissive area of the diaphragm in m^2
x is the excursion of the diaphragm in m
and w is radian frequency, i.e., 2 pi f
Note: assumed is that the diaphragm is mounted on a baffle
substantially larger than the wavelength, and itself is
substantially smaller than a wavelength.

So, yes, it suggests that all other things being equal, i.e.,
same frequency and same excursion, power output is proportional
to the square of the diaphragm area.

To the specific questions the original poster asked:

1. Is the area important (if amp and sensitivity is enough)

Yes, but the maximum linear excursion (Xmax) and the frequency
being produced are equally important, indeed, frequency is
"more" important since power output goes as frequency to the
fourth power, while power goes only as the square of area and
excursion.

2. Can the sound max strength be calculated for a mid at,
lets say 1Khz?

Most assuredly, it can. Since you have already determined one
parameter (frequency) all you need to know are the other two.
Say your midrange is a nominal 4" unit, whose typical Sd or
emissive area is about 0.008m^2. And let's say that its linear
excursion is, oh, 4 mm or 0.004m. By the above equation, we
find that the excursion-limited power output of such a driver
at 1000 Hz is around 500 watts. That would generate a sound
pressure level at 1 meter of about 136 dB SPL. Pretty damned
loud. Now, assume said driver has an electroacoustic efficiency
of 1%: to get that 500 watts, you'd need 50,000 watts of electrical
power going into the driver.

But, notice what the limits are from the same driver at different
frequencies:

Freq (Hz) Pa (W) Max SPL
20 6x10^-4 W 68.5
50 0.0035 84.4
100 0.055 96.4
200 0.88 108.4
500 34 124.4
800 226 132.5
1000 550 136.4

Now, mind you, these are the theoretical outputs. limited only by
area and excursion. They ignore completely efficiency and thermal
power dissipation issues.

3. The question is if 1x 4" mid (0,8-5khz) will be sufficient
to MATCH the [2 x 10"woofers] woofers?

Let's say your hypothetical system's woofers are capable of linear
excursions of 10mm (0.01m). The system's power output will look
like (6 dB higher than the single case due to coupling and such):

Freq (Hz) Pa (W) Max SPL
50 0.4 100.3 dB
100 2.2 112.3
200 34 124.4
500 1349 140.3
800 8841 148.5

Well, from this, i looks like your lowly 4" midrange could never
hope to match the two 10" woofer. However, this conclusion isgnores
the fact that the broadband limit of the woofers is set not at 800
Hz, but low frequencies, where the limit is much lower. Assume the
peak of the musical spectrum is 200 Hz: the woofer limit is 124 dB,
and is easily handled by the low end of the midrange, whose
excursion limit at the low end (132 dB @800 Hz) exceeds that of the
woofer (124.4 @200Hz).

Again, these are the excursion-limited power outputs ONLY. If we
assume the woofer and midrange are of equal efficiency and both are
a reasonable 1%, that 124 dB limit require the driver each handle
several thosand watts: certainly not the most realistic and practical
scenario.

Yes, and the tougher to handle is the woofer... (???) so the mid is even
more likely to keep up.... if serious boxing etc. Isn't it???

It would then be quite obvious that most 3w systems with 2x mids are
oversized, sometimes up to ((132dB+6dB)-124)/3 = 4,6 (~450%)

Conclusions (Right or wrong?)
1) 1x 4" mid is in most systems some overkill
2) Woofers importance regarded, but heavily undereestemated.
3) Only argument for 2x mids is to bring sensitivity up.

2x 6½ mids would only be matched with somthing like
64x 10" woofers ...or
~14x 12" (slightly bigger movement) ;o)

If this is even close to reality one would ask: Why aren't there more high
sensitive mid drivers designed and manufactured to match multiple woofers?


They are called horns.

I see in many PA type speakers, using too large of a driver, just
to increase sensitivity and handling, at the expense of beaming,

Beaming??? Spread, angle of correct sound?

instead of making better use of vertically placed smaller drivers.
Other than using a horn, making a higher efficiency midrange is
tough.

greg


Morgan O.

Morgan Ohlson wrote:
On 27 Sep 2005 11:48:08 -0700, wrote:
It would then be quite obvious that most 3w systems with 2x mids are
oversized, sometimes up to ((132dB+6dB)-124)/3 = 4,6 (~450%)

It does not seem obvious at all to me, perhaps you might
explain how you came to that conclusion.

Only based on moving air amount

But that is an extremely narrow criteria and leads to erroneous
conclusions, such as you have arrived at.

...and not taking heat and power limitations into it... ;o)

But driving a speaker to its power limits I guess that heat
is a matter... which I didn't think of at first.

And it's an important one at that.

Recall again that the excursion goes as the reciprocal of
the square of frequency. WHat this means is that, in general,
the low frequency limitation is set by mechanical limits, while
at high frequencies, it's set by thermal limits.

3) Only argument for 2x mids is to bring sensitivity up.

Wrong. It's a lot easire to make a more efficient midrange than
it is to make a more efficient woofer,

All the same, catalogues I have seen often have lots of highly sensitive
woofers, and only a limited, perhaps 1 or 2 Hifi mid drivers 91dB@1m1W

Look at more catalogs.

as the midrawnges are
often not contrained to work to their fundamental resonances,
moving mass is much less, and the resulting Bl/mass ratio works
much more favorably towards higher midrange efficiency. I don't
know where you got the idea that midranges are less efficient
inherently than woofers, but it is a conclusion not supported by
actual fact.

Must ask... The efficiency you talk about is basicly (exactly?)
the same as sensitivity?

No, it is not. Efficiency has a very specific and widely agreed
upon definition. It is the ratio of acoustic power out (measured
in Watts) to the electrical power in (also measured in Watts).

Sensitivity, on the other hand, has no real precise definition.
While typically it is specified as SPL measured at 1 meter with
and input of 2.83 volts (which would be 1 watt if the speaker
presented an 8 ohm resistive impedance), a lot of other parameters
are left unspecified, for example, baffle conditions, radiation
angle, and more.