# Cone Excursion

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Anonymous

If we know that a driver in a speaker system has a cone excursion of
0.5 inches peak-to-peak at 100 Hz, can we predict what the excursion
of this driver will be at lower frequencies, assuming that the
response is flat?

Keith Howard states in his article on Doppler distortion in the
November 2004 issue of Stereophile that a speaker which moves 0.5
inches peak-to-peak in order to reproduce a 100 Hz sine wave only has
a peak velocity of 9 miles/hour. (By my calculations, it's 19
miles/hour, but that's not all that relevent here.) I think Doppler is
clearly evident at, e.g., 40 Hz, and am interested in calculating the
cone velocity at this frequency.
--
% Randy Yates % "Though you ride on the wheels of tomorrow,
%% Fuquay-Varina, NC % you still wander the fields of your
%%% 919-577-9882 % sorrow."
%%%% <yates@ieee.org> % '21st Century Man', *Time*, ELO

In article <wtpnsyub.fsf@ieee.org>, Randy Yates <yates@ieee.org> wrote:

> If we know that a driver in a speaker system has a cone excursion of
> 0.5 inches peak-to-peak at 100 Hz, can we predict what the excursion
> of this driver will be at lower frequencies, assuming that the
> response is flat?
>

IMO It matters more what type of an enclosure a given driver is in. For
instance, a driver in a bass reflex enclosure may unload below its
tuning frequency.

Also IMO, there are many factors to consider and that calculations at
the driver itself is futile while not considering the space it will be
placed in as well.

> Keith Howard states in his article on Doppler distortion in the
> November 2004 issue of Stereophile that a speaker which moves 0.5
> inches peak-to-peak in order to reproduce a 100 Hz sine wave only has
> a peak velocity of 9 miles/hour. (By my calculations, it's 19
> miles/hour, but that's not all that relevent here.) I think Doppler is
> clearly evident at, e.g., 40 Hz, and am interested in calculating the
> cone velocity at this frequency.

hth,

--
Cyrus

*coughcasaucedoprodigynetcough*
Anonymous

>If we know that a driver in a speaker system has a cone
>excursion of 0.5 inches peak-to-peak at 100 Hz, can we
>predict what the excursion of this driver will be at lower
>frequencies, assuming that the response is flat?

All other things being equal, and assuming the cone in
question is the sole (or at least predominant) radiator
in the system, assuming we ignore mechanical limits, assuming
the cone is working at frequencies substantially smaller than
the wavelength being produced, and assuming we're attempting
to produce an SPL that is independent of frequency, then yes,
we can know the excursion at any frequency.

Very simply, a radiator's excursion, for a given SPL, goes
as the inverse square of the frequency. Differentiating to
obtain velocity, it goes as the inverse of frequency, and
once again to obtain acceleration, it's constant with
frequency for a given SPL.

Further, one can determine from frequency, excursion and
radiating area, what the SPL produced is. Let's assume your
0.5" pk-to-pk (.25" peak) at 100 Hz is being done by a 12"
driver: the SPL at 1 meter is a mere 122 dB SPL. To achieve
that same SPL at 50 Hz would require an excursion of 1" pk,
at 25 Hz, 4" peak.
Anonymous

>It will be the same, as this 'Xmax' value is a physical
>limit for the suspension. If we're talking about a constant
>SPL well below the actual Xmax value, then the excursion
>doubles for every halving of frequency

No, excursion goes as the inverse square of frequency.
>From Thiele, Small and others:

P = p c^2 Sd w^2 x

where P = sound pressure, p (rho, in fact) is the density
of air (1.18 kg/m^3), c = velocity of sound (3.42 m/s),
Sd is the emissive diameter in m^2, w is frequency in
radians/s, and x is excursion in meters. To convert that
to SPL:

SPL = 20 log(p c^2 Sd w^2 x)

For closed-box system, the generalized excursion vs frequency
transfer function looks like (view with fixed-width font):

1
X(s) = ------------------------
s^2 Tc^2 + s Tc/Qtc + 1

where X is normalized excursion, s is the complex frequency
variable, Tc is 1/2 pi Fc (system resonance) and Qtc is the
toal system Q.

>(a constant SPL
>implies that the same volume of air is moved per unit of
>time, so if you halve the number of excursions per
>second, you must double the volume per stroke).

No, that only is true if you assume the radiation load is independent
of frequency. The radiation impedance is, in
fact, frequency dependent

It turns out that the requirement for excursion to go as
the inverse square of frequency is exactly met by the fact
that a cone operating above its resonant frequency is in
the mass controlled region, and its excursion goes as the
inverse square of frequency.

Specifically, the total power radiated is:

Pa = |Uo|^2 Rar

where Pa is the total radiated power in watts, Uo is
the total system volume velocity, and Rar is the real part
of the radiation impedance which foes roughly as linear
function of frequency, as long as the wavelength is large compared to
Related ressources
Anonymous

"Randy Yates" <yates@ieee.org> wrote in message
news:is17sgd7.fsf@ieee.org...

> I'm fairly certain that it would be easy to demonstrate using a
> single, open-air driver with a combination of test tones, say, 40 Hz +
> 800 Hz. I'll have to get off my duff and do some experimenting.
> --

Why not use 40Hz & 3kHz? IIRC that's the frequency which used to be used
for wow & flutter testing (due to the ear's high sensitivity ?).

--
M Stewart
Milton Keynes, UK
http://www.megalith.freeserve.co.uk/oddimage.htm
Anonymous

Cone excursion is easily calculated from the basic simulation HP
transfer functions, regardless of type of design. If one knows the
basic T/S parameters and those parameters necessary to calculate
efficiency, you can pretty accurately calculate the excursion distance
given cabinet alignment, for any input frequency and Vin from the
corresponding transfer function. (assuming within normal operation
limits)

The limitations of excursion as expressed by Xmax are published to let
the designer know when the combination of stimulus frequency and input
signal amplitude come close to exceeding linear operation; the so
called large signal parameters. There are 2 common ones: Per and Par
which are highly related. Per is the maximum input wattage to remain
under Xmax. Par is the equivalent acoustic output corresponding to this
max input in dB SPL at threshold. The reason why Per and Par are so
neat and tidy is because there is a defined peak excursion defined by
the transfer function in all standard loudspeaker alignment types so
you really don't need to know the excursion all along the way. Thus the
peak is the place to watch.

BTW, one of the reasons why bass reflex systems remain popular is
because X max is less of a threat to linear high power operation. There
is a Xmin at Fb which the 2nd system doesn't have. Thus the large
signal parameter limits are more forgiving, depending on how they are
aligned of course.

Aside from numerous books, there very good classic and easy to read AES
papers (from wouldn't you have guessed Thiel and Small) which explain
this quite well.

Wessel
Anonymous

dpierce@cartchunk.org writes:

> >If we know that a driver in a speaker system has a cone
> >excursion of 0.5 inches peak-to-peak at 100 Hz, can we
> >predict what the excursion of this driver will be at lower
> >frequencies, assuming that the response is flat?
>
> All other things being equal, and assuming the cone in
> question is the sole (or at least predominant) radiator
> in the system, assuming we ignore mechanical limits, assuming
> the cone is working at frequencies substantially smaller than
> the wavelength being produced, and assuming we're attempting
> to produce an SPL that is independent of frequency, then yes,
> we can know the excursion at any frequency.
>
> Very simply, a radiator's excursion, for a given SPL, goes
> as the inverse square of the frequency.

OK, so

Xmax(f) = a / f^2

Right?

In our example,

0.25 [in] = a / 100^2
a = 2500 [in]

Thus

Xmax(f) = 2500 [in] / f^2.

Right?

Then

Xmax(f1) / Xmax(f2) = (2500 [in] / f1^2) / (2500 [in] / f2^2)
= (f2 / f1)^2.

Right?

So then a woofer at 40 Hz will excursion
(100/40)^2 = (5/2)^2 = 25/4 times as much as a woofer at 100 Hz. Thus
instead of 9 mph, the "train" is now running at over 54 mph - no longer
"slow" in my estimation, and thus this one piece of Mr. Howard's logic
falls apart.

How about 20 Hz? (100/20)^2 = 5^2 = 25. Thus the "train" is now moving
at 225 mph. Ya' think ya' just might hear a wee bit of doppler then,
Mr. Howard?
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124
Anonymous

Randy Yates wrote:

> dpierce@cartchunk.org writes:

>>Very simply, a radiator's excursion, for a given SPL, goes
>>as the inverse square of the frequency.
>
>
> OK, so

<snip>

>
> So then a woofer at 40 Hz will excursion
> (100/40)^2 = (5/2)^2 = 25/4 times as much as a woofer at 100 Hz. Thus
> instead of 9 mph, the "train" is now running at over 54 mph - no longer
> "slow" in my estimation, and thus this one piece of Mr. Howard's logic
> falls apart.
>
> How about 20 Hz? (100/20)^2 = 5^2 = 25. Thus the "train" is now moving
> at 225 mph. Ya' think ya' just might hear a wee bit of doppler then,
> Mr. Howard?

You are confusing position and velocity.

--
Eiron.
Anonymous

Eiron <e1ron@hotmail.com> writes:

> Randy Yates wrote:
>
> > dpierce@cartchunk.org writes:
>
> >>Very simply, a radiator's excursion, for a given SPL, goes
> >> as the inverse square of the frequency.
>
> > OK, so
>
>
> <snip>
>
> > So then a woofer at 40 Hz will excursion
>
> > (100/40)^2 = (5/2)^2 = 25/4 times as much as a woofer at 100 Hz. Thus
> > instead of 9 mph, the "train" is now running at over 54 mph - no longer
> > "slow" in my estimation, and thus this one piece of Mr. Howard's logic
> > falls apart.
> > How about 20 Hz? (100/20)^2 = 5^2 = 25. Thus the "train" is now
> > moving
>
> > at 225 mph. Ya' think ya' just might hear a wee bit of doppler then,
> > Mr. Howard?
>
> You are confusing position and velocity.

So you mean that d(a*x(t))/dt != a*dx(t)/dt?
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124
Anonymous

On 26 May 2005 10:07:03 -0400, Randy Yates
<randy.yates@sonyericsson.com> wrote:

>dpierce@cartchunk.org writes:
>
>> >If we know that a driver in a speaker system has a cone
>> >excursion of 0.5 inches peak-to-peak at 100 Hz, can we
>> >predict what the excursion of this driver will be at lower
>> >frequencies, assuming that the response is flat?
>>
>> All other things being equal, and assuming the cone in
>> question is the sole (or at least predominant) radiator
>> in the system, assuming we ignore mechanical limits, assuming
>> the cone is working at frequencies substantially smaller than
>> the wavelength being produced, and assuming we're attempting
>> to produce an SPL that is independent of frequency, then yes,
>> we can know the excursion at any frequency.
>>
>> Very simply, a radiator's excursion, for a given SPL, goes
>> as the inverse square of the frequency.
>
>OK, so
>
> Xmax(f) = a / f^2
>
>Right?
>
>In our example,
>
> 0.25 [in] = a / 100^2
> a = 2500 [in]
>
>Thus
>
> Xmax(f) = 2500 [in] / f^2.
>
>Right?
>
>Then
>
> Xmax(f1) / Xmax(f2) = (2500 [in] / f1^2) / (2500 [in] / f2^2)
> = (f2 / f1)^2.
>
>Right?
>
>So then a woofer at 40 Hz will excursion
>(100/40)^2 = (5/2)^2 = 25/4 times as much as a woofer at 100 Hz. Thus
>instead of 9 mph, the "train" is now running at over 54 mph - no longer
>"slow" in my estimation, and thus this one piece of Mr. Howard's logic
>falls apart.
>
>How about 20 Hz? (100/20)^2 = 5^2 = 25. Thus the "train" is now moving
>at 225 mph. Ya' think ya' just might hear a wee bit of doppler then,
>Mr. Howard?

You've gone a step too far here, as you have adjusted the 9mph peak
velocity figure for the increased excursion, but not for the increased
time per cycle. The piston *velocity* only goes up as the inverse of
frequency, not the inverse square. Hence, at 20Hz, the piston is only
peaking at 45mph. It would of course also have to have an Xmax of more
than 3 inches to achieve this, which is unlikely!
--

Stewart Pinkerton | Music is Art - Audio is Engineering
Anonymous

Stewart Pinkerton <patent3@dircon.co.uk> writes:
> [...]
> On 26 May 2005 10:07:03 -0400, Randy Yates
> <randy.yates@sonyericsson.com> wrote:
> >How about 20 Hz? (100/20)^2 = 5^2 = 25. Thus the "train" is now moving
> >at 225 mph. Ya' think ya' just might hear a wee bit of doppler then,
> >Mr. Howard?
>
> You've gone a step too far here, as you have adjusted the 9mph peak
> velocity figure for the increased excursion, but not for the increased
> time per cycle. The piston *velocity* only goes up as the inverse of
> frequency, not the inverse square.

Have I made a mistake? If we denote the piston position as a function
of time as x1(t), then the piston velocity is v1(t) = d x1(t) / dt. If we scale
the piston position by a factor a, so that x2(t) = a*x1(t), then

v2(t) = d x2(t) / dt
= d (a*x1(t)) / dt
= a * d x1(t) / dt
= a * v1(t).

Where have I gone wrong?
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124
Anonymous

Randy Yates wrote:
>OK, so
>
> Xmax(f) = a / f^2
>
>Right?

No.

First, it's not Xmax, it's simply X. Xmax is the total
available linear excursion independent of frequency.

Secondly, all I said was that for a given sound level,
excursion goes as the invers square of frequency, thus:

X prop 1/f^2

If what you're trying to do is derive peak cone velocity
from excursion and frequency, that's something different.

Since, assuming we're talking a 100 Hz sine, that:

x(t) = a sin(w t)

where t is time, a is the peak excursion, and w is radian
frequency.

Velocity is the first derivative of position WRT time,
thus:

v(t) = w a cos (w t)

Set t = 0, for maximum velocity, and we get:

v(0) = 2 pi 100 0.5 cos(0)

and thus v(0) = 314 in/s, 26 ft/s or 17.7 mph.

However, do harken back to the calculation of what a 12"
woofer moving 0.5" peak at 100 Hz will do for (or to, as
the case may be) to you: that's 122 dB SPL. That's
AWFULLY loud.
Anonymous

Randy Yates wrote:
>How about 20 Hz? (100/20)^2 = 5^2 = 25. Thus the "train"
>is now moving at 225 mph. Ya' think ya' just might hear
>a wee bit of doppler then, Mr. Howard?

No, because VELOCITY does NOT go as inverse frequency
squared, it only goes as inverse frequency.

Thus, take our original 17 MPH and multiple it by 5, not
25.

And, again, you're talking about ENORMOUS sound pressure
levels anyway. 1/2" of excursion of a 12" cone is still
110 dB at 50 Hz and 98 dB at 20 Hz. What sort of music
generates those sorts of sound pressure levels at those
frequencies?
Anonymous

Randy Yates <randy.yates@sonyericsson.com> writes:

> Stewart Pinkerton <patent3@dircon.co.uk> writes:
>> [...]
>> On 26 May 2005 10:07:03 -0400, Randy Yates
>> <randy.yates@sonyericsson.com> wrote:
>> >How about 20 Hz? (100/20)^2 = 5^2 = 25. Thus the "train" is now moving
>> >at 225 mph. Ya' think ya' just might hear a wee bit of doppler then,
>> >Mr. Howard?
>>
>> You've gone a step too far here, as you have adjusted the 9mph peak
>> velocity figure for the increased excursion, but not for the increased
>> time per cycle. The piston *velocity* only goes up as the inverse of
>> frequency, not the inverse square.
>
> Have I made a mistake? If we denote the piston position as a function
> of time as x1(t), then the piston velocity is v1(t) = d x1(t) / dt. If we scale
> the piston position by a factor a, so that x2(t) = a*x1(t), then
>
> v2(t) = d x2(t) / dt
> = d (a*x1(t)) / dt
> = a * d x1(t) / dt
> = a * v1(t).
>
> Where have I gone wrong?

x2(t) != a*x1(t).

Doh!
--
% Randy Yates % "How's life on earth?
%% Fuquay-Varina, NC % ... What is it worth?"
%%% 919-577-9882 % 'Mission (A World Record)',
%%%% <yates@ieee.org> % *A New World Record*, ELO
Anonymous

"Wessel Dirksen" <wdirksen@gmail.com> wrote in message
> BTW, one of the reasons why bass reflex systems remain popular is
> because X max is less of a threat to linear high power operation. There
> is a Xmin at Fb which the 2nd system doesn't have. Thus the large
> signal parameter limits are more forgiving, depending on how they are
> aligned of course.

Only if you make sure it is not driven too hard below Xmin. Even larger
excursion is required to maintain bass response if this is attempted. Many
do.

MrT.
Anonymous

Just the power input to achieve 1/2" excursion on a 12" woofer *at 100
Hz* is a fairly extreme scenario, as well, no? Is it possible the
power may even lie outside the comfortable handling limits of most
conventional woofers? Then you factor in suspension nonlinearities,
motor modulation, and thermal compression that would associate to such
a scenario, and probably there will be a few other "nasty things"
happening to the sound that make doppler concerns a might bit trivial?
Anonymous

Mr.T schreef:
> "Wessel Dirksen" <wdirksen@gmail.com> wrote in message
> > BTW, one of the reasons why bass reflex systems remain popular is
> > because X max is less of a threat to linear high power operation. There
> > is a Xmin at Fb which the 2nd system doesn't have. Thus the large
> > signal parameter limits are more forgiving, depending on how they are
> > aligned of course.
>
> Only if you make sure it is not driven too hard below Xmin. Even larger
> excursion is required to maintain bass response if this is attempted. Many
> do.
>

I don't quite get what you mean here, so perhaps you can elaborate.

Under normal operating conditions a standard electro magnetic driver
moves very predicably as it is being stimulated by its input signal.
There is no voodoo involved. Per also exists for bass reflex
alignments, so even if one designed a ported nightmare, the input
signal corresponding to the Xmax derived excursion limit would be
reduced as well. If you drive a driver "too" hard with a regular audio
signal it will have a longer excursion at low frequencies by definition
and thus will have already exceeded Xmax. X max is rating of linear
operation and not power handling. Most quality units can far exceed
Xmax in operation before getting into thermal trouble. Even applying a
very large input voltage at a single higher frequency yet under Xmax
would probably not be a problem for most woofers. I say under normal
operating conditions because could put a 200V RMS 8kHz signal into a
woofer and heat it up quite a bit, and you could block the cone from
moving, etc.

Your everyday quality woofer will under normal operating conditions
usually be able to handle more input power with a vented box than a
sealed one.
Anonymous

Dangling entity says:
> Just the power input to achieve 1/2" excursion on a
> 12" woofer *at 100 Hz* is a fairly extreme scenario,
> as well, no? Is it possible the power may even lie
> outside the comfortable handling limits of most
> conventional woofers?

Indeed. Assume a moderate efficiency of 1%, which corresponds
to a half-space sensitivity of 92 dB/w, 122 dB would require an
electric input of 1000 watts, which I would submit is a somewhat
absurd scenario.

> Then you factor in suspension nonlinearities, motor
> modulation, and thermal compression that would associate
> to such a scenario, and probably there will be a few
> other "nasty things" happening to the sound that make
> doppler concerns a might bit trivial?

Suspension and excursions related motor non-linearities are
assumed to be accounted for in the 1/2" basis for the argument,
i.e., it's assumed that the driver5 can, for the purpose of
discussion, move 1/2" linearly. The issue of thermal compression
can be ignored for the time being because the time constant of
the thermal compression is substantially longer than that of the
signal so is not a non-linearity per se.

However, the issue of magnet modulation is far from trivial.

This all supports the assertion that measurements and audibility
of doppler from practical experiments are seriously confounded
by all of these other factors.

One also might assert that at 1/2" excursion at 100 Hz, Doppler
is NOT audible because a after a little while listening at those
sorts of levels, nothing much will be audible due to hearing loss.
Anonymous

Mr.T schreef:
> "Wessel Dirksen" <wdirksen@gmail.com> wrote in message
> > I don't quite get what you mean here, so perhaps you can elaborate.
> > Under normal operating conditions a standard electro magnetic driver
> > moves very predicably as it is being stimulated by its input signal.
> > There is no voodoo involved. Per also exists for bass reflex
> > alignments, so even if one designed a ported nightmare, the input
> > signal corresponding to the Xmax derived excursion limit would be
> > reduced as well. If you drive a driver "too" hard with a regular audio
> > signal it will have a longer excursion at low frequencies by definition
> > and thus will have already exceeded Xmax. X max is rating of linear
> > operation and not power handling.
>
> Agreed, there should be no mechanical damage to a driver when operated at
> Xmax.
> The problem is that Xmax is more easily exceeded if one tries to get a given
> level of SPL at frequencies below the cut off of a ported box, compared to
> the same driver in a sealed box. A quick look at the slope of the response
> curves below tuned frequency will show you why.

Ok got ya, but you are refering to the compliance controlled region of
the alignment which is located on the slope itself thus out of the pass
band. As you state below this really only practically counts for
enclosures which don't go low at all. In most standard vented systems
(I would guess Fb <80 hz or so), the opposite is true as excursion has
a minimum at Fb. You would have to go quite a ways into the compliance
controlled region, but then not too far before the absence of subsonic
signal nullifies this impact.

> IME small ported enclosures suffer more mechanical damage from people
> winding up the bass control, than do sealed eclosures.
>
> >Most quality units can far exceed
> > Xmax in operation before getting into thermal trouble.
>
> Thermal limiting is rarely a problem when operating below the cut off
> frequency of a ported enclosure.
> Mechanical limits usually apply.
>
> >Even applying a
> > very large input voltage at a single higher frequency yet under Xmax
> > would probably not be a problem for most woofers. I say under normal
> > operating conditions because could put a 200V RMS 8kHz signal into a
> > woofer and heat it up quite a bit, and you could block the cone from
> > moving, etc.
>
> One would not put an 8kHz signal into a "woofer". Hardly "normal operating
> conditions"!
> ( I won't even comment on the 200V RMS stupidity, 5 kW at 8ohms)
> Please re-read my orignal statement with regard to operating BELOW the
> ported box tuned frequency. For many small speakers this is around 50 Hz or
> so.

I did not mean this as demeaning but I see that it kind reads that way.
This was just an extreme example to illustrate the point.

> > Your everyday quality woofer will under normal operating conditions
> > usually be able to handle more input power with a vented box than a
> > sealed one.
>
> In fact the same driver will handle EXACTLY the same power at nearly all
> frequencies above the box tuning frequency.
> In this area where thermal limiting applies, what magic would allow it to do
> otherwise?
>

I was still referring to usable linear excursion and x-max here, not
thermal limitations. You get more usable input power within the linear
operating range. You are right that when it comes to maximum limits,
stated above as the VC gets a wee bit hotter at Fb.

(OK I admit if it's a vented voice coil there may be a minor
improvement
> from better air flow)
>
> However drivers are usually designed for one box type or the other, so your
> statement is pointless.
>
> MrT.
Anonymous

Humm, I don't quite know what's going on here because we don't seem to
be understanding each other. I hope we are not just trying to win a
wits war here. (with the word "we" I also include myself)

I think we both agree that X-max is a constant within normal operation.

Fb is determined from data in Z plots where there is a Zmin dip close
to Rdc in magnitude crossing at 0 deg phase. The woofer barely moves
there because there is also an Xmin dip. Both of these are facts. Do we
agree on this?

If so, then it stands to reason that the stored energy in the resonant
acoustic system takes over temporarily at Fb. This is what you must be
referring to. But input power does go somewhere at this low Z, low neg
EMF frequency point while the cone is theoretically nearly standing
still and the cab volume and port are doing all the SPL work. The
region I believe you were originally refering to, the slope region, is
everything lower than Flow, the lower Zmax (0 deg phase) peak. This is
not Fb! This peak is from where cone compliance and a still active part
of the port mass come into resonance. Fb-Flow varies per alignment,
maybe an octave or so on average?

So the 80 hz was an off hand point in reference to the slope region
below Flow where the signal is being attenuated outside of its
passband. The part I believe you were talking about originally. It
seems to me that if you were to put Fb much lower, that this slope
range lower than Fl in most standard alignments would get into the thin
signal region below 30 or so hz. and thus not really influence the
limits of linear operation, X-max and such.
Anonymous

Suffice to say, this is still in a different realm than everyday
listening to just any 12" woofer you happen across. I guess the point
I was getting at was that the scenario to create an alleged doppler
effect is extreme enough as to be well outside the realm of casual
listening, and in those scenarios where those extreme measures are met,
it's highly possible there will be numerous other physical limitation
factors in play such that it is arguable if you would just hear
"Doppler" or the onset of just conventional distortion. Additionally,
someone else astutely pointed out that even assuming all other
conditions are perfect so as to soley demonstrate a Doppler effect, the
observer's own hearing would be at risk, if not at least momentarily.
Anonymous

I do take your point about time constant, but c'mon, the application of
a continuous 500-1000 W is surely to create some amount of parameter
change in a fairly short amount of time. Maybe not in msec, but surely
within seconds or 10's of seconds?   My point being, that not even
the woofer will be operating "in spec" with an input like that, to be
worthy of isolating/evaluating Doppler.
Anonymous

"Randy Yates" <yates@ieee.org> wrote in message
news:wtpnsyub.fsf@ieee.org...
> If we know that a driver in a speaker system has a cone excursion of
> 0.5 inches peak-to-peak at 100 Hz, can we predict what the excursion
> of this driver will be at lower frequencies, assuming that the
> response is flat?

Yes, it's Newtonian mechanics; the reason the cone moves further at lower
frequencies is because it's accelerating for longer in the same direction
before the signal reverses.

So excursion is inversely proportional to the square of the frequency.

Tim
Anonymous

"Randy Yates" <randy.yates@sonyericsson.com> wrote in message
news:xxpacmif5dk.fsf@usrts005.corpusers.net...

> So then a woofer at 40 Hz will excursion
> (100/40)^2 = (5/2)^2 = 25/4 times as much as a woofer at 100 Hz.

Correct.

> instead of 9 mph, the "train" is now running at over 54 mph - no longer
> "slow" in my estimation, and thus this one piece of Mr. Howard's logic
> falls apart.

Incorrect. the velocity is given by "v = ft" where f is acceleration and t
is time. The acceleration is proportional to the current, and is constant
for constant SPL. So the velocity is inversely proportional to the
frequency. At 40hz, the average and peak speeds will be 2.5 times those at
100Hz. I make the peak speeds at 100Hz and 40Hz 11.4mph and 28.4 mph.

Of course 0.5 inches is a pretty large excursion at 100Hz, as is 3.25 inches
at 40Hz. With normal-size drivers, the sound level at one metre would be
pretty loud, up around 115dB.

Tim

Tim
Anonymous

<dpierce@cartchunk.org> wrote in message
> Dangling entity says:
> > Just the power input to achieve 1/2" excursion on a
> > 12" woofer *at 100 Hz* is a fairly extreme scenario,
> > as well, no? Is it possible the power may even lie
> > outside the comfortable handling limits of most
> > conventional woofers?
>
> Indeed. Assume a moderate efficiency of 1%, which corresponds
> to a half-space sensitivity of 92 dB/w, 122 dB would require an
> electric input of 1000 watts, which I would submit is a somewhat
> absurd scenario.

Well, no, it can be done using PA speakers. These can certainly deliver
around 120-125dB at 100Hz, but they are generally more efficient than
92dB/w - closer to 98dB/w.

125dB is a reasonable normal *maximum* output for decent PA speakers; it's
enough to achieve audience sound levels of 100-105dB in a room 50x50x20
feet. But for this sound level you'd normally use 15-inch drivers rather
than 12-inch.

Generally, the low-end PA equipment (Mackie, JBL, Celestion etc) will use
ported enclosures tuned to about 50-55Hhz

Plot an Eminence Kappa 15 driven with 400 watts in a 100-litre enclosure
tuned to 55Hz, with a second-order highpass frequency with cutoff 55Hz.
you'll get a sound level of 125dB at 100hz, with an excursion of 0.5 inch.
however, because of the tuned enclosure, the excursion will fall at lower
frequencies, down to a minimum at 55hz, below which it will rise again.

Tim
Anonymous

On 27 May 2005 00:22:00 -0700, "dangling entity"

>Just the power input to achieve 1/2" excursion on a 12" woofer *at 100
>Hz* is a fairly extreme scenario, as well, no?

Not necessarily - this is perfectly reasonable for say an IB subwoofer
system.
--

Stewart Pinkerton | Music is Art - Audio is Engineering
Anonymous

"Wessel Dirksen" <wdirksen@gmail.com> wrote in message
> I don't quite get what you mean here, so perhaps you can elaborate.
> Under normal operating conditions a standard electro magnetic driver
> moves very predicably as it is being stimulated by its input signal.
> There is no voodoo involved. Per also exists for bass reflex
> alignments, so even if one designed a ported nightmare, the input
> signal corresponding to the Xmax derived excursion limit would be
> reduced as well. If you drive a driver "too" hard with a regular audio
> signal it will have a longer excursion at low frequencies by definition
> and thus will have already exceeded Xmax. X max is rating of linear
> operation and not power handling.

Agreed, there should be no mechanical damage to a driver when operated at
Xmax.
The problem is that Xmax is more easily exceeded if one tries to get a given
level of SPL at frequencies below the cut off of a ported box, compared to
the same driver in a sealed box. A quick look at the slope of the response
curves below tuned frequency will show you why.

IME small ported enclosures suffer more mechanical damage from people
winding up the bass control, than do sealed eclosures.

>Most quality units can far exceed
> Xmax in operation before getting into thermal trouble.

Thermal limiting is rarely a problem when operating below the cut off
frequency of a ported enclosure.
Mechanical limits usually apply.

>Even applying a
> very large input voltage at a single higher frequency yet under Xmax
> would probably not be a problem for most woofers. I say under normal
> operating conditions because could put a 200V RMS 8kHz signal into a
> woofer and heat it up quite a bit, and you could block the cone from
> moving, etc.

One would not put an 8kHz signal into a "woofer". Hardly "normal operating
conditions"!
( I won't even comment on the 200V RMS stupidity, 5 kW at 8ohms)
ported box tuned frequency. For many small speakers this is around 50 Hz or
so.

> Your everyday quality woofer will under normal operating conditions
> usually be able to handle more input power with a vented box than a
> sealed one.

In fact the same driver will handle EXACTLY the same power at nearly all
frequencies above the box tuning frequency.
In this area where thermal limiting applies, what magic would allow it to do
otherwise?
(OK I admit if it's a vented voice coil there may be a minor improvement
from better air flow)

However drivers are usually designed for one box type or the other, so your
statement is pointless.

MrT.
Anonymous

"Wessel Dirksen" <wdirksen@gmail.com> wrote in message
> Ok got ya, but you are refering to the compliance controlled region of
> the alignment which is located on the slope itself thus out of the pass
> band. As you state below this really only practically counts for
> enclosures which don't go low at all. In most standard vented systems
> (I would guess Fb <80 hz or so), the opposite is true as excursion has
> a minimum at Fb. You would have to go quite a ways into the compliance
> controlled region, but then not too far before the absence of subsonic
> signal nullifies this impact.

Are you claiming there is no significant signal below 80Hz on the vast
majority of available CD's?
If it were so, then I'd agree with you.
In fact even 40 Hz is already a full octave lower than this, and can be a
problem with smaller enclosures and drivers, as I said.

> I was still referring to usable linear excursion and x-max here, not
> thermal limitations. You get more usable input power within the linear
> operating range.

The driver linear excursion and Xmax doesn't change.
In fact the whole benefit of a ported enclosure is that the input power
DECREASES at the tuned frequency, for the same output SPL.

>You are right that when it comes to maximum limits,
> the vented design has one slight disadvantage from its very advantage
> stated above as the VC gets a wee bit hotter at Fb.

No it doesn't. Because the impedance rises at Fb, the current drawn, and
therefore the power dissapated, and heat generated, decreases.

MrT.
Anonymous

>> Indeed. Assume a moderate efficiency of 1%, which corresponds
>> to a half-space sensitivity of 92 dB/w, 122 dB would require an
>> electric input of 1000 watts, which I would submit is a somewhat
>> absurd scenario.
>
>I lost a connection here. Are you saying that a speaker of
>1% efficiency (i.e., 92 dB/w/m) would require 1000 watts
>to move the cone 1/2" peak-to-peak at 100 Hz?

No, not necessarily. I am simply showing what the sound pressure
level generated by a 12" cone moving 1/2" at 100 Hz produces.
Then, as a separate excercise, I am also showing the amount of
electrical power required by a not atypical 12" woofer to produce
that sort of sound level.

Look at it another way: 122 dB at 1 meter, assuming half-space
radiation, is 10 acoustic watts. That 10 acoustic watts has to come
from somewhere, and IF the driver has an efficiency of 1%, 1000
watts are neede for it. If it's 2%, 500 watts are needed. If it's
..2%, 5000 watts are needed.
Anonymous

dpierce@cartchunk.org writes:
> [...]
> And, again, you're talking about ENORMOUS sound pressure
> levels anyway. 1/2" of excursion of a 12" cone is still
> 110 dB at 50 Hz and 98 dB at 20 Hz. What sort of music
> generates those sorts of sound pressure levels at those
> frequencies?

I would not consider 110 dB SPL at 50 Hz highly unusual, Dick.
I've measured just those sorts of peaks on the low/mid-low
frequency content of my ELO CDs at listening-chair position
and it's not all that loud, given that the majority of that
was in the low frequency region. It is "good-and-strong," but
not ear-piercingly loud.
--
% Randy Yates % "Ticket to the moon, flight leaves here today
%% Fuquay-Varina, NC % from Satellite 2"
%%% 919-577-9882 % 'Ticket To The Moon'
%%%% <yates@ieee.org> % *Time*, Electric Light Orchestra
Anonymous

"Wessel Dirksen" <wdirksen@gmail.com> writes:

> Humm, I don't quite know what's going on here because we don't seem to
> be understanding each other. I hope we are not just trying to win a
> wits war here. (with the word "we" I also include myself)
>
> I think we both agree that X-max is a constant within normal operation.

I probably abused the terminology. When I said Xmax, I meant the maximum
the woofer will swing for a given frequency and SPL level, not the mechanical
limits of the woofer.
--
% Randy Yates % "With time with what you've learned,
%% Fuquay-Varina, NC % they'll kiss the ground you walk
%%% 919-577-9882 % upon."
%%%% <yates@ieee.org> % '21st Century Man', *Time*, ELO
Anonymous

dpierce@cartchunk.org writes:

> Dangling entity says:
>> Just the power input to achieve 1/2" excursion on a
>> 12" woofer *at 100 Hz* is a fairly extreme scenario,
>> as well, no? Is it possible the power may even lie
>> outside the comfortable handling limits of most
>> conventional woofers?
>
> Indeed. Assume a moderate efficiency of 1%, which corresponds
> to a half-space sensitivity of 92 dB/w, 122 dB would require an
> electric input of 1000 watts, which I would submit is a somewhat
> absurd scenario.

I lost a connection here. Are you saying that a speaker of
1% efficiency (i.e., 92 dB/w/m) would require 1000 watts
to move the cone 1/2" peak-to-peak at 100 Hz?
--
% Randy Yates % "Remember the good old 1980's, when
%% Fuquay-Varina, NC % things were so uncomplicated?"
%%% 919-577-9882 % 'Ticket To The Moon'
%%%% <yates@ieee.org> % *Time*, Electric Light Orchestra
Anonymous

dpierce@cartchunk.org writes:

>>> Indeed. Assume a moderate efficiency of 1%, which corresponds
>>> to a half-space sensitivity of 92 dB/w, 122 dB would require an
>>> electric input of 1000 watts, which I would submit is a somewhat
>>> absurd scenario.
>>
>>I lost a connection here. Are you saying that a speaker of
>>1% efficiency (i.e., 92 dB/w/m) would require 1000 watts
>>to move the cone 1/2" peak-to-peak at 100 Hz?
>
> No, not necessarily. I am simply showing what the sound pressure
> level generated by a 12" cone moving 1/2" at 100 Hz produces.
> Then, as a separate excercise, I am also showing the amount of
> electrical power required by a not atypical 12" woofer to produce
> that sort of sound level.
>
> Look at it another way: 122 dB at 1 meter, assuming half-space
> radiation, is 10 acoustic watts. That 10 acoustic watts has to come
> from somewhere, and IF the driver has an efficiency of 1%, 1000
> watts are neede for it. If it's 2%, 500 watts are needed. If it's
> .2%, 5000 watts are needed.

Ok, I see. I agree, 122 dB SPL at 100 Hz is an exercise for the laboratory,
not a typical listening situation (for home reproduction - professional
sound reinforcement is another issue).

I'll have to re-think Howard's scenario using more realistic numbers for
the normal listening situation.

BTW, thanks for the corrections on my initial miscalculation. I saw my error
about an hour after I posted, but I think my follow-up post glided past your
response post in some usenet news server.
--
% Randy Yates % "So now it's getting late,
%% Fuquay-Varina, NC % and those who hesitate
%%% 919-577-9882 % got no one..."
%%%% <yates@ieee.org> % 'Waterfall', *Face The Music*, ELO
Anonymous

dpierce@cartchunk.org wrote:
> Dangling entity says:

>> Just the power input to achieve 1/2" excursion on a
>> 12" woofer *at 100 Hz* is a fairly extreme scenario,
>> as well, no? Is it possible the power may even lie
>> outside the comfortable handling limits of most
>> conventional woofers?
>
> Indeed. Assume a moderate efficiency of 1%, which
corresponds
> to a half-space sensitivity of 92 dB/w, 122 dB would
require an
> electric input of 1000 watts, which I would submit is a
somewhat
> absurd scenario.

Either that or its a high performance stage monitor. ;-)
Anonymous

Randy Yates <yates@ieee.org> writes:

> dpierce@cartchunk.org writes:
> > [...]
> > And, again, you're talking about ENORMOUS sound pressure
> > levels anyway. 1/2" of excursion of a 12" cone is still
> > 110 dB at 50 Hz and 98 dB at 20 Hz. What sort of music
> > generates those sorts of sound pressure levels at those
> > frequencies?
>
> I would not consider 110 dB SPL at 50 Hz highly unusual, Dick.
> I've measured just those sorts of peaks on the low/mid-low
> frequency content of my ELO CDs at listening-chair position
> and it's not all that loud, given that the majority of that
> was in the low frequency region. It is "good-and-strong," but
> not ear-piercingly loud.

I should add that I made my measurements using the unweighted setting.
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124
Anonymous

dpierce@cartchunk.org writes:

> >> Indeed. Assume a moderate efficiency of 1%, which corresponds
> >> to a half-space sensitivity of 92 dB/w, 122 dB would require an
> >> electric input of 1000 watts, which I would submit is a somewhat
> >> absurd scenario.
> >
> >I lost a connection here. Are you saying that a speaker of
> >1% efficiency (i.e., 92 dB/w/m) would require 1000 watts
> >to move the cone 1/2" peak-to-peak at 100 Hz?
>
> No, not necessarily. I am simply showing what the sound pressure
> level generated by a 12" cone moving 1/2" at 100 Hz produces.

How do you know that a 12" cone moving 1/2" at 100 Hz produces
122 dB SPL?

(I'm countering my sister post in which I said I understood - I
guess I don't.)
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124
Anonymous

>> >I lost a connection here. Are you saying that a speaker of
>> >1% efficiency (i.e., 92 dB/w/m) would require 1000 watts
>> >to move the cone 1/2" peak-to-peak at 100 Hz?
>> No, not necessarily. I am simply showing what the sound pressure
>> level generated by a 12" cone moving 1/2" at 100 Hz produces.
>
>How do you know that a 12" cone moving 1/2" at 100 Hz produces
>122 dB SPL?

The total sound power (from a variety of source e.g. Small, etc)
from a piston radiating at frequencies where the wavelength is
substantially larger than the piston is:

Pa = p/(2 pi c) (Sd w^2 x)^2

where

Pa - acoustic power, in watts
p - density of air, 1.18 kg/m^3
c - speed of sound, 343 m/s
Sd - emissive area of piston, m^2
w - radian frequency = 2 pi f
x - excursion

p/(2 pi c) is, in effect, a constant for STP conditions, and reduces
to a value of about 5.48x10^-4.

So, in the example given, a typical 12" woofer has an emissive area
of about 0.073 m^2, w = 2 pi * 100 = 628 r/s, x is 0.0063 m, thus
Pa ~= 18 watts.

An omnidirectional source radiating 1 watt produces a sound pressure
1 meter away of 109 dB SPL. The SPL relative to 1 watt is thus:

SPL = 109 + 10 log (Pa)

For 18 watts, this is

SPL = 109 + 10 log 18

SPL = 109 + 10 * 1.26

SPL = 109 + 12.6

SPL = 121.6 dB
Anonymous

On 01 Jun 2005 14:02:11 -0400, Randy Yates
<randy.yates@sonyericsson.com> wrote:

>dpierce@cartchunk.org writes:
>
>> >> Indeed. Assume a moderate efficiency of 1%, which corresponds
>> >> to a half-space sensitivity of 92 dB/w, 122 dB would require an
>> >> electric input of 1000 watts, which I would submit is a somewhat
>> >> absurd scenario.
>> >
>> >I lost a connection here. Are you saying that a speaker of
>> >1% efficiency (i.e., 92 dB/w/m) would require 1000 watts
>> >to move the cone 1/2" peak-to-peak at 100 Hz?
>>
>> No, not necessarily. I am simply showing what the sound pressure
>> level generated by a 12" cone moving 1/2" at 100 Hz produces.
>
>How do you know that a 12" cone moving 1/2" at 100 Hz produces
>122 dB SPL?

Because it just does, when you work out the radiated acoustic power
(into half-space here, I think). Even Dick canna change the laws o'
physics, cap'n.................

--

Stewart Pinkerton | Music is Art - Audio is Engineering
Anonymous

dpierce@cartchunk.org writes:
> [...]
> Indeed. Assume a moderate efficiency of 1%, which corresponds
> to a half-space sensitivity of 92 dB/w, 122 dB would require an
> electric input of 1000 watts, which I would submit is a somewhat
> absurd scenario.

An efficiency of 1% would require an electrical input of 1800 watts
to provide the 18 acoustic watts computed in your subsequent post,
would it not?

Yes, this actually strenghens your argument and weakens mine, but
I'm just trying to get the computations consistent at this point.
--
% Randy Yates % "With time with what you've learned,
%% Fuquay-Varina, NC % they'll kiss the ground you walk
%%% 919-577-9882 % upon."
%%%% <yates@ieee.org> % '21st Century Man', *Time*, ELO
Anonymous

dpierce@cartchunk.org writes:

>>> >I lost a connection here. Are you saying that a speaker of
>>> >1% efficiency (i.e., 92 dB/w/m) would require 1000 watts
>>> >to move the cone 1/2" peak-to-peak at 100 Hz?
>>> No, not necessarily. I am simply showing what the sound pressure
>>> level generated by a 12" cone moving 1/2" at 100 Hz produces.
>>
>>How do you know that a 12" cone moving 1/2" at 100 Hz produces
>>122 dB SPL?
>
> The total sound power (from a variety of source e.g. Small, etc)
> from a piston radiating at frequencies where the wavelength is
> substantially larger than the piston is:
>
> Pa = p/(2 pi c) (Sd w^2 x)^2
>
> where
>
> Pa - acoustic power, in watts
> p - density of air, 1.18 kg/m^3
> c - speed of sound, 343 m/s
> Sd - emissive area of piston, m^2
> w - radian frequency = 2 pi f
> x - excursion
>
> p/(2 pi c) is, in effect, a constant for STP conditions, and reduces
> to a value of about 5.48x10^-4.
>
> So, in the example given, a typical 12" woofer has an emissive area
> of about 0.073 m^2, w = 2 pi * 100 = 628 r/s, x is 0.0063 m, thus
> Pa ~= 18 watts.
>
> An omnidirectional source radiating 1 watt produces a sound pressure
> 1 meter away of 109 dB SPL. The SPL relative to 1 watt is thus:
>
> SPL = 109 + 10 log (Pa)
>
> For 18 watts, this is
>
> SPL = 109 + 10 log 18
>
> SPL = 109 + 10 * 1.26
>
> SPL = 109 + 12.6
>
> SPL = 121.6 dB

OK, thanks Dick. I'm assuming that "x" is the "peak" excursion and not
"peak-to-peak"? Under this assumption, my computations match yours.
--
% Randy Yates % "Remember the good old 1980's, when
%% Fuquay-Varina, NC % things were so uncomplicated?"
%%% 919-577-9882 % 'Ticket To The Moon'
%%%% <yates@ieee.org> % *Time*, Electric Light Orchestra
Anonymous

Randy Yates <yates@ieee.org> writes:

> dpierce@cartchunk.org writes:
>
>>>> >I lost a connection here. Are you saying that a speaker of
>>>> >1% efficiency (i.e., 92 dB/w/m) would require 1000 watts
>>>> >to move the cone 1/2" peak-to-peak at 100 Hz?
>>>> No, not necessarily. I am simply showing what the sound pressure
>>>> level generated by a 12" cone moving 1/2" at 100 Hz produces.
>>>
>>>How do you know that a 12" cone moving 1/2" at 100 Hz produces
>>>122 dB SPL?
>>
>> The total sound power (from a variety of source e.g. Small, etc)
>> from a piston radiating at frequencies where the wavelength is
>> substantially larger than the piston is:
>>
>> Pa = p/(2 pi c) (Sd w^2 x)^2
>>
>> where
>>
>> Pa - acoustic power, in watts
>> p - density of air, 1.18 kg/m^3
>> c - speed of sound, 343 m/s
>> Sd - emissive area of piston, m^2
>> w - radian frequency = 2 pi f
>> x - excursion
>>
>> p/(2 pi c) is, in effect, a constant for STP conditions, and reduces
>> to a value of about 5.48x10^-4.
>>
>> So, in the example given, a typical 12" woofer has an emissive area
>> of about 0.073 m^2, w = 2 pi * 100 = 628 r/s, x is 0.0063 m, thus
>> Pa ~= 18 watts.
>>
>> An omnidirectional source radiating 1 watt produces a sound pressure
>> 1 meter away of 109 dB SPL. The SPL relative to 1 watt is thus:
>>
>> SPL = 109 + 10 log (Pa)
>>
>> For 18 watts, this is
>>
>> SPL = 109 + 10 log 18
>>
>> SPL = 109 + 10 * 1.26
>>
>> SPL = 109 + 12.6
>>
>> SPL = 121.6 dB
>
> OK, thanks Dick. I'm assuming that "x" is the "peak" excursion and not
> "peak-to-peak"? Under this assumption, my computations match yours.

They were generated with this Matlab m-file:

function Pa = pa(speakerdiameterininches, peakconeexcursionininches, linearfrequency)

p = 1.18; % density of air, in kg/m^3
c = 343; % speed of sound, in m/s
metersperinch = 1 / 39.4;
d = speakerdiameterininches * metersperinch;
x = peakconeexcursionininches * metersperinch;
w = 2 * pi * linearfrequency;
Sd = pi * (d/2)^2;
Pa = (p/(2*pi*c)) * (Sd * w^2 * x)^2;

which results in

» pa(12, 0.25, 100)

ans =

18.23622431673969

»
--
% Randy Yates % "Remember the good old 1980's, when
%% Fuquay-Varina, NC % things were so uncomplicated?"
%%% 919-577-9882 % 'Ticket To The Moon'
%%%% <yates@ieee.org> % *Time*, Electric Light Orchestra
Anonymous

dpierce@cartchunk.org writes:
> [...]
> The total sound power (from a variety of source e.g. Small, etc)
> from a piston radiating at frequencies where the wavelength is
> substantially larger than the piston is:
>
> Pa = p/(2 pi c) (Sd w^2 x)^2
>
> where
>
> Pa - acoustic power, in watts
> p - density of air, 1.18 kg/m^3
> c - speed of sound, 343 m/s
> Sd - emissive area of piston, m^2
> w - radian frequency = 2 pi f
> x - excursion

There must be some assumptions made that I'm not aware of here. Surely
not ALL speakers produce 18 acoustic watts when their woofers are
traveling 0.5" p-p, are they? For example, a horn woofer won't
be travelling as far as an infinite-baffle direct radiator for
a given acoustic output power, will it? If not, then where is
the discrepancy in this equation? Is there an assumption on
--
% Randy Yates % "Maybe one day I'll feel her cold embrace,
%% Fuquay-Varina, NC % and kiss her interface,
%%% 919-577-9882 % til then, I'll leave her alone."
%%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO
Anonymous

Randy Yates wrote:
>There must be some assumptions made that I'm not aware of here.
>Surely not ALL speakers produce 18 acoustic watts when their
>woofers are traveling 0.5" p-p, are they?

into the same solid angle, they MUST.

>For example, a horn woofer won't be travelling as far as an
>infinite-baffle direct radiator for a given acoustic output power,
>will it?

For the same ACOUSTIC power or the same SPL?

>If not, then where is the discrepancy in this equation? Is there

The basis for the equation is as follows (from Small):

"The acoustic power radiated by the system is:

Pa = | Uo |^2 Rar (2)

where

PA acoustic output power
Uo total system volume velocity

"Eq. (2) is generally valid to the upper limit of the
driver piston range because the driver is normally the
only significant radiator at frequencies high enough
for the aperture spacings to become important."

"In a recent paper [5], Allison and Berkovitz have
demonstrated that the low-frequency load on a loud-
speaker system in a typical listening room is essen-
tially that for one side of a piston mounted in an
infinite baffle. The resistive part of this radiation

Rar = p0 w^2 / (2 pi c) (3)

where

p0 density of air
c velocity of sound in air.

"Eq. (3) is valid only in the system piston range, but
within this range the value of Rar is independent of
the size of the enclosure or its apertures."

JAES, 1972 June

So the "hidden" variable here is the real part of the radiation
impedance, which is VERY different for a horn loaded system than
Anonymous

On Thu, 02 Jun 2005 02:43:48 GMT, Randy Yates <yates@ieee.org> wrote:

>dpierce@cartchunk.org writes:
>> [...]
>> The total sound power (from a variety of source e.g. Small, etc)
>> from a piston radiating at frequencies where the wavelength is
>> substantially larger than the piston is:
>>
>> Pa = p/(2 pi c) (Sd w^2 x)^2
>>
>> where
>>
>> Pa - acoustic power, in watts
>> p - density of air, 1.18 kg/m^3
>> c - speed of sound, 343 m/s
>> Sd - emissive area of piston, m^2
>> w - radian frequency = 2 pi f
>> x - excursion
>
>There must be some assumptions made that I'm not aware of here. Surely
>not ALL speakers produce 18 acoustic watts when their woofers are
>traveling 0.5" p-p, are they? For example, a horn woofer won't
>be travelling as far as an infinite-baffle direct radiator for
>a given acoustic output power, will it? If not, then where is
>the discrepancy in this equation? Is there an assumption on

The horn will not be radiating into half-space, but into a smaller
steroid, depending on its directivity. You could argue that it's still
producing 18 watts total radiated power, but the SPL in front of the
horn will be quite different.
--

Stewart Pinkerton | Music is Art - Audio is Engineering
Anonymous

dpierce@cartchunk.org writes:

> Randy Yates wrote:
>>There must be some assumptions made that I'm not aware of here.
>>Surely not ALL speakers produce 18 acoustic watts when their
>>woofers are traveling 0.5" p-p, are they?
>
> If they are direct radiators with the same emissive area radiating
> into the same solid angle, they MUST.
>
>>For example, a horn woofer won't be travelling as far as an
>>infinite-baffle direct radiator for a given acoustic output power,
>>will it?
>
> For the same ACOUSTIC power or the same SPL?
>
>>If not, then where is the discrepancy in this equation? Is there
>
> The basis for the equation is as follows (from Small):
>
> "The acoustic power radiated by the system is:
>
> Pa = | Uo |^2 Rar (2)
>
> where
>
> PA acoustic output power
> Uo total system volume velocity

and also previously wrote:

> The total sound power (from a variety of source e.g. Small, etc)
> from a piston radiating at frequencies where the wavelength is
> substantially larger than the piston is:
>
> Pa = p/(2 pi c) (Sd w^2 x)^2
>
> where
>
> Pa - acoustic power, in watts
> p - density of air, 1.18 kg/m^3
> c - speed of sound, 343 m/s
> Sd - emissive area of piston, m^2
> w - radian frequency = 2 pi f
> x - excursion

OK so obviously

| Uo | = (Sd * w^2 * x),

but I don't see how you get this. First of all, the (peak) velocity of
a driver producing a sinusoid at radian frequency w is going to be w,
not w^2. Then also why use the peak excursion rather than
peak-to-peak? I'm shooting in the dark here since I don't really know
how "total system volume velocity" is defined.

What I mean to say, Dick, is that I trust your equations, but I
understand?
--
% Randy Yates % "I met someone who looks alot like you,
%% Fuquay-Varina, NC % she does the things you do,
%%% 919-577-9882 % but she is an IBM."
%%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO
Anonymous

"Randy Yates" <yates@ieee.org> wrote in message
news:1x7l7aqo.fsf@ieee.org...
> dpierce@cartchunk.org writes:
> > [...]
> > Indeed. Assume a moderate efficiency of 1%, which corresponds
> > to a half-space sensitivity of 92 dB/w, 122 dB would require an
> > electric input of 1000 watts, which I would submit is a somewhat
> > absurd scenario.
>
> An efficiency of 1% would require an electrical input of 1800 watts
> to provide the 18 acoustic watts computed in your subsequent post,
> would it not?

Obviously 122dB is 30 dB above 92 dB and would require 1000* 1Watt.
It seems the "Assumption" of 1% efficiency is only approximate. 0.5% would
be a more usual figure from the subsequent calculations.

MrT.