On 01 Jun 2005 14:02:11 -0400, Randy Yates
wrote:

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

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

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."
%%%% % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr

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'
%%%% % *Time*, Electric Light Orchestra
http://home.earthlink.net/~yatescr

Randy Yates writes:

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'
%%%% % *Time*, Electric Light Orchestra
http://home.earthlink.net/~yatescr

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
radiation impedance?
--
% 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."
%%%% % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

On Thu, 02 Jun 2005 02:43:48 GMT, Randy Yates wrote:

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
radiation impedance?

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

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
an assumption on radiation impedance?

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
Rar resistive part of radiation load on system.
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
load [3, p. 216] is

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

where

p0 density of air
w steady-state radian frequency
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."

"Direct Radiator Loudspeaker System Analysis,"
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
it is for a direct-radiator.

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
an assumption on radiation impedance?

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
Rar resistive part of radiation load on system.
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
don't understand how they were derived. Can you please help me
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."
%%%% % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

"Randy Yates" wrote in message
...
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.