"Rusty Boudreaux" wrote in message ...
"Svante" wrote in message
om...
Sigh... If you were able to put the flashlight within a
fraction of a
wavelength from the mirror the power WOULD increase. In the
audio case
this is easily accomplished. I think it is frequently used with
antennas too (correct me if I am wrong). Power DOES increase.
Read the
parallel thread, in particular

Really? Adding a mirror changes the power output of the
filament? So all we have to do is take a few mirrors and we can
transform a tiny flashlight into a death ray to knock airplanes
out of the sky. Better yet, couple it with some solar cells and
we could power the Earth off a flashlight.

The problem with this analogy is that it is very hard to put the
filament within a fraction of a wavelength to the mirror.


Antennas: An isotropic radiator is an antenna that radiates
equally in all directions. Antenna gain is the improvement above
an isotropic radiator. However, the gain comes at a price -
directivity. Look at common antenna radiation patterns. The
higher the gain the narrower the radiation pattern. Total
radiation power is the same...it's just squeezed into a narrower
beam. Study the photos he
http://www.hdtvprimer.com/ANTENNAS/types.html

Agreed. Antennas already have a high efficiency, and obviously it can
never go above 100%. Speakers don't, so there is plenty of power left
over. Increasing the radiation resistance will make use of some of
that. I'm sorry about making the antenna analogy, it does not hold
when it comes to efficiency.

It the same thing with audio. Gain can increase at the price of
directivity. Power can not increase. There is no free lunch in
the physical world.

Consider this parallel; an electrical voltage source has an output
impedance of 10 kohms. You connect a load of 100 ohms to this source.
A certain (small) amount of power will be dissipated in the 100 ohm
resistor when you turn the source on. Now change the load resistor to
200 ohms. What will happen to the power in the load resistor? It will
almost double. Efficiency will almost double. This is what happens in
the loudspeaker, only the 10 kohm resistor corresponds to the
mechanical impedances (mass etc) of the cone, and the load resistor
corresponds to the radiation resistance. In the antenna case the
source impedance would be ~0 ohms, and no efficiency change would
occur.

"Rusty Boudreaux" wrote in message ...
"Svante" wrote in message
om...
Sigh... If you were able to put the flashlight within a
fraction of a
wavelength from the mirror the power WOULD increase. In the
audio case
this is easily accomplished. I think it is frequently used with
antennas too (correct me if I am wrong). Power DOES increase.
Read the
parallel thread, in particular

Really? Adding a mirror changes the power output of the
filament? So all we have to do is take a few mirrors and we can
transform a tiny flashlight into a death ray to knock airplanes
out of the sky. Better yet, couple it with some solar cells and
we could power the Earth off a flashlight.

The problem with this analogy is that it is very hard to put the
filament within a fraction of a wavelength to the mirror.


Antennas: An isotropic radiator is an antenna that radiates
equally in all directions. Antenna gain is the improvement above
an isotropic radiator. However, the gain comes at a price -
directivity. Look at common antenna radiation patterns. The
higher the gain the narrower the radiation pattern. Total
radiation power is the same...it's just squeezed into a narrower
beam. Study the photos he
http://www.hdtvprimer.com/ANTENNAS/types.html

Agreed. Antennas already have a high efficiency, and obviously it can
never go above 100%. Speakers don't, so there is plenty of power left
over. Increasing the radiation resistance will make use of some of
that. I'm sorry about making the antenna analogy, it does not hold
when it comes to efficiency.

It the same thing with audio. Gain can increase at the price of
directivity. Power can not increase. There is no free lunch in
the physical world.

Consider this parallel; an electrical voltage source has an output
impedance of 10 kohms. You connect a load of 100 ohms to this source.
A certain (small) amount of power will be dissipated in the 100 ohm
resistor when you turn the source on. Now change the load resistor to
200 ohms. What will happen to the power in the load resistor? It will
almost double. Efficiency will almost double. This is what happens in
the loudspeaker, only the 10 kohm resistor corresponds to the
mechanical impedances (mass etc) of the cone, and the load resistor
corresponds to the radiation resistance. In the antenna case the
source impedance would be ~0 ohms, and no efficiency change would
occur.

"Rusty Boudreaux" wrote in message ...
"Svante" wrote in message
om...
Sigh... If you were able to put the flashlight within a
fraction of a
wavelength from the mirror the power WOULD increase. In the
audio case
this is easily accomplished. I think it is frequently used with
antennas too (correct me if I am wrong). Power DOES increase.
Read the
parallel thread, in particular

Really? Adding a mirror changes the power output of the
filament? So all we have to do is take a few mirrors and we can
transform a tiny flashlight into a death ray to knock airplanes
out of the sky. Better yet, couple it with some solar cells and
we could power the Earth off a flashlight.

The problem with this analogy is that it is very hard to put the
filament within a fraction of a wavelength to the mirror.


Antennas: An isotropic radiator is an antenna that radiates
equally in all directions. Antenna gain is the improvement above
an isotropic radiator. However, the gain comes at a price -
directivity. Look at common antenna radiation patterns. The
higher the gain the narrower the radiation pattern. Total
radiation power is the same...it's just squeezed into a narrower
beam. Study the photos he
http://www.hdtvprimer.com/ANTENNAS/types.html

Agreed. Antennas already have a high efficiency, and obviously it can
never go above 100%. Speakers don't, so there is plenty of power left
over. Increasing the radiation resistance will make use of some of
that. I'm sorry about making the antenna analogy, it does not hold
when it comes to efficiency.

It the same thing with audio. Gain can increase at the price of
directivity. Power can not increase. There is no free lunch in
the physical world.

Consider this parallel; an electrical voltage source has an output
impedance of 10 kohms. You connect a load of 100 ohms to this source.
A certain (small) amount of power will be dissipated in the 100 ohm
resistor when you turn the source on. Now change the load resistor to
200 ohms. What will happen to the power in the load resistor? It will
almost double. Efficiency will almost double. This is what happens in
the loudspeaker, only the 10 kohm resistor corresponds to the
mechanical impedances (mass etc) of the cone, and the load resistor
corresponds to the radiation resistance. In the antenna case the
source impedance would be ~0 ohms, and no efficiency change would
occur.

"Rusty Boudreaux" wrote in message ...
"Svante" wrote in message
om...
Sigh... If you were able to put the flashlight within a
fraction of a
wavelength from the mirror the power WOULD increase. In the
audio case
this is easily accomplished. I think it is frequently used with
antennas too (correct me if I am wrong). Power DOES increase.
Read the
parallel thread, in particular

Really? Adding a mirror changes the power output of the
filament? So all we have to do is take a few mirrors and we can
transform a tiny flashlight into a death ray to knock airplanes
out of the sky. Better yet, couple it with some solar cells and
we could power the Earth off a flashlight.

The problem with this analogy is that it is very hard to put the
filament within a fraction of a wavelength to the mirror.


Antennas: An isotropic radiator is an antenna that radiates
equally in all directions. Antenna gain is the improvement above
an isotropic radiator. However, the gain comes at a price -
directivity. Look at common antenna radiation patterns. The
higher the gain the narrower the radiation pattern. Total
radiation power is the same...it's just squeezed into a narrower
beam. Study the photos he
http://www.hdtvprimer.com/ANTENNAS/types.html

Agreed. Antennas already have a high efficiency, and obviously it can
never go above 100%. Speakers don't, so there is plenty of power left
over. Increasing the radiation resistance will make use of some of
that. I'm sorry about making the antenna analogy, it does not hold
when it comes to efficiency.

It the same thing with audio. Gain can increase at the price of
directivity. Power can not increase. There is no free lunch in
the physical world.

Consider this parallel; an electrical voltage source has an output
impedance of 10 kohms. You connect a load of 100 ohms to this source.
A certain (small) amount of power will be dissipated in the 100 ohm
resistor when you turn the source on. Now change the load resistor to
200 ohms. What will happen to the power in the load resistor? It will
almost double. Efficiency will almost double. This is what happens in
the loudspeaker, only the 10 kohm resistor corresponds to the
mechanical impedances (mass etc) of the cone, and the load resistor
corresponds to the radiation resistance. In the antenna case the
source impedance would be ~0 ohms, and no efficiency change would
occur.

On Thu, 8 Jan 2004 00:29:16 -0600, "Rusty Boudreaux"
wrote:

"Svante" wrote in message
. com...
Sigh... If you were able to put the flashlight within a
fraction of a
wavelength from the mirror the power WOULD increase. In the
audio case
this is easily accomplished. I think it is frequently used with
antennas too (correct me if I am wrong). Power DOES increase.
Read the
parallel thread, in particular

Really? Adding a mirror changes the power output of the
filament?

Well at least it will make it hotter, if focussed back on itself and
guess what would happen then.

So all we have to do is take a few mirrors and we can
transform a tiny flashlight into a death ray to knock airplanes
out of the sky. Better yet, couple it with some solar cells and
we could power the Earth off a flashlight.

Don't be silly now, nobody is claiming perpetual motion.

As it happens. . . one already has 100MW pulses but it is not nearly
enough for a "death ray to knock airplanes out of the sky". For that
you need energy and making pulses shorter does not help there. Try to
distinguish between energy and power, and amplitude and power, and
coherent and incoherent summation, and effeciency changes and
perpetual motion machines, if you are going to get sarcastic.

On Thu, 8 Jan 2004 00:29:16 -0600, "Rusty Boudreaux"
wrote:

"Svante" wrote in message
. com...
Sigh... If you were able to put the flashlight within a
fraction of a
wavelength from the mirror the power WOULD increase. In the
audio case
this is easily accomplished. I think it is frequently used with
antennas too (correct me if I am wrong). Power DOES increase.
Read the
parallel thread, in particular

Really? Adding a mirror changes the power output of the
filament?

Well at least it will make it hotter, if focussed back on itself and
guess what would happen then.

So all we have to do is take a few mirrors and we can
transform a tiny flashlight into a death ray to knock airplanes
out of the sky. Better yet, couple it with some solar cells and
we could power the Earth off a flashlight.

Don't be silly now, nobody is claiming perpetual motion.

As it happens. . . one already has 100MW pulses but it is not nearly
enough for a "death ray to knock airplanes out of the sky". For that
you need energy and making pulses shorter does not help there. Try to
distinguish between energy and power, and amplitude and power, and
coherent and incoherent summation, and effeciency changes and
perpetual motion machines, if you are going to get sarcastic.

On Thu, 8 Jan 2004 00:29:16 -0600, "Rusty Boudreaux"
wrote:

"Svante" wrote in message
. com...
Sigh... If you were able to put the flashlight within a
fraction of a
wavelength from the mirror the power WOULD increase. In the
audio case
this is easily accomplished. I think it is frequently used with
antennas too (correct me if I am wrong). Power DOES increase.
Read the
parallel thread, in particular

Really? Adding a mirror changes the power output of the
filament?

Well at least it will make it hotter, if focussed back on itself and
guess what would happen then.

So all we have to do is take a few mirrors and we can
transform a tiny flashlight into a death ray to knock airplanes
out of the sky. Better yet, couple it with some solar cells and
we could power the Earth off a flashlight.

Don't be silly now, nobody is claiming perpetual motion.

As it happens. . . one already has 100MW pulses but it is not nearly
enough for a "death ray to knock airplanes out of the sky". For that
you need energy and making pulses shorter does not help there. Try to
distinguish between energy and power, and amplitude and power, and
coherent and incoherent summation, and effeciency changes and
perpetual motion machines, if you are going to get sarcastic.

On Thu, 8 Jan 2004 00:29:16 -0600, "Rusty Boudreaux"
wrote:

"Svante" wrote in message
. com...
Sigh... If you were able to put the flashlight within a
fraction of a
wavelength from the mirror the power WOULD increase. In the
audio case
this is easily accomplished. I think it is frequently used with
antennas too (correct me if I am wrong). Power DOES increase.
Read the
parallel thread, in particular

Really? Adding a mirror changes the power output of the
filament?

Well at least it will make it hotter, if focussed back on itself and
guess what would happen then.

So all we have to do is take a few mirrors and we can
transform a tiny flashlight into a death ray to knock airplanes
out of the sky. Better yet, couple it with some solar cells and
we could power the Earth off a flashlight.

Don't be silly now, nobody is claiming perpetual motion.

As it happens. . . one already has 100MW pulses but it is not nearly
enough for a "death ray to knock airplanes out of the sky". For that
you need energy and making pulses shorter does not help there. Try to
distinguish between energy and power, and amplitude and power, and
coherent and incoherent summation, and effeciency changes and
perpetual motion machines, if you are going to get sarcastic.

Svante wrote:

efficiency. This is not only an issue of directivity, it is also an
issue of radiation resistance (which is doubled).

Aha.

Please READ the parallel thread I referred to!

OK

I can back this up with some math if you are interested.


Kind regards

Peter Larsen


--
*******************************************
* My site is at: http://www.muyiovatki.dk *
*******************************************

Svante wrote:

efficiency. This is not only an issue of directivity, it is also an
issue of radiation resistance (which is doubled).

Aha.

Please READ the parallel thread I referred to!

OK

I can back this up with some math if you are interested.


Kind regards

Peter Larsen


--
*******************************************
* My site is at: http://www.muyiovatki.dk *
*******************************************

Svante wrote:

efficiency. This is not only an issue of directivity, it is also an
issue of radiation resistance (which is doubled).

Aha.

Please READ the parallel thread I referred to!

OK

I can back this up with some math if you are interested.


Kind regards

Peter Larsen


--
*******************************************
* My site is at: http://www.muyiovatki.dk *
*******************************************

Svante wrote:

efficiency. This is not only an issue of directivity, it is also an
issue of radiation resistance (which is doubled).

Aha.

Please READ the parallel thread I referred to!

OK

I can back this up with some math if you are interested.


Kind regards

Peter Larsen


--
*******************************************
* My site is at: http://www.muyiovatki.dk *
*******************************************

Peter Larsen wrote:


Intensity on some locations increase at the expense of intensity at
other locations, total power from anything, be it loudspeaker-units or
shocking pink lasers does not change becuase of what their surroundings
do with their output


I'm not so sure this is a universal truth. I think it depends
somewhat on where one draws the boundary between the
"anything" under consideration and the surroundings.

For example, if we considered the "anything" to be an op amp,
and the "surroundings" to include its feedback network,
wouldn't the output power of the op amp depend on what the
feedback network does with the op amp's output?

Peter Larsen wrote:


Intensity on some locations increase at the expense of intensity at
other locations, total power from anything, be it loudspeaker-units or
shocking pink lasers does not change becuase of what their surroundings
do with their output


I'm not so sure this is a universal truth. I think it depends
somewhat on where one draws the boundary between the
"anything" under consideration and the surroundings.

For example, if we considered the "anything" to be an op amp,
and the "surroundings" to include its feedback network,
wouldn't the output power of the op amp depend on what the
feedback network does with the op amp's output?

Peter Larsen wrote:


Intensity on some locations increase at the expense of intensity at
other locations, total power from anything, be it loudspeaker-units or
shocking pink lasers does not change becuase of what their surroundings
do with their output


I'm not so sure this is a universal truth. I think it depends
somewhat on where one draws the boundary between the
"anything" under consideration and the surroundings.

For example, if we considered the "anything" to be an op amp,
and the "surroundings" to include its feedback network,
wouldn't the output power of the op amp depend on what the
feedback network does with the op amp's output?

Peter Larsen wrote:


Intensity on some locations increase at the expense of intensity at
other locations, total power from anything, be it loudspeaker-units or
shocking pink lasers does not change becuase of what their surroundings
do with their output


I'm not so sure this is a universal truth. I think it depends
somewhat on where one draws the boundary between the
"anything" under consideration and the surroundings.

For example, if we considered the "anything" to be an op amp,
and the "surroundings" to include its feedback network,
wouldn't the output power of the op amp depend on what the
feedback network does with the op amp's output?

On Thu, 08 Jan 2004 16:29:58 GMT, henryf
wrote:

Peter Larsen wrote:


Intensity on some locations increase at the expense of intensity at
other locations, total power from anything, be it loudspeaker-units or
shocking pink lasers does not change becuase of what their surroundings
do with their output


I'm not so sure this is a universal truth. I think it depends
somewhat on where one draws the boundary between the
"anything" under consideration and the surroundings.

For example, if we considered the "anything" to be an op amp,
and the "surroundings" to include its feedback network,
wouldn't the output power of the op amp depend on what the
feedback network does with the op amp's output?

No, that only affects the gain.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Thu, 08 Jan 2004 16:29:58 GMT, henryf
wrote:

Peter Larsen wrote:


Intensity on some locations increase at the expense of intensity at
other locations, total power from anything, be it loudspeaker-units or
shocking pink lasers does not change becuase of what their surroundings
do with their output


I'm not so sure this is a universal truth. I think it depends
somewhat on where one draws the boundary between the
"anything" under consideration and the surroundings.

For example, if we considered the "anything" to be an op amp,
and the "surroundings" to include its feedback network,
wouldn't the output power of the op amp depend on what the
feedback network does with the op amp's output?

No, that only affects the gain.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Thu, 08 Jan 2004 16:29:58 GMT, henryf
wrote:

Peter Larsen wrote:


Intensity on some locations increase at the expense of intensity at
other locations, total power from anything, be it loudspeaker-units or
shocking pink lasers does not change becuase of what their surroundings
do with their output


I'm not so sure this is a universal truth. I think it depends
somewhat on where one draws the boundary between the
"anything" under consideration and the surroundings.

For example, if we considered the "anything" to be an op amp,
and the "surroundings" to include its feedback network,
wouldn't the output power of the op amp depend on what the
feedback network does with the op amp's output?

No, that only affects the gain.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Thu, 08 Jan 2004 16:29:58 GMT, henryf
wrote:

Peter Larsen wrote:


Intensity on some locations increase at the expense of intensity at
other locations, total power from anything, be it loudspeaker-units or
shocking pink lasers does not change becuase of what their surroundings
do with their output


I'm not so sure this is a universal truth. I think it depends
somewhat on where one draws the boundary between the
"anything" under consideration and the surroundings.

For example, if we considered the "anything" to be an op amp,
and the "surroundings" to include its feedback network,
wouldn't the output power of the op amp depend on what the
feedback network does with the op amp's output?

No, that only affects the gain.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On 8 Jan 2004 00:45:28 -0800, (Svante) wrote:

Sigh... If you were able to put the flashlight within a fraction of a
wavelength from the mirror the power WOULD increase. In the audio case
this is easily accomplished. I think it is frequently used with
antennas too (correct me if I am wrong). Power DOES increase. Read the
parallel thread, in particular

It's apparent that your definition of power is not the
same as is commonly used.

In the case of antennas, for example, a theoretical isotropic
source has efficiency of 100%, and gain of zero.

A real dipole might have 90% efficiency, but real gain
of 1.8 dB as measured far-field broadside by virtue of
nulls off the ends of the dipole. The measured response
has increased, but 'power' has decreased.

Add a reflector and director, and you may have 7 dB
gain, but still only 90% efficiency - IOW, the power
is the same.

Stack a second three element yagi, and you will have
maybe 10 dB gain, but STILL only 90% efficiency, and
still LESS power than the zero gain but 100% efficient
isotropic source.

'IF' measured signal in front of the array had anything
to do with power, than efficiencies would routinely
be way over 100%, which is impossible I believe.

Ok, so my reference to antennas maybe wasn't a good one, at least not
when it comes to efficiency. I presume that the reason for that is
that antennas have such a high efficiency from the start, and
obviously, efficiency can never exceed 100%. In the loudspeaker case,
the radiation resistance is a terribly small part of the total
impedance seen by the voice coil. Doubling the radiaton resistance
will double the power output, since the velocity of the cone
essentially will remain the same. Introducing a wall (ground plane, in
the analogy) would double the radiaton resistance. Maybe it would work
for antennas as well if you see it like this (I'm on thin ice here...
:-) ) :
Take a stick of metal and say it is an antenna. This antenna has an
electric impedance from the radiaton impedance. Let's say it is 100
ohms. Let's also say that this antenna has no ground plane and that
the antenna is short compared to the wavelength. Take a resistor of 10
kohms and connect it in series with the antenna and connect the whole
thing to a radio transmitter. Only a fraction of the power delivered
by the source will actually be radiated, most of the power will be
dissipated in the resistor. Now introduce a ground plane. I bet (but
not very much, since I'm not really into antennas) that the impedance
of the antenna would increase, probably it would be doubled. Now,
since the current through the antenna remains essentially the same,
the radiated power would double (if the impedance was doubled).
I know this is not how antennas usually are connected, but it is an
illustration to how the acoustic case works. The guesstimate of 100
ohms is probably also very wrong, but it does not matter for the
principle.
So, I DO think we have the same definition of power. :-)

OK. It's easy to agree that anything that improves efficiency
will increase radiated power. Small (inefficient) antennas can
be easily improved by reducing losses - normally by making
them larger, like adding a ground plane or the other half of
the dipole. As soon as the antenna gets close to being a
half wavelength in size, efficiencies in the 90% range are
easily obtained. However, an array of inefficient antennas
will still be inefficient and will exhibit only the array
gain.

--
Steve Maki

On 8 Jan 2004 00:45:28 -0800, (Svante) wrote:

Sigh... If you were able to put the flashlight within a fraction of a
wavelength from the mirror the power WOULD increase. In the audio case
this is easily accomplished. I think it is frequently used with
antennas too (correct me if I am wrong). Power DOES increase. Read the
parallel thread, in particular

It's apparent that your definition of power is not the
same as is commonly used.

In the case of antennas, for example, a theoretical isotropic
source has efficiency of 100%, and gain of zero.

A real dipole might have 90% efficiency, but real gain
of 1.8 dB as measured far-field broadside by virtue of
nulls off the ends of the dipole. The measured response
has increased, but 'power' has decreased.

Add a reflector and director, and you may have 7 dB
gain, but still only 90% efficiency - IOW, the power
is the same.

Stack a second three element yagi, and you will have
maybe 10 dB gain, but STILL only 90% efficiency, and
still LESS power than the zero gain but 100% efficient
isotropic source.

'IF' measured signal in front of the array had anything
to do with power, than efficiencies would routinely
be way over 100%, which is impossible I believe.

Ok, so my reference to antennas maybe wasn't a good one, at least not
when it comes to efficiency. I presume that the reason for that is
that antennas have such a high efficiency from the start, and
obviously, efficiency can never exceed 100%. In the loudspeaker case,
the radiation resistance is a terribly small part of the total
impedance seen by the voice coil. Doubling the radiaton resistance
will double the power output, since the velocity of the cone
essentially will remain the same. Introducing a wall (ground plane, in
the analogy) would double the radiaton resistance. Maybe it would work
for antennas as well if you see it like this (I'm on thin ice here...
:-) ) :
Take a stick of metal and say it is an antenna. This antenna has an
electric impedance from the radiaton impedance. Let's say it is 100
ohms. Let's also say that this antenna has no ground plane and that
the antenna is short compared to the wavelength. Take a resistor of 10
kohms and connect it in series with the antenna and connect the whole
thing to a radio transmitter. Only a fraction of the power delivered
by the source will actually be radiated, most of the power will be
dissipated in the resistor. Now introduce a ground plane. I bet (but
not very much, since I'm not really into antennas) that the impedance
of the antenna would increase, probably it would be doubled. Now,
since the current through the antenna remains essentially the same,
the radiated power would double (if the impedance was doubled).
I know this is not how antennas usually are connected, but it is an
illustration to how the acoustic case works. The guesstimate of 100
ohms is probably also very wrong, but it does not matter for the
principle.
So, I DO think we have the same definition of power. :-)

OK. It's easy to agree that anything that improves efficiency
will increase radiated power. Small (inefficient) antennas can
be easily improved by reducing losses - normally by making
them larger, like adding a ground plane or the other half of
the dipole. As soon as the antenna gets close to being a
half wavelength in size, efficiencies in the 90% range are
easily obtained. However, an array of inefficient antennas
will still be inefficient and will exhibit only the array
gain.

--
Steve Maki

On 8 Jan 2004 00:45:28 -0800, (Svante) wrote:

Sigh... If you were able to put the flashlight within a fraction of a
wavelength from the mirror the power WOULD increase. In the audio case
this is easily accomplished. I think it is frequently used with
antennas too (correct me if I am wrong). Power DOES increase. Read the
parallel thread, in particular

It's apparent that your definition of power is not the
same as is commonly used.

In the case of antennas, for example, a theoretical isotropic
source has efficiency of 100%, and gain of zero.

A real dipole might have 90% efficiency, but real gain
of 1.8 dB as measured far-field broadside by virtue of
nulls off the ends of the dipole. The measured response
has increased, but 'power' has decreased.

Add a reflector and director, and you may have 7 dB
gain, but still only 90% efficiency - IOW, the power
is the same.

Stack a second three element yagi, and you will have
maybe 10 dB gain, but STILL only 90% efficiency, and
still LESS power than the zero gain but 100% efficient
isotropic source.

'IF' measured signal in front of the array had anything
to do with power, than efficiencies would routinely
be way over 100%, which is impossible I believe.

Ok, so my reference to antennas maybe wasn't a good one, at least not
when it comes to efficiency. I presume that the reason for that is
that antennas have such a high efficiency from the start, and
obviously, efficiency can never exceed 100%. In the loudspeaker case,
the radiation resistance is a terribly small part of the total
impedance seen by the voice coil. Doubling the radiaton resistance
will double the power output, since the velocity of the cone
essentially will remain the same. Introducing a wall (ground plane, in
the analogy) would double the radiaton resistance. Maybe it would work
for antennas as well if you see it like this (I'm on thin ice here...
:-) ) :
Take a stick of metal and say it is an antenna. This antenna has an
electric impedance from the radiaton impedance. Let's say it is 100
ohms. Let's also say that this antenna has no ground plane and that
the antenna is short compared to the wavelength. Take a resistor of 10
kohms and connect it in series with the antenna and connect the whole
thing to a radio transmitter. Only a fraction of the power delivered
by the source will actually be radiated, most of the power will be
dissipated in the resistor. Now introduce a ground plane. I bet (but
not very much, since I'm not really into antennas) that the impedance
of the antenna would increase, probably it would be doubled. Now,
since the current through the antenna remains essentially the same,
the radiated power would double (if the impedance was doubled).
I know this is not how antennas usually are connected, but it is an
illustration to how the acoustic case works. The guesstimate of 100
ohms is probably also very wrong, but it does not matter for the
principle.
So, I DO think we have the same definition of power. :-)

OK. It's easy to agree that anything that improves efficiency
will increase radiated power. Small (inefficient) antennas can
be easily improved by reducing losses - normally by making
them larger, like adding a ground plane or the other half of
the dipole. As soon as the antenna gets close to being a
half wavelength in size, efficiencies in the 90% range are
easily obtained. However, an array of inefficient antennas
will still be inefficient and will exhibit only the array
gain.

--
Steve Maki

On 8 Jan 2004 00:45:28 -0800, (Svante) wrote:

Sigh... If you were able to put the flashlight within a fraction of a
wavelength from the mirror the power WOULD increase. In the audio case
this is easily accomplished. I think it is frequently used with
antennas too (correct me if I am wrong). Power DOES increase. Read the
parallel thread, in particular

It's apparent that your definition of power is not the
same as is commonly used.

In the case of antennas, for example, a theoretical isotropic
source has efficiency of 100%, and gain of zero.

A real dipole might have 90% efficiency, but real gain
of 1.8 dB as measured far-field broadside by virtue of
nulls off the ends of the dipole. The measured response
has increased, but 'power' has decreased.

Add a reflector and director, and you may have 7 dB
gain, but still only 90% efficiency - IOW, the power
is the same.

Stack a second three element yagi, and you will have
maybe 10 dB gain, but STILL only 90% efficiency, and
still LESS power than the zero gain but 100% efficient
isotropic source.

'IF' measured signal in front of the array had anything
to do with power, than efficiencies would routinely
be way over 100%, which is impossible I believe.

Ok, so my reference to antennas maybe wasn't a good one, at least not
when it comes to efficiency. I presume that the reason for that is
that antennas have such a high efficiency from the start, and
obviously, efficiency can never exceed 100%. In the loudspeaker case,
the radiation resistance is a terribly small part of the total
impedance seen by the voice coil. Doubling the radiaton resistance
will double the power output, since the velocity of the cone
essentially will remain the same. Introducing a wall (ground plane, in
the analogy) would double the radiaton resistance. Maybe it would work
for antennas as well if you see it like this (I'm on thin ice here...
:-) ) :
Take a stick of metal and say it is an antenna. This antenna has an
electric impedance from the radiaton impedance. Let's say it is 100
ohms. Let's also say that this antenna has no ground plane and that
the antenna is short compared to the wavelength. Take a resistor of 10
kohms and connect it in series with the antenna and connect the whole
thing to a radio transmitter. Only a fraction of the power delivered
by the source will actually be radiated, most of the power will be
dissipated in the resistor. Now introduce a ground plane. I bet (but
not very much, since I'm not really into antennas) that the impedance
of the antenna would increase, probably it would be doubled. Now,
since the current through the antenna remains essentially the same,
the radiated power would double (if the impedance was doubled).
I know this is not how antennas usually are connected, but it is an
illustration to how the acoustic case works. The guesstimate of 100
ohms is probably also very wrong, but it does not matter for the
principle.
So, I DO think we have the same definition of power. :-)

OK. It's easy to agree that anything that improves efficiency
will increase radiated power. Small (inefficient) antennas can
be easily improved by reducing losses - normally by making
them larger, like adding a ground plane or the other half of
the dipole. As soon as the antenna gets close to being a
half wavelength in size, efficiencies in the 90% range are
easily obtained. However, an array of inefficient antennas
will still be inefficient and will exhibit only the array
gain.

--
Steve Maki

ow (Goofball_star_dot_etal) wrote in message ...
On 8 Jan 2004 00:59:37 -0800, (Svante)
wrote:

(Goofball_star_dot_etal) wrote in message ...
On 7 Jan 2004 10:33:01 -0800, (Svante)
wrote:


Do you also have a feeling of déjà vu? ;-)

Yes, but this time I don't see the fine print. :-)

Errh... The parallel would be that two speakers in free space
essentially is the same as one speaker in half space. Was that what
you meant by "fine print"?

I was thinking of the phrase:
"Do you see how I EXPLICITLY chose to NOT talk about the acoustical
effects? The reason being is precisely BECAUSE of the complex
depednecny on frequency"
This no longer applies.


Ahh... You're right! Maybe I should withdraw my statement below... :-)

I have a lot of faith in academic researchers. You should know your
stuff and perhaps people should listen more.

I don't want people accepting what I say without understanding. If
someone thinks I'm wrong I want to know that. When I understand that I
am wrong I say so, and I am mostly happy about that because it means
that I have learned something. The goal of a discussion is, IMO, an
ending where everyone agree, and everyone has learned something. But
you are right, listening is of course a prerequisite for this to
happen. Entering a discussion to prove that one is right, however,
easily ends up in the sand box, and nobody learns anything.

ow (Goofball_star_dot_etal) wrote in message ...
On 8 Jan 2004 00:59:37 -0800, (Svante)
wrote:

(Goofball_star_dot_etal) wrote in message ...
On 7 Jan 2004 10:33:01 -0800, (Svante)
wrote:


Do you also have a feeling of déjà vu? ;-)

Yes, but this time I don't see the fine print. :-)

Errh... The parallel would be that two speakers in free space
essentially is the same as one speaker in half space. Was that what
you meant by "fine print"?

I was thinking of the phrase:
"Do you see how I EXPLICITLY chose to NOT talk about the acoustical
effects? The reason being is precisely BECAUSE of the complex
depednecny on frequency"
This no longer applies.


Ahh... You're right! Maybe I should withdraw my statement below... :-)

I have a lot of faith in academic researchers. You should know your
stuff and perhaps people should listen more.

I don't want people accepting what I say without understanding. If
someone thinks I'm wrong I want to know that. When I understand that I
am wrong I say so, and I am mostly happy about that because it means
that I have learned something. The goal of a discussion is, IMO, an
ending where everyone agree, and everyone has learned something. But
you are right, listening is of course a prerequisite for this to
happen. Entering a discussion to prove that one is right, however,
easily ends up in the sand box, and nobody learns anything.

ow (Goofball_star_dot_etal) wrote in message ...
On 8 Jan 2004 00:59:37 -0800, (Svante)
wrote:

(Goofball_star_dot_etal) wrote in message ...
On 7 Jan 2004 10:33:01 -0800, (Svante)
wrote:


Do you also have a feeling of déjà vu? ;-)

Yes, but this time I don't see the fine print. :-)

Errh... The parallel would be that two speakers in free space
essentially is the same as one speaker in half space. Was that what
you meant by "fine print"?

I was thinking of the phrase:
"Do you see how I EXPLICITLY chose to NOT talk about the acoustical
effects? The reason being is precisely BECAUSE of the complex
depednecny on frequency"
This no longer applies.


Ahh... You're right! Maybe I should withdraw my statement below... :-)

I have a lot of faith in academic researchers. You should know your
stuff and perhaps people should listen more.

I don't want people accepting what I say without understanding. If
someone thinks I'm wrong I want to know that. When I understand that I
am wrong I say so, and I am mostly happy about that because it means
that I have learned something. The goal of a discussion is, IMO, an
ending where everyone agree, and everyone has learned something. But
you are right, listening is of course a prerequisite for this to
happen. Entering a discussion to prove that one is right, however,
easily ends up in the sand box, and nobody learns anything.

ow (Goofball_star_dot_etal) wrote in message ...
On 8 Jan 2004 00:59:37 -0800, (Svante)
wrote:

(Goofball_star_dot_etal) wrote in message ...
On 7 Jan 2004 10:33:01 -0800, (Svante)
wrote:


Do you also have a feeling of déjà vu? ;-)

Yes, but this time I don't see the fine print. :-)

Errh... The parallel would be that two speakers in free space
essentially is the same as one speaker in half space. Was that what
you meant by "fine print"?

I was thinking of the phrase:
"Do you see how I EXPLICITLY chose to NOT talk about the acoustical
effects? The reason being is precisely BECAUSE of the complex
depednecny on frequency"
This no longer applies.


Ahh... You're right! Maybe I should withdraw my statement below... :-)

I have a lot of faith in academic researchers. You should know your
stuff and perhaps people should listen more.

I don't want people accepting what I say without understanding. If
someone thinks I'm wrong I want to know that. When I understand that I
am wrong I say so, and I am mostly happy about that because it means
that I have learned something. The goal of a discussion is, IMO, an
ending where everyone agree, and everyone has learned something. But
you are right, listening is of course a prerequisite for this to
happen. Entering a discussion to prove that one is right, however,
easily ends up in the sand box, and nobody learns anything.

"Rusty Boudreaux" wrote in message ...
"Goofball_star_dot_etal" wrote in
message ...
You can argue all you want but you will be wrong.

Betting is still open. . .

Sorry, but the race is over.

White light is incoherent and the wavlength of green, say 532
nm is
considerably shorter than, say, a 1cm filament. You analogy is
********.

Sorry, the analogy wasn't perfect. I was trying to explain in
terms a moron could understand. Usually a visible example helps
but I guess not for you.

It amazes me you can't understand why changing the radiation
pattern can not increase the output power of a driver. Goofball.

I don't think it is the "Golfball" that is stuck in the sandpit... :-)

Changing the radiation pattern (by changing from free space to half
space) doubles the intensity (=+3dB), right. But in the case when the
source is near its mirror image (compared to wavelength) the radiation
resistance also doubles. Given that the cone velocity remains the same
twice the power is produced in the radiation resistance, which yields
another intensity doubling (=+3 dB). So, if a speaker is mounted in a
wall, the intensity is raised by +6 dB (you may have heard that number
(+ SIX dB) in this context before?), of which 3 dB comes from
directivity, and 3 dB comes from increased efficiency.

And please... Try to understand what I write, follow the links I
presented before to a previous thread and at least CONSIDER that you
might be wrong, instead of just deciding that you are right. If you,
after that, still think the above is wrong, then please enlighten us
on how and why. And please try to leave out the word "moron" in your
explanation.
Otherwise this discussion is going straight to the bunker.

"Rusty Boudreaux" wrote in message ...
"Goofball_star_dot_etal" wrote in
message ...
You can argue all you want but you will be wrong.

Betting is still open. . .

Sorry, but the race is over.

White light is incoherent and the wavlength of green, say 532
nm is
considerably shorter than, say, a 1cm filament. You analogy is
********.

Sorry, the analogy wasn't perfect. I was trying to explain in
terms a moron could understand. Usually a visible example helps
but I guess not for you.

It amazes me you can't understand why changing the radiation
pattern can not increase the output power of a driver. Goofball.

I don't think it is the "Golfball" that is stuck in the sandpit... :-)

Changing the radiation pattern (by changing from free space to half
space) doubles the intensity (=+3dB), right. But in the case when the
source is near its mirror image (compared to wavelength) the radiation
resistance also doubles. Given that the cone velocity remains the same
twice the power is produced in the radiation resistance, which yields
another intensity doubling (=+3 dB). So, if a speaker is mounted in a
wall, the intensity is raised by +6 dB (you may have heard that number
(+ SIX dB) in this context before?), of which 3 dB comes from
directivity, and 3 dB comes from increased efficiency.

And please... Try to understand what I write, follow the links I
presented before to a previous thread and at least CONSIDER that you
might be wrong, instead of just deciding that you are right. If you,
after that, still think the above is wrong, then please enlighten us
on how and why. And please try to leave out the word "moron" in your
explanation.
Otherwise this discussion is going straight to the bunker.

"Rusty Boudreaux" wrote in message ...
"Goofball_star_dot_etal" wrote in
message ...
You can argue all you want but you will be wrong.

Betting is still open. . .

Sorry, but the race is over.

White light is incoherent and the wavlength of green, say 532
nm is
considerably shorter than, say, a 1cm filament. You analogy is
********.

Sorry, the analogy wasn't perfect. I was trying to explain in
terms a moron could understand. Usually a visible example helps
but I guess not for you.

It amazes me you can't understand why changing the radiation
pattern can not increase the output power of a driver. Goofball.

I don't think it is the "Golfball" that is stuck in the sandpit... :-)

Changing the radiation pattern (by changing from free space to half
space) doubles the intensity (=+3dB), right. But in the case when the
source is near its mirror image (compared to wavelength) the radiation
resistance also doubles. Given that the cone velocity remains the same
twice the power is produced in the radiation resistance, which yields
another intensity doubling (=+3 dB). So, if a speaker is mounted in a
wall, the intensity is raised by +6 dB (you may have heard that number
(+ SIX dB) in this context before?), of which 3 dB comes from
directivity, and 3 dB comes from increased efficiency.

And please... Try to understand what I write, follow the links I
presented before to a previous thread and at least CONSIDER that you
might be wrong, instead of just deciding that you are right. If you,
after that, still think the above is wrong, then please enlighten us
on how and why. And please try to leave out the word "moron" in your
explanation.
Otherwise this discussion is going straight to the bunker.

"Rusty Boudreaux" wrote in message ...
"Goofball_star_dot_etal" wrote in
message ...
You can argue all you want but you will be wrong.

Betting is still open. . .

Sorry, but the race is over.

White light is incoherent and the wavlength of green, say 532
nm is
considerably shorter than, say, a 1cm filament. You analogy is
********.

Sorry, the analogy wasn't perfect. I was trying to explain in
terms a moron could understand. Usually a visible example helps
but I guess not for you.

It amazes me you can't understand why changing the radiation
pattern can not increase the output power of a driver. Goofball.

I don't think it is the "Golfball" that is stuck in the sandpit... :-)

Changing the radiation pattern (by changing from free space to half
space) doubles the intensity (=+3dB), right. But in the case when the
source is near its mirror image (compared to wavelength) the radiation
resistance also doubles. Given that the cone velocity remains the same
twice the power is produced in the radiation resistance, which yields
another intensity doubling (=+3 dB). So, if a speaker is mounted in a
wall, the intensity is raised by +6 dB (you may have heard that number
(+ SIX dB) in this context before?), of which 3 dB comes from
directivity, and 3 dB comes from increased efficiency.

And please... Try to understand what I write, follow the links I
presented before to a previous thread and at least CONSIDER that you
might be wrong, instead of just deciding that you are right. If you,
after that, still think the above is wrong, then please enlighten us
on how and why. And please try to leave out the word "moron" in your
explanation.
Otherwise this discussion is going straight to the bunker.

Stewart Pinkerton wrote:

On Thu, 08 Jan 2004 16:29:58 GMT, henryf
wrote:


Peter Larsen wrote:


Intensity on some locations increase at the expense of intensity at
other locations, total power from anything, be it loudspeaker-units or
shocking pink lasers does not change becuase of what their surroundings
do with their output


I'm not so sure this is a universal truth. I think it depends
somewhat on where one draws the boundary between the
"anything" under consideration and the surroundings.

For example, if we considered the "anything" to be an op amp,
and the "surroundings" to include its feedback network,
wouldn't the output power of the op amp depend on what the
feedback network does with the op amp's output?


No, that only affects the gain.

You're kidding, right? OK, then a simpler example. The
"anything" is an audio amplifier, and the "surroundings"
consists of a multi-pushbutton selector switch and a few sets
of speakers. You know, the type of demo setup typically used
at the local speaker supermarket.

The selector switch can connect the output of the amplifier to
4 Ohm speakers, 8 Ohm speakers, 16 Ohm speakers, or to nothing
at all. I think it's obvious to most readers of this group
that the power delivered by the amplifier will depend more or
less on which load gets connected to the amp's output.

Stewart Pinkerton wrote:

On Thu, 08 Jan 2004 16:29:58 GMT, henryf
wrote:


Peter Larsen wrote:


Intensity on some locations increase at the expense of intensity at
other locations, total power from anything, be it loudspeaker-units or
shocking pink lasers does not change becuase of what their surroundings
do with their output


I'm not so sure this is a universal truth. I think it depends
somewhat on where one draws the boundary between the
"anything" under consideration and the surroundings.

For example, if we considered the "anything" to be an op amp,
and the "surroundings" to include its feedback network,
wouldn't the output power of the op amp depend on what the
feedback network does with the op amp's output?


No, that only affects the gain.

You're kidding, right? OK, then a simpler example. The
"anything" is an audio amplifier, and the "surroundings"
consists of a multi-pushbutton selector switch and a few sets
of speakers. You know, the type of demo setup typically used
at the local speaker supermarket.

The selector switch can connect the output of the amplifier to
4 Ohm speakers, 8 Ohm speakers, 16 Ohm speakers, or to nothing
at all. I think it's obvious to most readers of this group
that the power delivered by the amplifier will depend more or
less on which load gets connected to the amp's output.

Stewart Pinkerton wrote:

On Thu, 08 Jan 2004 16:29:58 GMT, henryf
wrote:


Peter Larsen wrote:


Intensity on some locations increase at the expense of intensity at
other locations, total power from anything, be it loudspeaker-units or
shocking pink lasers does not change becuase of what their surroundings
do with their output


I'm not so sure this is a universal truth. I think it depends
somewhat on where one draws the boundary between the
"anything" under consideration and the surroundings.

For example, if we considered the "anything" to be an op amp,
and the "surroundings" to include its feedback network,
wouldn't the output power of the op amp depend on what the
feedback network does with the op amp's output?


No, that only affects the gain.

You're kidding, right? OK, then a simpler example. The
"anything" is an audio amplifier, and the "surroundings"
consists of a multi-pushbutton selector switch and a few sets
of speakers. You know, the type of demo setup typically used
at the local speaker supermarket.

The selector switch can connect the output of the amplifier to
4 Ohm speakers, 8 Ohm speakers, 16 Ohm speakers, or to nothing
at all. I think it's obvious to most readers of this group
that the power delivered by the amplifier will depend more or
less on which load gets connected to the amp's output.

Stewart Pinkerton wrote:

On Thu, 08 Jan 2004 16:29:58 GMT, henryf
wrote:


Peter Larsen wrote:


Intensity on some locations increase at the expense of intensity at
other locations, total power from anything, be it loudspeaker-units or
shocking pink lasers does not change becuase of what their surroundings
do with their output


I'm not so sure this is a universal truth. I think it depends
somewhat on where one draws the boundary between the
"anything" under consideration and the surroundings.

For example, if we considered the "anything" to be an op amp,
and the "surroundings" to include its feedback network,
wouldn't the output power of the op amp depend on what the
feedback network does with the op amp's output?


No, that only affects the gain.

You're kidding, right? OK, then a simpler example. The
"anything" is an audio amplifier, and the "surroundings"
consists of a multi-pushbutton selector switch and a few sets
of speakers. You know, the type of demo setup typically used
at the local speaker supermarket.

The selector switch can connect the output of the amplifier to
4 Ohm speakers, 8 Ohm speakers, 16 Ohm speakers, or to nothing
at all. I think it's obvious to most readers of this group
that the power delivered by the amplifier will depend more or
less on which load gets connected to the amp's output.

On 8 Jan 2004 15:30:55 -0800, (Svante)
wrote:

(Goofball_star_dot_etal) wrote in message ...
On 8 Jan 2004 00:59:37 -0800, (Svante)
wrote:

(Goofball_star_dot_etal) wrote in message ...
On 7 Jan 2004 10:33:01 -0800, (Svante)
wrote:


Do you also have a feeling of déjà vu? ;-)

Yes, but this time I don't see the fine print. :-)

Errh... The parallel would be that two speakers in free space
essentially is the same as one speaker in half space. Was that what
you meant by "fine print"?

I was thinking of the phrase:
"Do you see how I EXPLICITLY chose to NOT talk about the acoustical
effects? The reason being is precisely BECAUSE of the complex
depednecny on frequency"
This no longer applies.


Ahh... You're right! Maybe I should withdraw my statement below... :-)

I feel I may have done some people an injustice in my my own mind. I
had come to the conclusion that they lacked the "balls" to admit they
were wrong (except one, I think), whereas, in fact, they had just not
realized that they could be wrong.

It is difficult to conclude, after your explanations, that the
statement:

"Actually, no, the efficiency, specifically, the ratio of total
radiated acoustic power to total input electrical power does NOT
change. The sound pressure level on axis MAY change, but the
efficiency does not. The same power is distributed over a smaller
area, and since SPL is a measure of power per unit area, the SPL
changes, but the power output does not."

is other than wrong.



I have a lot of faith in academic researchers. You should know your
stuff and perhaps people should listen more.

I don't want people accepting what I say without understanding. If
someone thinks I'm wrong I want to know that. When I understand that I
am wrong I say so, and I am mostly happy about that because it means
that I have learned something. The goal of a discussion is, IMO, an
ending where everyone agree, and everyone has learned something. But
you are right, listening is of course a prerequisite for this to
happen. Entering a discussion to prove that one is right, however,
easily ends up in the sand box, and nobody learns anything.

On 8 Jan 2004 15:30:55 -0800, (Svante)
wrote:

(Goofball_star_dot_etal) wrote in message ...
On 8 Jan 2004 00:59:37 -0800, (Svante)
wrote:

(Goofball_star_dot_etal) wrote in message ...
On 7 Jan 2004 10:33:01 -0800, (Svante)
wrote:


Do you also have a feeling of déjà vu? ;-)

Yes, but this time I don't see the fine print. :-)

Errh... The parallel would be that two speakers in free space
essentially is the same as one speaker in half space. Was that what
you meant by "fine print"?

I was thinking of the phrase:
"Do you see how I EXPLICITLY chose to NOT talk about the acoustical
effects? The reason being is precisely BECAUSE of the complex
depednecny on frequency"
This no longer applies.


Ahh... You're right! Maybe I should withdraw my statement below... :-)

I feel I may have done some people an injustice in my my own mind. I
had come to the conclusion that they lacked the "balls" to admit they
were wrong (except one, I think), whereas, in fact, they had just not
realized that they could be wrong.

It is difficult to conclude, after your explanations, that the
statement:

"Actually, no, the efficiency, specifically, the ratio of total
radiated acoustic power to total input electrical power does NOT
change. The sound pressure level on axis MAY change, but the
efficiency does not. The same power is distributed over a smaller
area, and since SPL is a measure of power per unit area, the SPL
changes, but the power output does not."

is other than wrong.



I have a lot of faith in academic researchers. You should know your
stuff and perhaps people should listen more.

I don't want people accepting what I say without understanding. If
someone thinks I'm wrong I want to know that. When I understand that I
am wrong I say so, and I am mostly happy about that because it means
that I have learned something. The goal of a discussion is, IMO, an
ending where everyone agree, and everyone has learned something. But
you are right, listening is of course a prerequisite for this to
happen. Entering a discussion to prove that one is right, however,
easily ends up in the sand box, and nobody learns anything.

On 8 Jan 2004 15:30:55 -0800, (Svante)
wrote:

(Goofball_star_dot_etal) wrote in message ...
On 8 Jan 2004 00:59:37 -0800, (Svante)
wrote:

(Goofball_star_dot_etal) wrote in message ...
On 7 Jan 2004 10:33:01 -0800, (Svante)
wrote:


Do you also have a feeling of déjà vu? ;-)

Yes, but this time I don't see the fine print. :-)

Errh... The parallel would be that two speakers in free space
essentially is the same as one speaker in half space. Was that what
you meant by "fine print"?

I was thinking of the phrase:
"Do you see how I EXPLICITLY chose to NOT talk about the acoustical
effects? The reason being is precisely BECAUSE of the complex
depednecny on frequency"
This no longer applies.


Ahh... You're right! Maybe I should withdraw my statement below... :-)

I feel I may have done some people an injustice in my my own mind. I
had come to the conclusion that they lacked the "balls" to admit they
were wrong (except one, I think), whereas, in fact, they had just not
realized that they could be wrong.

It is difficult to conclude, after your explanations, that the
statement:

"Actually, no, the efficiency, specifically, the ratio of total
radiated acoustic power to total input electrical power does NOT
change. The sound pressure level on axis MAY change, but the
efficiency does not. The same power is distributed over a smaller
area, and since SPL is a measure of power per unit area, the SPL
changes, but the power output does not."

is other than wrong.



I have a lot of faith in academic researchers. You should know your
stuff and perhaps people should listen more.

I don't want people accepting what I say without understanding. If
someone thinks I'm wrong I want to know that. When I understand that I
am wrong I say so, and I am mostly happy about that because it means
that I have learned something. The goal of a discussion is, IMO, an
ending where everyone agree, and everyone has learned something. But
you are right, listening is of course a prerequisite for this to
happen. Entering a discussion to prove that one is right, however,
easily ends up in the sand box, and nobody learns anything.

On 8 Jan 2004 15:30:55 -0800, (Svante)
wrote:

(Goofball_star_dot_etal) wrote in message ...
On 8 Jan 2004 00:59:37 -0800, (Svante)
wrote:

(Goofball_star_dot_etal) wrote in message ...
On 7 Jan 2004 10:33:01 -0800, (Svante)
wrote:


Do you also have a feeling of déjà vu? ;-)

Yes, but this time I don't see the fine print. :-)

Errh... The parallel would be that two speakers in free space
essentially is the same as one speaker in half space. Was that what
you meant by "fine print"?

I was thinking of the phrase:
"Do you see how I EXPLICITLY chose to NOT talk about the acoustical
effects? The reason being is precisely BECAUSE of the complex
depednecny on frequency"
This no longer applies.


Ahh... You're right! Maybe I should withdraw my statement below... :-)

I feel I may have done some people an injustice in my my own mind. I
had come to the conclusion that they lacked the "balls" to admit they
were wrong (except one, I think), whereas, in fact, they had just not
realized that they could be wrong.

It is difficult to conclude, after your explanations, that the
statement:

"Actually, no, the efficiency, specifically, the ratio of total
radiated acoustic power to total input electrical power does NOT
change. The sound pressure level on axis MAY change, but the
efficiency does not. The same power is distributed over a smaller
area, and since SPL is a measure of power per unit area, the SPL
changes, but the power output does not."

is other than wrong.



I have a lot of faith in academic researchers. You should know your
stuff and perhaps people should listen more.

I don't want people accepting what I say without understanding. If
someone thinks I'm wrong I want to know that. When I understand that I
am wrong I say so, and I am mostly happy about that because it means
that I have learned something. The goal of a discussion is, IMO, an
ending where everyone agree, and everyone has learned something. But
you are right, listening is of course a prerequisite for this to
happen. Entering a discussion to prove that one is right, however,
easily ends up in the sand box, and nobody learns anything.