"Rusty Boudreaux" wrote in message
"Svante" wrote in message
Actually I argue the the output power DOES increase. The wall acts as

You can argue all you want but you will be wrong.

I agree that me arguing will not make it true. But what more can I do?
Explain?



a mirror (in the optical analogy) and the source and its mirror image
together double the sound pressure, and thus quadruples the intensity.
This leads to a level increase of +6dB. However, since the area hit by
this radiation is of half the size, the power is "only" increased by a
factor 2. Another way of seing it is that the radiation impedance is
doubled.

No. The output power of the driver does not change. The
efficiency does not change.

It's similar to changing a flashlight focus from a widebeam flood
to a narrow beam spot. The light may be directed to a smaller
area but the power output and efficiency of the bulb does not
change.

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

http://groups.google.com/groups?hl=e...com%26rnum%3D3

where Bob Stanton proved this experimentally. His experiment was with
four speakers connected in series-parallel, compared to a single
speaker. The electrical impedance for this combination is the same as
for a single speaker, and thus the same electrical power was drawn
from the amplifier. Yet the level became 6 dB higher when all four
speakers were connected. This is analog to a single speaker playing in
quarter-space.

(Sorry about the lengthy link, you might have to remove line breaks)
You may alternatively want to look for "Speaker sensitivity and fs in
multiples." in rec.audio.tech. Bob Stanton's post occured 5 Dec 2003.


There is a specific definition for efficiency, which is very clear
on the meaning, that is, the efficiency above the system cutoff
but below the frequency where the wavelength is equal to the
circumference of the radiator. SMall and Thiele and others are quite
clear on this definition, and I defer to their definition.

OK, that's what most people do, and I see the benefits of doing so.
However, the original definition of efficiency (which can be applied
to many things, such as car motors etc) is the ratio between output
and input power.

As it is in this case. Acoustical output of the driver in watts
divided by electrical input in watts. You may be changing the
radiation pattern but the power output of the driver does not
change.

Given that my arguments and Bob Stanton's experiment are valid (please
read above), the output power DOES actuelly increase. For half-space,
the level goes up 6 dB (intensity is quadrupled) but only half of the
space is hit by the radiation, net power gain factor = 4 * 1/2 = 2.

Any person who has heard the word "efficiency" in other contexts may
be confused IMO. And I guess those are not few.

The context is the same in this situation.

Yes, it is in my mind too. But I think you are confused... :-)

"Rusty Boudreaux" wrote in message
"Svante" wrote in message
Actually I argue the the output power DOES increase. The wall acts as

You can argue all you want but you will be wrong.

I agree that me arguing will not make it true. But what more can I do?
Explain?



a mirror (in the optical analogy) and the source and its mirror image
together double the sound pressure, and thus quadruples the intensity.
This leads to a level increase of +6dB. However, since the area hit by
this radiation is of half the size, the power is "only" increased by a
factor 2. Another way of seing it is that the radiation impedance is
doubled.

No. The output power of the driver does not change. The
efficiency does not change.

It's similar to changing a flashlight focus from a widebeam flood
to a narrow beam spot. The light may be directed to a smaller
area but the power output and efficiency of the bulb does not
change.

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

http://groups.google.com/groups?hl=e...com%26rnum%3D3

where Bob Stanton proved this experimentally. His experiment was with
four speakers connected in series-parallel, compared to a single
speaker. The electrical impedance for this combination is the same as
for a single speaker, and thus the same electrical power was drawn
from the amplifier. Yet the level became 6 dB higher when all four
speakers were connected. This is analog to a single speaker playing in
quarter-space.

(Sorry about the lengthy link, you might have to remove line breaks)
You may alternatively want to look for "Speaker sensitivity and fs in
multiples." in rec.audio.tech. Bob Stanton's post occured 5 Dec 2003.


There is a specific definition for efficiency, which is very clear
on the meaning, that is, the efficiency above the system cutoff
but below the frequency where the wavelength is equal to the
circumference of the radiator. SMall and Thiele and others are quite
clear on this definition, and I defer to their definition.

OK, that's what most people do, and I see the benefits of doing so.
However, the original definition of efficiency (which can be applied
to many things, such as car motors etc) is the ratio between output
and input power.

As it is in this case. Acoustical output of the driver in watts
divided by electrical input in watts. You may be changing the
radiation pattern but the power output of the driver does not
change.

Given that my arguments and Bob Stanton's experiment are valid (please
read above), the output power DOES actuelly increase. For half-space,
the level goes up 6 dB (intensity is quadrupled) but only half of the
space is hit by the radiation, net power gain factor = 4 * 1/2 = 2.

Any person who has heard the word "efficiency" in other contexts may
be confused IMO. And I guess those are not few.

The context is the same in this situation.

Yes, it is in my mind too. But I think you are confused... :-)

"Rusty Boudreaux" wrote in message
"Svante" wrote in message
Actually I argue the the output power DOES increase. The wall acts as

You can argue all you want but you will be wrong.

I agree that me arguing will not make it true. But what more can I do?
Explain?



a mirror (in the optical analogy) and the source and its mirror image
together double the sound pressure, and thus quadruples the intensity.
This leads to a level increase of +6dB. However, since the area hit by
this radiation is of half the size, the power is "only" increased by a
factor 2. Another way of seing it is that the radiation impedance is
doubled.

No. The output power of the driver does not change. The
efficiency does not change.

It's similar to changing a flashlight focus from a widebeam flood
to a narrow beam spot. The light may be directed to a smaller
area but the power output and efficiency of the bulb does not
change.

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

http://groups.google.com/groups?hl=e...com%26rnum%3D3

where Bob Stanton proved this experimentally. His experiment was with
four speakers connected in series-parallel, compared to a single
speaker. The electrical impedance for this combination is the same as
for a single speaker, and thus the same electrical power was drawn
from the amplifier. Yet the level became 6 dB higher when all four
speakers were connected. This is analog to a single speaker playing in
quarter-space.

(Sorry about the lengthy link, you might have to remove line breaks)
You may alternatively want to look for "Speaker sensitivity and fs in
multiples." in rec.audio.tech. Bob Stanton's post occured 5 Dec 2003.


There is a specific definition for efficiency, which is very clear
on the meaning, that is, the efficiency above the system cutoff
but below the frequency where the wavelength is equal to the
circumference of the radiator. SMall and Thiele and others are quite
clear on this definition, and I defer to their definition.

OK, that's what most people do, and I see the benefits of doing so.
However, the original definition of efficiency (which can be applied
to many things, such as car motors etc) is the ratio between output
and input power.

As it is in this case. Acoustical output of the driver in watts
divided by electrical input in watts. You may be changing the
radiation pattern but the power output of the driver does not
change.

Given that my arguments and Bob Stanton's experiment are valid (please
read above), the output power DOES actuelly increase. For half-space,
the level goes up 6 dB (intensity is quadrupled) but only half of the
space is hit by the radiation, net power gain factor = 4 * 1/2 = 2.

Any person who has heard the word "efficiency" in other contexts may
be confused IMO. And I guess those are not few.

The context is the same in this situation.

Yes, it is in my mind too. But I think you are confused... :-)

"Rusty Boudreaux" wrote in message
"Svante" wrote in message
Actually I argue the the output power DOES increase. The wall acts as

You can argue all you want but you will be wrong.

I agree that me arguing will not make it true. But what more can I do?
Explain?



a mirror (in the optical analogy) and the source and its mirror image
together double the sound pressure, and thus quadruples the intensity.
This leads to a level increase of +6dB. However, since the area hit by
this radiation is of half the size, the power is "only" increased by a
factor 2. Another way of seing it is that the radiation impedance is
doubled.

No. The output power of the driver does not change. The
efficiency does not change.

It's similar to changing a flashlight focus from a widebeam flood
to a narrow beam spot. The light may be directed to a smaller
area but the power output and efficiency of the bulb does not
change.

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

http://groups.google.com/groups?hl=e...com%26rnum%3D3

where Bob Stanton proved this experimentally. His experiment was with
four speakers connected in series-parallel, compared to a single
speaker. The electrical impedance for this combination is the same as
for a single speaker, and thus the same electrical power was drawn
from the amplifier. Yet the level became 6 dB higher when all four
speakers were connected. This is analog to a single speaker playing in
quarter-space.

(Sorry about the lengthy link, you might have to remove line breaks)
You may alternatively want to look for "Speaker sensitivity and fs in
multiples." in rec.audio.tech. Bob Stanton's post occured 5 Dec 2003.


There is a specific definition for efficiency, which is very clear
on the meaning, that is, the efficiency above the system cutoff
but below the frequency where the wavelength is equal to the
circumference of the radiator. SMall and Thiele and others are quite
clear on this definition, and I defer to their definition.

OK, that's what most people do, and I see the benefits of doing so.
However, the original definition of efficiency (which can be applied
to many things, such as car motors etc) is the ratio between output
and input power.

As it is in this case. Acoustical output of the driver in watts
divided by electrical input in watts. You may be changing the
radiation pattern but the power output of the driver does not
change.

Given that my arguments and Bob Stanton's experiment are valid (please
read above), the output power DOES actuelly increase. For half-space,
the level goes up 6 dB (intensity is quadrupled) but only half of the
space is hit by the radiation, net power gain factor = 4 * 1/2 = 2.

Any person who has heard the word "efficiency" in other contexts may
be confused IMO. And I guess those are not few.

The context is the same in this situation.

Yes, it is in my mind too. But I think you are confused... :-)

On 7 Jan 2004 09:00:38 -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.

--
Steve Maki

On 7 Jan 2004 09:00:38 -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.

--
Steve Maki

On 7 Jan 2004 09:00:38 -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.

--
Steve Maki

On 7 Jan 2004 09:00:38 -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.

--
Steve Maki

ow (Goofball_star_dot_etal) wrote in message

On Wed, 7 Jan 2004 03:58:09 -0600, "Rusty Boudreaux"

"Svante" wrote in message

Actually I argue the the output power DOES increase. The wall acts as

You can argue all you want but you will be wrong.


Betting is still open. . .

a mirror (in the optical analogy) and the source and its mirror image
together double the sound pressure, and thus quadruples the intensity.
This leads to a level increase of +6dB. However, since the area hit by
this radiation is of half the size, the power is "only" increased by a
factor 2. Another way of seing it is that the radiation impedance is
doubled.

No. The output power of the driver does not change. The
efficiency does not change.

It's similar to changing a flashlight focus from a widebeam flood
to a narrow beam spot. The light may be directed to a smaller
area but the power output and efficiency of the bulb does not
change.


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

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

ow (Goofball_star_dot_etal) wrote in message

On Wed, 7 Jan 2004 03:58:09 -0600, "Rusty Boudreaux"

"Svante" wrote in message

Actually I argue the the output power DOES increase. The wall acts as

You can argue all you want but you will be wrong.


Betting is still open. . .

a mirror (in the optical analogy) and the source and its mirror image
together double the sound pressure, and thus quadruples the intensity.
This leads to a level increase of +6dB. However, since the area hit by
this radiation is of half the size, the power is "only" increased by a
factor 2. Another way of seing it is that the radiation impedance is
doubled.

No. The output power of the driver does not change. The
efficiency does not change.

It's similar to changing a flashlight focus from a widebeam flood
to a narrow beam spot. The light may be directed to a smaller
area but the power output and efficiency of the bulb does not
change.


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

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

ow (Goofball_star_dot_etal) wrote in message

On Wed, 7 Jan 2004 03:58:09 -0600, "Rusty Boudreaux"

"Svante" wrote in message

Actually I argue the the output power DOES increase. The wall acts as

You can argue all you want but you will be wrong.


Betting is still open. . .

a mirror (in the optical analogy) and the source and its mirror image
together double the sound pressure, and thus quadruples the intensity.
This leads to a level increase of +6dB. However, since the area hit by
this radiation is of half the size, the power is "only" increased by a
factor 2. Another way of seing it is that the radiation impedance is
doubled.

No. The output power of the driver does not change. The
efficiency does not change.

It's similar to changing a flashlight focus from a widebeam flood
to a narrow beam spot. The light may be directed to a smaller
area but the power output and efficiency of the bulb does not
change.


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

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

ow (Goofball_star_dot_etal) wrote in message

On Wed, 7 Jan 2004 03:58:09 -0600, "Rusty Boudreaux"

"Svante" wrote in message

Actually I argue the the output power DOES increase. The wall acts as

You can argue all you want but you will be wrong.


Betting is still open. . .

a mirror (in the optical analogy) and the source and its mirror image
together double the sound pressure, and thus quadruples the intensity.
This leads to a level increase of +6dB. However, since the area hit by
this radiation is of half the size, the power is "only" increased by a
factor 2. Another way of seing it is that the radiation impedance is
doubled.

No. The output power of the driver does not change. The
efficiency does not change.

It's similar to changing a flashlight focus from a widebeam flood
to a narrow beam spot. The light may be directed to a smaller
area but the power output and efficiency of the bulb does not
change.


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

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

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. :-)
I have a lot of faith in academic researchers. You should know your
stuff and perhaps people should listen more.
http://www.speech.kth.se/~svante/
Anyhow, this ain't rocket science. . .

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. :-)
I have a lot of faith in academic researchers. You should know your
stuff and perhaps people should listen more.
http://www.speech.kth.se/~svante/
Anyhow, this ain't rocket science. . .

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. :-)
I have a lot of faith in academic researchers. You should know your
stuff and perhaps people should listen more.
http://www.speech.kth.se/~svante/
Anyhow, this ain't rocket science. . .

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. :-)
I have a lot of faith in academic researchers. You should know your
stuff and perhaps people should listen more.
http://www.speech.kth.se/~svante/
Anyhow, this ain't rocket science. . .

"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.

"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.

"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.

"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.

Svante wrote:

"Svante" wrote in message

Actually I argue the the output power DOES increase. The wall ...

You can argue all you want but you will be wrong.

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


Kind regards

Peter Larsen

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

Svante wrote:

"Svante" wrote in message

Actually I argue the the output power DOES increase. The wall ...

You can argue all you want but you will be wrong.

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


Kind regards

Peter Larsen

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

Svante wrote:

"Svante" wrote in message

Actually I argue the the output power DOES increase. The wall ...

You can argue all you want but you will be wrong.

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


Kind regards

Peter Larsen

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

Svante wrote:

"Svante" wrote in message

Actually I argue the the output power DOES increase. The wall ...

You can argue all you want but you will be wrong.

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


Kind regards

Peter Larsen

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

Steve Maki wrote in message . ..
On 7 Jan 2004 09:00:38 -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. :-)

Steve Maki wrote in message . ..
On 7 Jan 2004 09:00:38 -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. :-)

Steve Maki wrote in message . ..
On 7 Jan 2004 09:00:38 -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. :-)

Steve Maki wrote in message . ..
On 7 Jan 2004 09:00:38 -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. :-)

ow (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 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.

http://www.speech.kth.se/~svante/

You found me!

Anyhow, this ain't rocket science. . .

Depends on the size of the speakers... :-)

ow (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 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.

http://www.speech.kth.se/~svante/

You found me!

Anyhow, this ain't rocket science. . .

Depends on the size of the speakers... :-)

ow (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 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.

http://www.speech.kth.se/~svante/

You found me!

Anyhow, this ain't rocket science. . .

Depends on the size of the speakers... :-)

ow (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 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.

http://www.speech.kth.se/~svante/

You found me!

Anyhow, this ain't rocket science. . .

Depends on the size of the speakers... :-)

On Thu, 8 Jan 2004 00:36:06 -0600, "Rusty Boudreaux"
wrote:

"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.

Very true.

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.


This would be sad indeed if it was true. I am trying to simulate and
improve the radiation pattern of the antenna of a research "clear air"
radar to reduce ground clutter, right now. If I was totally clueless
it might be a waste of time. Usually I mess with Lidar, so Radar is a
bit new to me.

On Thu, 8 Jan 2004 00:36:06 -0600, "Rusty Boudreaux"
wrote:

"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.

Very true.

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.


This would be sad indeed if it was true. I am trying to simulate and
improve the radiation pattern of the antenna of a research "clear air"
radar to reduce ground clutter, right now. If I was totally clueless
it might be a waste of time. Usually I mess with Lidar, so Radar is a
bit new to me.

On Thu, 8 Jan 2004 00:36:06 -0600, "Rusty Boudreaux"
wrote:

"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.

Very true.

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.


This would be sad indeed if it was true. I am trying to simulate and
improve the radiation pattern of the antenna of a research "clear air"
radar to reduce ground clutter, right now. If I was totally clueless
it might be a waste of time. Usually I mess with Lidar, so Radar is a
bit new to me.

On Thu, 8 Jan 2004 00:36:06 -0600, "Rusty Boudreaux"
wrote:

"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.

Very true.

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.


This would be sad indeed if it was true. I am trying to simulate and
improve the radiation pattern of the antenna of a research "clear air"
radar to reduce ground clutter, right now. If I was totally clueless
it might be a waste of time. Usually I mess with Lidar, so Radar is a
bit new to me.

Peter Larsen wrote in message ...
Svante wrote:

"Svante" wrote in message

Actually I argue the the output power DOES increase. The wall ...

You can argue all you want but you will be wrong.

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 am aware that the directivity will contribute with a level gain of 3
dB when the space is halved (measured anywhere in the half-space).
However, the level gain is 6 dB, the 3 extra dBs come from increased
efficiency. This is not only an issue of directivity, it is also an
issue of radiation resistance (which is doubled).

Please READ the parallel thread I referred to!

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

Peter Larsen wrote in message ...
Svante wrote:

"Svante" wrote in message

Actually I argue the the output power DOES increase. The wall ...

You can argue all you want but you will be wrong.

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 am aware that the directivity will contribute with a level gain of 3
dB when the space is halved (measured anywhere in the half-space).
However, the level gain is 6 dB, the 3 extra dBs come from increased
efficiency. This is not only an issue of directivity, it is also an
issue of radiation resistance (which is doubled).

Please READ the parallel thread I referred to!

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

Peter Larsen wrote in message ...
Svante wrote:

"Svante" wrote in message

Actually I argue the the output power DOES increase. The wall ...

You can argue all you want but you will be wrong.

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 am aware that the directivity will contribute with a level gain of 3
dB when the space is halved (measured anywhere in the half-space).
However, the level gain is 6 dB, the 3 extra dBs come from increased
efficiency. This is not only an issue of directivity, it is also an
issue of radiation resistance (which is doubled).

Please READ the parallel thread I referred to!

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

Peter Larsen wrote in message ...
Svante wrote:

"Svante" wrote in message

Actually I argue the the output power DOES increase. The wall ...

You can argue all you want but you will be wrong.

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 am aware that the directivity will contribute with a level gain of 3
dB when the space is halved (measured anywhere in the half-space).
However, the level gain is 6 dB, the 3 extra dBs come from increased
efficiency. This is not only an issue of directivity, it is also an
issue of radiation resistance (which is doubled).

Please READ the parallel thread I referred to!

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