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#41
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Speakers Then and Now
"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... :-) |
#42
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Speakers Then and Now
"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... :-) |
#43
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Speakers Then and Now
"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... :-) |
#44
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Speakers Then and Now
"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... :-) |
#46
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#47
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#48
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#49
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#51
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#52
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#53
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Speakers Then and Now
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. . . |
#54
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Speakers Then and Now
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. . . |
#55
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Speakers Then and Now
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. . . |
#56
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Speakers Then and Now
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. . . |
#57
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Speakers Then and Now
"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. |
#58
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Speakers Then and Now
"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. |
#59
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Speakers Then and Now
"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. |
#60
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Speakers Then and Now
"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. |
#61
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Speakers Then and Now
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 * ******************************************* |
#62
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Speakers Then and Now
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 * ******************************************* |
#63
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Speakers Then and Now
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 * ******************************************* |
#64
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Speakers Then and Now
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 * ******************************************* |
#65
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Speakers Then and Now
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. :-) |
#66
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Speakers Then and Now
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. :-) |
#67
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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. :-) |
#68
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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. :-) |
#70
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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... :-) |
#71
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Speakers Then and Now
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... :-) |
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Speakers Then and Now
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... :-) |
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Speakers Then and Now
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. |
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Speakers Then and Now
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. |
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Speakers Then and Now
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. |
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Speakers Then and Now
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. |
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Speakers Then and Now
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. |
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Speakers Then and Now
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. |
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Speakers Then and Now
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. |
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Speakers Then and Now
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. |