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#241
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Old speaker - See what you did Dick - lol
I'm still here Dick,
It is just taking me more time than I thought it would to digest the info you gave me. The original form is equally useful: if you know the design goals for enclosure size, system tuning and bandwidth, the relation will tell you the maximum efficiency you can expect out of the system. The relationship is, indeed, a beginning point in the system design process: it sets the minimum set of basic parameters needed to achieve the broadest of system constraints. It will tell you, for example, that a driver that's MORE efficienct that what the equation specifies WILL NOT WORK to achieve that system specification, something will have to be sacrificed: either the system will have to be tuned differently, the cutoff frequency has to be higher or the box has to be larger, or some combination thereof. As I remember, the cutoff for the low end was higher, so it fits the equation you gave. Did his design actually put out a higher SPL output level per wattage in than if in the "right" size box that the speaker was designed for. Are speaker parameters (mass/spider/surround/etc) designed for particular box sizes, or is this where the volume of the port comes into play for different size boxes? Also, does it matter if a port is round or rectangular? I have heard both and if anyone could tell me the real truth, I figure it will be you. Thanks, John |
#242
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Old speaker - See what you did Dick - lol
Dick, would you care to share some general comments/insight regarding the behavior of spkr(s) in an open-back enclosure, with respect to the various basic formulae & behaviors you've mentioned for suspended or loaded ones? Dumb question I am sure, but do you mean by open back one that is not in a sealed or ported enclosure, and as an example, rear deck speakers going into a trunk? |
#243
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Old speaker - See what you did Dick - lol
Dick, would you care to share some general comments/insight regarding the behavior of spkr(s) in an open-back enclosure, with respect to the various basic formulae & behaviors you've mentioned for suspended or loaded ones? Dumb question I am sure, but do you mean by open back one that is not in a sealed or ported enclosure, and as an example, rear deck speakers going into a trunk? |
#244
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Old speaker - See what you did Dick - lol
Dick, would you care to share some general comments/insight regarding the behavior of spkr(s) in an open-back enclosure, with respect to the various basic formulae & behaviors you've mentioned for suspended or loaded ones? Dumb question I am sure, but do you mean by open back one that is not in a sealed or ported enclosure, and as an example, rear deck speakers going into a trunk? |
#245
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
The original form is equally useful: if you know the design goals for enclosure size, system tuning and bandwidth, the relation will tell you the maximum efficiency you can expect out of the system. The relationship is, indeed, a beginning point in the system design process: it sets the minimum set of basic parameters needed to achieve the broadest of system constraints. It will tell you, for example, that a driver that's MORE efficienct that what the equation specifies WILL NOT WORK to achieve that system specification, something will have to be sacrificed: either the system will have to be tuned differently, the cutoff frequency has to be higher or the box has to be larger, or some combination thereof. As I remember, the cutoff for the low end was higher, so it fits the equation you gave. Did his design actually put out a higher SPL output level per wattage in than if in the "right" size box that the speaker was designed for. Who is "he" in the above and what design are we talkgin about. Are speaker parameters (mass/spider/surround/etc) designed for particular box sizes, In an ideal world, marketing tells engineering the most basic of parameters (enclosure size, intended range of receiver/amplifiers, tells you how much power, which might clue you as to required efficiency, etc.). From that, engineering the develops a detailed system specification and FROM THAT, driver parameters are derived, the driver is designed and the whole thing integrated together. It seldom happens that way, as the vats majority of speaker system manufacturers do not have the ability to custom design drivers, so they are forced to redesign the system to accomodate the available drivers. or is this where the volume of the port comes into play for different size boxes? A common misconception: it is NOT the volume of the port that is important. It's something a little more complex and obtuse than that. A vented enclosure is, in and of itself, a resonant acoustic circuit, whose reactive elements consist of the acoustic stiffness of the air in the box and the acoustic "inertance" or mass of the vent or port. The acoustic inertance IS NOT THE SAME THING as the physical mass of the air in the port. The air mass is basically the density of air times the volume of the port, Mp = p l pi d^2 / 4 where p is the density of air (about 1.18 kg/m^3), l is the length of the port in meters and d is the diameter of the port in meters (all this assume a cylindrical port). The mass is in units of kilograms. However, the acoustical mass is actually quite different: it's equal to the density of air times the effective length of the port DIVIDED BY the cross-sectional area of the port, or: Map = p l'/(pi d^2/4) and has the strange units of kg/m^4. This is what leads to the rather counter-intuitive property that if tow ports have the same length, but one is smaller in diameter than the other, the smaller port has the GREATER effective acoustic mass. Why is this so? well, let's engage in a little gedanken (a "thought experiment"). Consider two tubes, on is 1 cm (0.01m) in diameter and 10 m long, the other is 10 cm in diameter and 10 cm long. They both have precisely the same internal volume: V = 10m * pi * .01m^2/4 = 7.85 * 10^-4 m^3 V = .1m * pi * .1m^2/4 = 7.85 * 10^-4 m^3 and thus will have the same physical mass of about 9 * 10^-4 kg, of just under 1 gram. Now, assume both have a suitable mouthpiece at one end of the cylinder. Imagine trying the following experiment: using the mouthpiece, try alternatively sucking and blowing air into each cylinder repeating as rapidly as you possibly can. What you will find is that it is VERY esay to do with the short, fat cylinder and much harder to do with the long skinny tube. The reasona as mentioned is a little obtuse, but, simplifying it as much as possible, here's what you'll find, and the clue is in the term "intertance," or the resistance to CHANGE in motion. When you alternatively such and blow in the short fat cylinder, because of it's larger width, the volume of air your lungs are pumping in and out results in a very small change in velocity in the cylinder, whereas in the long skinny tube, the change in volume of air intriduced by your lungs results in a much higher velocity. SO the difference is that a narrow tube requires much larger chnages in velocity in the air inside the tube, and the higher velocity requires higher acceleration, and higher acceleration requires higher forces. Indeed, the differences in velocity in the air are propertional to the RECIPROCAL of the area of the port, or the reciprocal of the square of tne diameter, thus the seemingly strange result that making a port narrower INCREASES it's effective acoustic mass. Also, does it matter if a port is round or rectangular? I have heard both and if anyone could tell me the real truth, I figure it will be you. Over the range of port configurations where the ratio of width to height is small, in other words, it's near a square in cross section, there's little if any difference. The major effect at or near the enclosure reconance will be due to the differences in friction due to the fact that the wall area of a cylinder is as small as it can get, and no rectangular port can have an area as small. Assuming the same wall friction, a cylindrical port will have less friction than a square port of the same area by a small factor, about 12%. The difference will be in the frictional port losses, which may not be a big contributor to verall enclosure losses to begin with. So, the answer is that it COULD, but it need not necessarily be so. |
#246
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
The original form is equally useful: if you know the design goals for enclosure size, system tuning and bandwidth, the relation will tell you the maximum efficiency you can expect out of the system. The relationship is, indeed, a beginning point in the system design process: it sets the minimum set of basic parameters needed to achieve the broadest of system constraints. It will tell you, for example, that a driver that's MORE efficienct that what the equation specifies WILL NOT WORK to achieve that system specification, something will have to be sacrificed: either the system will have to be tuned differently, the cutoff frequency has to be higher or the box has to be larger, or some combination thereof. As I remember, the cutoff for the low end was higher, so it fits the equation you gave. Did his design actually put out a higher SPL output level per wattage in than if in the "right" size box that the speaker was designed for. Who is "he" in the above and what design are we talkgin about. Are speaker parameters (mass/spider/surround/etc) designed for particular box sizes, In an ideal world, marketing tells engineering the most basic of parameters (enclosure size, intended range of receiver/amplifiers, tells you how much power, which might clue you as to required efficiency, etc.). From that, engineering the develops a detailed system specification and FROM THAT, driver parameters are derived, the driver is designed and the whole thing integrated together. It seldom happens that way, as the vats majority of speaker system manufacturers do not have the ability to custom design drivers, so they are forced to redesign the system to accomodate the available drivers. or is this where the volume of the port comes into play for different size boxes? A common misconception: it is NOT the volume of the port that is important. It's something a little more complex and obtuse than that. A vented enclosure is, in and of itself, a resonant acoustic circuit, whose reactive elements consist of the acoustic stiffness of the air in the box and the acoustic "inertance" or mass of the vent or port. The acoustic inertance IS NOT THE SAME THING as the physical mass of the air in the port. The air mass is basically the density of air times the volume of the port, Mp = p l pi d^2 / 4 where p is the density of air (about 1.18 kg/m^3), l is the length of the port in meters and d is the diameter of the port in meters (all this assume a cylindrical port). The mass is in units of kilograms. However, the acoustical mass is actually quite different: it's equal to the density of air times the effective length of the port DIVIDED BY the cross-sectional area of the port, or: Map = p l'/(pi d^2/4) and has the strange units of kg/m^4. This is what leads to the rather counter-intuitive property that if tow ports have the same length, but one is smaller in diameter than the other, the smaller port has the GREATER effective acoustic mass. Why is this so? well, let's engage in a little gedanken (a "thought experiment"). Consider two tubes, on is 1 cm (0.01m) in diameter and 10 m long, the other is 10 cm in diameter and 10 cm long. They both have precisely the same internal volume: V = 10m * pi * .01m^2/4 = 7.85 * 10^-4 m^3 V = .1m * pi * .1m^2/4 = 7.85 * 10^-4 m^3 and thus will have the same physical mass of about 9 * 10^-4 kg, of just under 1 gram. Now, assume both have a suitable mouthpiece at one end of the cylinder. Imagine trying the following experiment: using the mouthpiece, try alternatively sucking and blowing air into each cylinder repeating as rapidly as you possibly can. What you will find is that it is VERY esay to do with the short, fat cylinder and much harder to do with the long skinny tube. The reasona as mentioned is a little obtuse, but, simplifying it as much as possible, here's what you'll find, and the clue is in the term "intertance," or the resistance to CHANGE in motion. When you alternatively such and blow in the short fat cylinder, because of it's larger width, the volume of air your lungs are pumping in and out results in a very small change in velocity in the cylinder, whereas in the long skinny tube, the change in volume of air intriduced by your lungs results in a much higher velocity. SO the difference is that a narrow tube requires much larger chnages in velocity in the air inside the tube, and the higher velocity requires higher acceleration, and higher acceleration requires higher forces. Indeed, the differences in velocity in the air are propertional to the RECIPROCAL of the area of the port, or the reciprocal of the square of tne diameter, thus the seemingly strange result that making a port narrower INCREASES it's effective acoustic mass. Also, does it matter if a port is round or rectangular? I have heard both and if anyone could tell me the real truth, I figure it will be you. Over the range of port configurations where the ratio of width to height is small, in other words, it's near a square in cross section, there's little if any difference. The major effect at or near the enclosure reconance will be due to the differences in friction due to the fact that the wall area of a cylinder is as small as it can get, and no rectangular port can have an area as small. Assuming the same wall friction, a cylindrical port will have less friction than a square port of the same area by a small factor, about 12%. The difference will be in the frictional port losses, which may not be a big contributor to verall enclosure losses to begin with. So, the answer is that it COULD, but it need not necessarily be so. |
#247
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
The original form is equally useful: if you know the design goals for enclosure size, system tuning and bandwidth, the relation will tell you the maximum efficiency you can expect out of the system. The relationship is, indeed, a beginning point in the system design process: it sets the minimum set of basic parameters needed to achieve the broadest of system constraints. It will tell you, for example, that a driver that's MORE efficienct that what the equation specifies WILL NOT WORK to achieve that system specification, something will have to be sacrificed: either the system will have to be tuned differently, the cutoff frequency has to be higher or the box has to be larger, or some combination thereof. As I remember, the cutoff for the low end was higher, so it fits the equation you gave. Did his design actually put out a higher SPL output level per wattage in than if in the "right" size box that the speaker was designed for. Who is "he" in the above and what design are we talkgin about. Are speaker parameters (mass/spider/surround/etc) designed for particular box sizes, In an ideal world, marketing tells engineering the most basic of parameters (enclosure size, intended range of receiver/amplifiers, tells you how much power, which might clue you as to required efficiency, etc.). From that, engineering the develops a detailed system specification and FROM THAT, driver parameters are derived, the driver is designed and the whole thing integrated together. It seldom happens that way, as the vats majority of speaker system manufacturers do not have the ability to custom design drivers, so they are forced to redesign the system to accomodate the available drivers. or is this where the volume of the port comes into play for different size boxes? A common misconception: it is NOT the volume of the port that is important. It's something a little more complex and obtuse than that. A vented enclosure is, in and of itself, a resonant acoustic circuit, whose reactive elements consist of the acoustic stiffness of the air in the box and the acoustic "inertance" or mass of the vent or port. The acoustic inertance IS NOT THE SAME THING as the physical mass of the air in the port. The air mass is basically the density of air times the volume of the port, Mp = p l pi d^2 / 4 where p is the density of air (about 1.18 kg/m^3), l is the length of the port in meters and d is the diameter of the port in meters (all this assume a cylindrical port). The mass is in units of kilograms. However, the acoustical mass is actually quite different: it's equal to the density of air times the effective length of the port DIVIDED BY the cross-sectional area of the port, or: Map = p l'/(pi d^2/4) and has the strange units of kg/m^4. This is what leads to the rather counter-intuitive property that if tow ports have the same length, but one is smaller in diameter than the other, the smaller port has the GREATER effective acoustic mass. Why is this so? well, let's engage in a little gedanken (a "thought experiment"). Consider two tubes, on is 1 cm (0.01m) in diameter and 10 m long, the other is 10 cm in diameter and 10 cm long. They both have precisely the same internal volume: V = 10m * pi * .01m^2/4 = 7.85 * 10^-4 m^3 V = .1m * pi * .1m^2/4 = 7.85 * 10^-4 m^3 and thus will have the same physical mass of about 9 * 10^-4 kg, of just under 1 gram. Now, assume both have a suitable mouthpiece at one end of the cylinder. Imagine trying the following experiment: using the mouthpiece, try alternatively sucking and blowing air into each cylinder repeating as rapidly as you possibly can. What you will find is that it is VERY esay to do with the short, fat cylinder and much harder to do with the long skinny tube. The reasona as mentioned is a little obtuse, but, simplifying it as much as possible, here's what you'll find, and the clue is in the term "intertance," or the resistance to CHANGE in motion. When you alternatively such and blow in the short fat cylinder, because of it's larger width, the volume of air your lungs are pumping in and out results in a very small change in velocity in the cylinder, whereas in the long skinny tube, the change in volume of air intriduced by your lungs results in a much higher velocity. SO the difference is that a narrow tube requires much larger chnages in velocity in the air inside the tube, and the higher velocity requires higher acceleration, and higher acceleration requires higher forces. Indeed, the differences in velocity in the air are propertional to the RECIPROCAL of the area of the port, or the reciprocal of the square of tne diameter, thus the seemingly strange result that making a port narrower INCREASES it's effective acoustic mass. Also, does it matter if a port is round or rectangular? I have heard both and if anyone could tell me the real truth, I figure it will be you. Over the range of port configurations where the ratio of width to height is small, in other words, it's near a square in cross section, there's little if any difference. The major effect at or near the enclosure reconance will be due to the differences in friction due to the fact that the wall area of a cylinder is as small as it can get, and no rectangular port can have an area as small. Assuming the same wall friction, a cylindrical port will have less friction than a square port of the same area by a small factor, about 12%. The difference will be in the frictional port losses, which may not be a big contributor to verall enclosure losses to begin with. So, the answer is that it COULD, but it need not necessarily be so. |
#248
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Old speaker - See what you did Dick - lol
Just another tech who worked at the store.
Thanks for your answer even if it means more time to digest the math involved. A sealed 12" in a small box. John Who is "he" in the above and what design are we talkgin about. |
#249
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Old speaker - See what you did Dick - lol
Just another tech who worked at the store.
Thanks for your answer even if it means more time to digest the math involved. A sealed 12" in a small box. John Who is "he" in the above and what design are we talkgin about. |
#250
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Old speaker - See what you did Dick - lol
Just another tech who worked at the store.
Thanks for your answer even if it means more time to digest the math involved. A sealed 12" in a small box. John Who is "he" in the above and what design are we talkgin about. |
#251
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
Dick, would you care to share some general comments/insight regarding the behavior of spkr(s) in an open-back enclosure, with respect to the various basic formulae & behaviors you've mentioned for suspended or loaded ones? Dumb question I am sure, but do you mean by open back one that is not in a sealed or ported enclosure, and as an example, rear deck speakers going into a trunk? Partially, though the trunk of a car is not really big eneough to be considered truly open-backed, unless the suspension of the driver is very stiff. Here again, we come to the matter of what defines a particular enclosure or baffle type. We saw before that "acoustic suspension" speakers are those where the enclosure stiffness dominates the system stiffness, i.e., where the enclosure stiffness is at least 3-4 times that of the driver suspension stiffness. Well, in the case of an open-back speaker, a variant of which might be considered the "infinite baffle," it is the stiffness of the driver suspension that dominates and that provided by the enclosure is in- significant. As a rough estimate, I would say the total stiffness provided by the enclosure should be less than 1/10th that of the driver suspension. Generally, the behavior of an open back system, to a first approximation, is that of the driver itself in an infinite baffle. Thus, the response of the system, its efficiency and all that is determined by the raw driver parameters, for the most part. We saw earlier that there was a relationship between a speaker system's efficiency, it's enclosure volume and bandwidth. And we also learned that the efficiency of the system is pretty much set by that of the driver. We can now explore what determines the efficiency of the driver and, in this case, it IS a dependent variable as we would expect. That efficiency is, in terms of the driver's electromechanical units, is: p B^2 l^2 Sd^2 n0 = ------ * ------------ 2 pi c Re Mms^2 where p is the density of air, about 1.18 kg/m^3, c is the speed of sound, about 342 m/s, B is the flux density in the gap, in Tesla, l is the length of wire in the gap, in meters, Sd is the emissive area of the cone in square meters, Re is the DC resistance of the voice coil, in ohms and Mms is the effective moving mass of the cone, in kilograms. Again, to a first approximation, the response of such a system is that of the driver itself. What do I mean by "to a first approximation?" Well, in any open-back speaker, you have the complicating factor of an additional zero (a null, a cancellation) in the response at DC, where the front and rear waves of the driver are 180 out of phase and the resulting baffle delay is sufficient to cause complete cancellation. Above that point, the size and shape of the baffle determine the degree if cancellation in a complex fashion. This clearly modifies the response of the system such that it is no longer a simple 2nd-order response. The resulting 180 degree signal and it's attendant (complex) delay then causes the response to roll of in, essentially a 3rd or higher order fashion. |
#252
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
Dick, would you care to share some general comments/insight regarding the behavior of spkr(s) in an open-back enclosure, with respect to the various basic formulae & behaviors you've mentioned for suspended or loaded ones? Dumb question I am sure, but do you mean by open back one that is not in a sealed or ported enclosure, and as an example, rear deck speakers going into a trunk? Partially, though the trunk of a car is not really big eneough to be considered truly open-backed, unless the suspension of the driver is very stiff. Here again, we come to the matter of what defines a particular enclosure or baffle type. We saw before that "acoustic suspension" speakers are those where the enclosure stiffness dominates the system stiffness, i.e., where the enclosure stiffness is at least 3-4 times that of the driver suspension stiffness. Well, in the case of an open-back speaker, a variant of which might be considered the "infinite baffle," it is the stiffness of the driver suspension that dominates and that provided by the enclosure is in- significant. As a rough estimate, I would say the total stiffness provided by the enclosure should be less than 1/10th that of the driver suspension. Generally, the behavior of an open back system, to a first approximation, is that of the driver itself in an infinite baffle. Thus, the response of the system, its efficiency and all that is determined by the raw driver parameters, for the most part. We saw earlier that there was a relationship between a speaker system's efficiency, it's enclosure volume and bandwidth. And we also learned that the efficiency of the system is pretty much set by that of the driver. We can now explore what determines the efficiency of the driver and, in this case, it IS a dependent variable as we would expect. That efficiency is, in terms of the driver's electromechanical units, is: p B^2 l^2 Sd^2 n0 = ------ * ------------ 2 pi c Re Mms^2 where p is the density of air, about 1.18 kg/m^3, c is the speed of sound, about 342 m/s, B is the flux density in the gap, in Tesla, l is the length of wire in the gap, in meters, Sd is the emissive area of the cone in square meters, Re is the DC resistance of the voice coil, in ohms and Mms is the effective moving mass of the cone, in kilograms. Again, to a first approximation, the response of such a system is that of the driver itself. What do I mean by "to a first approximation?" Well, in any open-back speaker, you have the complicating factor of an additional zero (a null, a cancellation) in the response at DC, where the front and rear waves of the driver are 180 out of phase and the resulting baffle delay is sufficient to cause complete cancellation. Above that point, the size and shape of the baffle determine the degree if cancellation in a complex fashion. This clearly modifies the response of the system such that it is no longer a simple 2nd-order response. The resulting 180 degree signal and it's attendant (complex) delay then causes the response to roll of in, essentially a 3rd or higher order fashion. |
#253
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
Dick, would you care to share some general comments/insight regarding the behavior of spkr(s) in an open-back enclosure, with respect to the various basic formulae & behaviors you've mentioned for suspended or loaded ones? Dumb question I am sure, but do you mean by open back one that is not in a sealed or ported enclosure, and as an example, rear deck speakers going into a trunk? Partially, though the trunk of a car is not really big eneough to be considered truly open-backed, unless the suspension of the driver is very stiff. Here again, we come to the matter of what defines a particular enclosure or baffle type. We saw before that "acoustic suspension" speakers are those where the enclosure stiffness dominates the system stiffness, i.e., where the enclosure stiffness is at least 3-4 times that of the driver suspension stiffness. Well, in the case of an open-back speaker, a variant of which might be considered the "infinite baffle," it is the stiffness of the driver suspension that dominates and that provided by the enclosure is in- significant. As a rough estimate, I would say the total stiffness provided by the enclosure should be less than 1/10th that of the driver suspension. Generally, the behavior of an open back system, to a first approximation, is that of the driver itself in an infinite baffle. Thus, the response of the system, its efficiency and all that is determined by the raw driver parameters, for the most part. We saw earlier that there was a relationship between a speaker system's efficiency, it's enclosure volume and bandwidth. And we also learned that the efficiency of the system is pretty much set by that of the driver. We can now explore what determines the efficiency of the driver and, in this case, it IS a dependent variable as we would expect. That efficiency is, in terms of the driver's electromechanical units, is: p B^2 l^2 Sd^2 n0 = ------ * ------------ 2 pi c Re Mms^2 where p is the density of air, about 1.18 kg/m^3, c is the speed of sound, about 342 m/s, B is the flux density in the gap, in Tesla, l is the length of wire in the gap, in meters, Sd is the emissive area of the cone in square meters, Re is the DC resistance of the voice coil, in ohms and Mms is the effective moving mass of the cone, in kilograms. Again, to a first approximation, the response of such a system is that of the driver itself. What do I mean by "to a first approximation?" Well, in any open-back speaker, you have the complicating factor of an additional zero (a null, a cancellation) in the response at DC, where the front and rear waves of the driver are 180 out of phase and the resulting baffle delay is sufficient to cause complete cancellation. Above that point, the size and shape of the baffle determine the degree if cancellation in a complex fashion. This clearly modifies the response of the system such that it is no longer a simple 2nd-order response. The resulting 180 degree signal and it's attendant (complex) delay then causes the response to roll of in, essentially a 3rd or higher order fashion. |
#254
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
As I remember, the cutoff for the low end was higher, so it fits the equation you gave. Did his design actually put out a higher SPL output level per wattage in than if in the "right" size box that the speaker was designed for. Who is "he" in the above and what design are we talking about. Just another tech who worked at the store. Thanks for your answer even if it means more time to digest the math involved. A sealed 12" in a small box. No, that's what I'm trying to get people to understand. If all you are doing is changing the box size, the reference efficiency of the system will NOT change, because it CAN'T change: it's determined by the basic electromechanical properties of the driver. What changes is the cutoff frequency and, to a greater extent, the response shape, which is changing the system efficiency constant kn in the relation: n0 = kn Vb F3^3 For a sealed box, as an example, kn is governed by a variety of factors including the ratio of total damping to electrical damping, the cube of the ratio of the system resonant frequency to driver free air resonant frequency, and others. So changing the box size WILL change these ratios, and thus change kn. Changing the box size WILL change the response of resulting system, it will NOT change the reference efficiency. Again, a reasonable definition of efficiency involves the electrical parameters, such as the electrical resistance of the voice coil, the magnetic flux density and the amount of wire immersed in the magnetic field, as well as mechanical parameters, such as the mass of the cone and it's diameter. NONE of these parameters are changed by changing the box size, so efficiency does not change. |
#255
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
As I remember, the cutoff for the low end was higher, so it fits the equation you gave. Did his design actually put out a higher SPL output level per wattage in than if in the "right" size box that the speaker was designed for. Who is "he" in the above and what design are we talking about. Just another tech who worked at the store. Thanks for your answer even if it means more time to digest the math involved. A sealed 12" in a small box. No, that's what I'm trying to get people to understand. If all you are doing is changing the box size, the reference efficiency of the system will NOT change, because it CAN'T change: it's determined by the basic electromechanical properties of the driver. What changes is the cutoff frequency and, to a greater extent, the response shape, which is changing the system efficiency constant kn in the relation: n0 = kn Vb F3^3 For a sealed box, as an example, kn is governed by a variety of factors including the ratio of total damping to electrical damping, the cube of the ratio of the system resonant frequency to driver free air resonant frequency, and others. So changing the box size WILL change these ratios, and thus change kn. Changing the box size WILL change the response of resulting system, it will NOT change the reference efficiency. Again, a reasonable definition of efficiency involves the electrical parameters, such as the electrical resistance of the voice coil, the magnetic flux density and the amount of wire immersed in the magnetic field, as well as mechanical parameters, such as the mass of the cone and it's diameter. NONE of these parameters are changed by changing the box size, so efficiency does not change. |
#256
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
As I remember, the cutoff for the low end was higher, so it fits the equation you gave. Did his design actually put out a higher SPL output level per wattage in than if in the "right" size box that the speaker was designed for. Who is "he" in the above and what design are we talking about. Just another tech who worked at the store. Thanks for your answer even if it means more time to digest the math involved. A sealed 12" in a small box. No, that's what I'm trying to get people to understand. If all you are doing is changing the box size, the reference efficiency of the system will NOT change, because it CAN'T change: it's determined by the basic electromechanical properties of the driver. What changes is the cutoff frequency and, to a greater extent, the response shape, which is changing the system efficiency constant kn in the relation: n0 = kn Vb F3^3 For a sealed box, as an example, kn is governed by a variety of factors including the ratio of total damping to electrical damping, the cube of the ratio of the system resonant frequency to driver free air resonant frequency, and others. So changing the box size WILL change these ratios, and thus change kn. Changing the box size WILL change the response of resulting system, it will NOT change the reference efficiency. Again, a reasonable definition of efficiency involves the electrical parameters, such as the electrical resistance of the voice coil, the magnetic flux density and the amount of wire immersed in the magnetic field, as well as mechanical parameters, such as the mass of the cone and it's diameter. NONE of these parameters are changed by changing the box size, so efficiency does not change. |
#257
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Old speaker - See what you did Dick - lol
No, that's what I'm trying to get people to understand. If all you are doing is changing the box size, the reference efficiency of the system will NOT change, because it CAN'T change: it's determined by the basic electromechanical properties of the driver. What changes is the cutoff frequency and, to a greater extent, the response shape, which is changing the system efficiency constant kn in the relation: n0 = kn Vb F3^3 For a sealed box, as an example, kn is governed by a variety of factors including the ratio of total damping to electrical damping, the cube of the ratio of the system resonant frequency to driver free air resonant frequency, and others. So changing the box size WILL change these ratios, and thus change kn. Changing the box size WILL change the response of resulting system, it will NOT change the reference efficiency. Again, a reasonable definition of efficiency involves the electrical parameters, such as the electrical resistance of the voice coil, the magnetic flux density and the amount of wire immersed in the magnetic field, as well as mechanical parameters, such as the mass of the cone and it's diameter. NONE of these parameters are changed by changing the box size, so efficiency does not change. If I understand this right, the relative cut-off freq. and box volume are more "related" than having influence of cabinet volume on the total SPL of the cabinet. The efficiency of a system is driver related and not cabinet related? I think his reasoning was that by limiting the throw of the speaker by having a higher cut-off meant that it could take more wattage before the voice coil jumped out of the gap. Wouldn't more watts to get more volume at the new higher cut-off more likely fry/warp the coil before the spider lets it jump out of the gap? |
#258
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Old speaker - See what you did Dick - lol
No, that's what I'm trying to get people to understand. If all you are doing is changing the box size, the reference efficiency of the system will NOT change, because it CAN'T change: it's determined by the basic electromechanical properties of the driver. What changes is the cutoff frequency and, to a greater extent, the response shape, which is changing the system efficiency constant kn in the relation: n0 = kn Vb F3^3 For a sealed box, as an example, kn is governed by a variety of factors including the ratio of total damping to electrical damping, the cube of the ratio of the system resonant frequency to driver free air resonant frequency, and others. So changing the box size WILL change these ratios, and thus change kn. Changing the box size WILL change the response of resulting system, it will NOT change the reference efficiency. Again, a reasonable definition of efficiency involves the electrical parameters, such as the electrical resistance of the voice coil, the magnetic flux density and the amount of wire immersed in the magnetic field, as well as mechanical parameters, such as the mass of the cone and it's diameter. NONE of these parameters are changed by changing the box size, so efficiency does not change. If I understand this right, the relative cut-off freq. and box volume are more "related" than having influence of cabinet volume on the total SPL of the cabinet. The efficiency of a system is driver related and not cabinet related? I think his reasoning was that by limiting the throw of the speaker by having a higher cut-off meant that it could take more wattage before the voice coil jumped out of the gap. Wouldn't more watts to get more volume at the new higher cut-off more likely fry/warp the coil before the spider lets it jump out of the gap? |
#259
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Old speaker - See what you did Dick - lol
No, that's what I'm trying to get people to understand. If all you are doing is changing the box size, the reference efficiency of the system will NOT change, because it CAN'T change: it's determined by the basic electromechanical properties of the driver. What changes is the cutoff frequency and, to a greater extent, the response shape, which is changing the system efficiency constant kn in the relation: n0 = kn Vb F3^3 For a sealed box, as an example, kn is governed by a variety of factors including the ratio of total damping to electrical damping, the cube of the ratio of the system resonant frequency to driver free air resonant frequency, and others. So changing the box size WILL change these ratios, and thus change kn. Changing the box size WILL change the response of resulting system, it will NOT change the reference efficiency. Again, a reasonable definition of efficiency involves the electrical parameters, such as the electrical resistance of the voice coil, the magnetic flux density and the amount of wire immersed in the magnetic field, as well as mechanical parameters, such as the mass of the cone and it's diameter. NONE of these parameters are changed by changing the box size, so efficiency does not change. If I understand this right, the relative cut-off freq. and box volume are more "related" than having influence of cabinet volume on the total SPL of the cabinet. The efficiency of a system is driver related and not cabinet related? I think his reasoning was that by limiting the throw of the speaker by having a higher cut-off meant that it could take more wattage before the voice coil jumped out of the gap. Wouldn't more watts to get more volume at the new higher cut-off more likely fry/warp the coil before the spider lets it jump out of the gap? |
#260
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
No, that's what I'm trying to get people to understand. If all you are doing is changing the box size, the reference efficiency of the system will NOT change, because it CAN'T change: it's determined by the basic electromechanical properties of the driver. frequency, and others. So changing the box size WILL change these ratios, and thus change kn. Changing the box size WILL change the response of resulting system, it will NOT change the reference efficiency. Again, a reasonable definition of efficiency involves the electrical parameters, such as the electrical resistance of the voice coil, the magnetic flux density and the amount of wire immersed in the magnetic field, as well as mechanical parameters, such as the mass of the cone and it's diameter. NONE of these parameters are changed by changing the box size, so efficiency does not change. If I understand this right, the relative cut-off freq. and box volume are more "related" than having influence of cabinet volume on the total SPL of the cabinet. The efficiency of a system is driver related and not cabinet related? Okay, let's try this again with a PRECISE definition of "efficiency." There is a specirfic parameter called "reference efficiency." designated as n0. What this means, in short, is simply the ratio of the acoustical output power to the electrical input power at frequecnies above the low-frequency cutoff and below the radiation-load imposed high frequency limit. For a 12" woofer that might mean the efficiency in the bandwidth of about 100 Hz to 500 Hz or so. Given that definition, no, the box size for any given driver WILL NOT change the efficiency. It can't. Period. I think his reasoning was that by limiting the throw of the speaker by having a higher cut-off meant that it could take more wattage before the voice coil jumped out of the gap. Regrettably, his reasoning is simply wrong. It's as simple and as straightforward as that. Above the cutoff frequency of the system, the total stiffness of the enclosure DOES NOT AFFECT EFFICIENCY. Wouldn't more watts to get more volume at the new higher cut-off more likely fry/warp the coil before the spider lets it jump out of the gap? Once you are above the cutoff frequency, it takes EXACTLY the same input power to produce the same sound pressure level NO MATTER WHAT THE CABINET SIZE. More fundamentally, if you want to produce a specific sound pressure level at a given frequency, you MUST have a certain excursion on the woofer NO MATTER WHAT. Changin the box size CAN make a difference in power handling BELOW system cutoff and system resonance, NOT ABOVE. That's because, as you recall, the system is stiffness controlled BELOW resonance, and thus suspension stiffness and enclosure size matter BELOW resonance. But ABOVE resonance, the system is mass-controlled and, once again, the system stiffness IS NOT the deternmining factor. Thus, once again, ABOVE SYSTEM CUTOFF, CABINET SIZE WILL NOT AND CANNOT CHANGE THE SYSTEM EFFICIENCY OR SYSTEM POWER HANDLING. This may all seem counterintuitive, and that's simply because the common intuition about these things is just plain wrong. People push on the cone, feel the stiffness and thus conclude things about how it operates. Those conclusions are wrong because the speaker, in operation, is moving MUCH faster than you can push with your hand. You might be able to move the cone back and forth two or three times a second, but that's FAR below the cutoff frequency and, yes, YOU are operating the speaker in its stiffness controlled region. But that's NOT where the speaker operates: it operates at MUCH higher frequencies. Here's a real practical experiment that illustrates the point. Go out and find a brick. Yes, an ordinary brick. Put it in your hand and move it up and down, once a second or so. Not a lot of effort required, eh? Try moving it up and down the same distance twice a second. It's harder, about twice as hard. Now try it faster still. It's starting to get hard. You will QUICKLY reach a frequency where you simply do NOT have enough muscle power to move the brick "efficiently." Now, according to the premise under which you've been operating, since our little experiment has NO stiffness, it should be VERY EASY to move that brick. Without any stiffness, the efficiency should be high. But our experiment DIRECTLY contradicts that premise. That contradiction suggests very strongly that something is wrong with the premise that the efficiency of the system is determined by box size (which controls the system stiffness). Had your friend done the simple experiment of taking the same driver, putting it in two different boxes, driving each with the same level of pink noise limited to the bandwidth of the system, and measuring the output on a calibrated microphone placed at a fixed distance away, he would have found NO difference in the efficiency of these two systems, despite the difference in the size of the cabinets. Had your friend taken that same driver in those two different boxes, driven each one with an appropriately high-level signal above cutoff, and measured the actual excursion of each, he would have found that the excursion was exactly the same, despite the difference in the size of the cabinets. |
#261
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
No, that's what I'm trying to get people to understand. If all you are doing is changing the box size, the reference efficiency of the system will NOT change, because it CAN'T change: it's determined by the basic electromechanical properties of the driver. frequency, and others. So changing the box size WILL change these ratios, and thus change kn. Changing the box size WILL change the response of resulting system, it will NOT change the reference efficiency. Again, a reasonable definition of efficiency involves the electrical parameters, such as the electrical resistance of the voice coil, the magnetic flux density and the amount of wire immersed in the magnetic field, as well as mechanical parameters, such as the mass of the cone and it's diameter. NONE of these parameters are changed by changing the box size, so efficiency does not change. If I understand this right, the relative cut-off freq. and box volume are more "related" than having influence of cabinet volume on the total SPL of the cabinet. The efficiency of a system is driver related and not cabinet related? Okay, let's try this again with a PRECISE definition of "efficiency." There is a specirfic parameter called "reference efficiency." designated as n0. What this means, in short, is simply the ratio of the acoustical output power to the electrical input power at frequecnies above the low-frequency cutoff and below the radiation-load imposed high frequency limit. For a 12" woofer that might mean the efficiency in the bandwidth of about 100 Hz to 500 Hz or so. Given that definition, no, the box size for any given driver WILL NOT change the efficiency. It can't. Period. I think his reasoning was that by limiting the throw of the speaker by having a higher cut-off meant that it could take more wattage before the voice coil jumped out of the gap. Regrettably, his reasoning is simply wrong. It's as simple and as straightforward as that. Above the cutoff frequency of the system, the total stiffness of the enclosure DOES NOT AFFECT EFFICIENCY. Wouldn't more watts to get more volume at the new higher cut-off more likely fry/warp the coil before the spider lets it jump out of the gap? Once you are above the cutoff frequency, it takes EXACTLY the same input power to produce the same sound pressure level NO MATTER WHAT THE CABINET SIZE. More fundamentally, if you want to produce a specific sound pressure level at a given frequency, you MUST have a certain excursion on the woofer NO MATTER WHAT. Changin the box size CAN make a difference in power handling BELOW system cutoff and system resonance, NOT ABOVE. That's because, as you recall, the system is stiffness controlled BELOW resonance, and thus suspension stiffness and enclosure size matter BELOW resonance. But ABOVE resonance, the system is mass-controlled and, once again, the system stiffness IS NOT the deternmining factor. Thus, once again, ABOVE SYSTEM CUTOFF, CABINET SIZE WILL NOT AND CANNOT CHANGE THE SYSTEM EFFICIENCY OR SYSTEM POWER HANDLING. This may all seem counterintuitive, and that's simply because the common intuition about these things is just plain wrong. People push on the cone, feel the stiffness and thus conclude things about how it operates. Those conclusions are wrong because the speaker, in operation, is moving MUCH faster than you can push with your hand. You might be able to move the cone back and forth two or three times a second, but that's FAR below the cutoff frequency and, yes, YOU are operating the speaker in its stiffness controlled region. But that's NOT where the speaker operates: it operates at MUCH higher frequencies. Here's a real practical experiment that illustrates the point. Go out and find a brick. Yes, an ordinary brick. Put it in your hand and move it up and down, once a second or so. Not a lot of effort required, eh? Try moving it up and down the same distance twice a second. It's harder, about twice as hard. Now try it faster still. It's starting to get hard. You will QUICKLY reach a frequency where you simply do NOT have enough muscle power to move the brick "efficiently." Now, according to the premise under which you've been operating, since our little experiment has NO stiffness, it should be VERY EASY to move that brick. Without any stiffness, the efficiency should be high. But our experiment DIRECTLY contradicts that premise. That contradiction suggests very strongly that something is wrong with the premise that the efficiency of the system is determined by box size (which controls the system stiffness). Had your friend done the simple experiment of taking the same driver, putting it in two different boxes, driving each with the same level of pink noise limited to the bandwidth of the system, and measuring the output on a calibrated microphone placed at a fixed distance away, he would have found NO difference in the efficiency of these two systems, despite the difference in the size of the cabinets. Had your friend taken that same driver in those two different boxes, driven each one with an appropriately high-level signal above cutoff, and measured the actual excursion of each, he would have found that the excursion was exactly the same, despite the difference in the size of the cabinets. |
#262
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
No, that's what I'm trying to get people to understand. If all you are doing is changing the box size, the reference efficiency of the system will NOT change, because it CAN'T change: it's determined by the basic electromechanical properties of the driver. frequency, and others. So changing the box size WILL change these ratios, and thus change kn. Changing the box size WILL change the response of resulting system, it will NOT change the reference efficiency. Again, a reasonable definition of efficiency involves the electrical parameters, such as the electrical resistance of the voice coil, the magnetic flux density and the amount of wire immersed in the magnetic field, as well as mechanical parameters, such as the mass of the cone and it's diameter. NONE of these parameters are changed by changing the box size, so efficiency does not change. If I understand this right, the relative cut-off freq. and box volume are more "related" than having influence of cabinet volume on the total SPL of the cabinet. The efficiency of a system is driver related and not cabinet related? Okay, let's try this again with a PRECISE definition of "efficiency." There is a specirfic parameter called "reference efficiency." designated as n0. What this means, in short, is simply the ratio of the acoustical output power to the electrical input power at frequecnies above the low-frequency cutoff and below the radiation-load imposed high frequency limit. For a 12" woofer that might mean the efficiency in the bandwidth of about 100 Hz to 500 Hz or so. Given that definition, no, the box size for any given driver WILL NOT change the efficiency. It can't. Period. I think his reasoning was that by limiting the throw of the speaker by having a higher cut-off meant that it could take more wattage before the voice coil jumped out of the gap. Regrettably, his reasoning is simply wrong. It's as simple and as straightforward as that. Above the cutoff frequency of the system, the total stiffness of the enclosure DOES NOT AFFECT EFFICIENCY. Wouldn't more watts to get more volume at the new higher cut-off more likely fry/warp the coil before the spider lets it jump out of the gap? Once you are above the cutoff frequency, it takes EXACTLY the same input power to produce the same sound pressure level NO MATTER WHAT THE CABINET SIZE. More fundamentally, if you want to produce a specific sound pressure level at a given frequency, you MUST have a certain excursion on the woofer NO MATTER WHAT. Changin the box size CAN make a difference in power handling BELOW system cutoff and system resonance, NOT ABOVE. That's because, as you recall, the system is stiffness controlled BELOW resonance, and thus suspension stiffness and enclosure size matter BELOW resonance. But ABOVE resonance, the system is mass-controlled and, once again, the system stiffness IS NOT the deternmining factor. Thus, once again, ABOVE SYSTEM CUTOFF, CABINET SIZE WILL NOT AND CANNOT CHANGE THE SYSTEM EFFICIENCY OR SYSTEM POWER HANDLING. This may all seem counterintuitive, and that's simply because the common intuition about these things is just plain wrong. People push on the cone, feel the stiffness and thus conclude things about how it operates. Those conclusions are wrong because the speaker, in operation, is moving MUCH faster than you can push with your hand. You might be able to move the cone back and forth two or three times a second, but that's FAR below the cutoff frequency and, yes, YOU are operating the speaker in its stiffness controlled region. But that's NOT where the speaker operates: it operates at MUCH higher frequencies. Here's a real practical experiment that illustrates the point. Go out and find a brick. Yes, an ordinary brick. Put it in your hand and move it up and down, once a second or so. Not a lot of effort required, eh? Try moving it up and down the same distance twice a second. It's harder, about twice as hard. Now try it faster still. It's starting to get hard. You will QUICKLY reach a frequency where you simply do NOT have enough muscle power to move the brick "efficiently." Now, according to the premise under which you've been operating, since our little experiment has NO stiffness, it should be VERY EASY to move that brick. Without any stiffness, the efficiency should be high. But our experiment DIRECTLY contradicts that premise. That contradiction suggests very strongly that something is wrong with the premise that the efficiency of the system is determined by box size (which controls the system stiffness). Had your friend done the simple experiment of taking the same driver, putting it in two different boxes, driving each with the same level of pink noise limited to the bandwidth of the system, and measuring the output on a calibrated microphone placed at a fixed distance away, he would have found NO difference in the efficiency of these two systems, despite the difference in the size of the cabinets. Had your friend taken that same driver in those two different boxes, driven each one with an appropriately high-level signal above cutoff, and measured the actual excursion of each, he would have found that the excursion was exactly the same, despite the difference in the size of the cabinets. |
#263
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Old speaker - See what you did Dick - lol
To get more air moving, or SPL volume (as compared to cabinet size
volume to differentiate between the two types of volumes), one must either have more speaker cone area or more power to the speaker - which would create a longer throw. This is what I was taught about speakers. ( Same ol' teachers - lol ) From what you are saying, the excursion of the throw parameters of the speaker have no effect on volume? Logic seems to say that the more power, the further the voice coil would have to move in/out. More power to the coil should repel/attract the voice coil in the gap? The only thing I could think of would be that the cone movement would be faster with a higher current going through it? But that would change the frequency If my reasoning is off on this (and I am sure it is), how does a speaker get more SPL volume? Had your friend taken that same driver in those two different boxes, driven each one with an appropriately high-level signal above cutoff, and measured the actual excursion of each, he would have found that the excursion was exactly the same, despite the difference in the size of the cabinets. |
#264
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Old speaker - See what you did Dick - lol
To get more air moving, or SPL volume (as compared to cabinet size
volume to differentiate between the two types of volumes), one must either have more speaker cone area or more power to the speaker - which would create a longer throw. This is what I was taught about speakers. ( Same ol' teachers - lol ) From what you are saying, the excursion of the throw parameters of the speaker have no effect on volume? Logic seems to say that the more power, the further the voice coil would have to move in/out. More power to the coil should repel/attract the voice coil in the gap? The only thing I could think of would be that the cone movement would be faster with a higher current going through it? But that would change the frequency If my reasoning is off on this (and I am sure it is), how does a speaker get more SPL volume? Had your friend taken that same driver in those two different boxes, driven each one with an appropriately high-level signal above cutoff, and measured the actual excursion of each, he would have found that the excursion was exactly the same, despite the difference in the size of the cabinets. |
#265
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Old speaker - See what you did Dick - lol
To get more air moving, or SPL volume (as compared to cabinet size
volume to differentiate between the two types of volumes), one must either have more speaker cone area or more power to the speaker - which would create a longer throw. This is what I was taught about speakers. ( Same ol' teachers - lol ) From what you are saying, the excursion of the throw parameters of the speaker have no effect on volume? Logic seems to say that the more power, the further the voice coil would have to move in/out. More power to the coil should repel/attract the voice coil in the gap? The only thing I could think of would be that the cone movement would be faster with a higher current going through it? But that would change the frequency If my reasoning is off on this (and I am sure it is), how does a speaker get more SPL volume? Had your friend taken that same driver in those two different boxes, driven each one with an appropriately high-level signal above cutoff, and measured the actual excursion of each, he would have found that the excursion was exactly the same, despite the difference in the size of the cabinets. |
#266
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
To get more air moving, or SPL volume (as compared to cabinet size volume to differentiate between the two types of volumes), one must either have more speaker cone area or more power to the speaker - which would create a longer throw. This is what I was taught about speakers. ( Same ol' teachers - lol ) From what you are saying, the excursion of the throw parameters of the speaker have no effect on volume? Sorry, but it's difficult to parse this sentence, what with something about "excursion of throw parameters," but I absolutely did NOT say that excursion has not effect on volume. I made TWO statements: that the efficiency of the two example in the pass band of the system would have been the same for the same driver level AND the excursion would have been the same. Now, combine those two, you'll come to the conclusion that for a given sound pressure level at a given frequency, a specific excursion is required NO MATTER WHAT the cabinet size is. Logic seems to say that the more power, the further the voice coil would have to move in/out. More power to the coil should repel/attract the voice coil in the gap? I didn't say that, did I. Please read what I wrote: I SPECIFICALLY talked about THE SAME DRIVE LEVEL, didin't I? The only thing I could think of would be that the cone movement would be faster with a higher current going through it? But that would change the frequency Wrong. Absolutely wrong. There's a difference between the peak or average velocity of the cone and its frequency. If my reasoning is off on this (and I am sure it is), how does a speaker get more SPL volume? Well, I don't know what your reasoning is, but you're seriously misinterpreting what I said. Sound pressure level IS most definitely linked to excursion, I never said otherwise. Indeed, one finds that in the piston band, there is a direct link between the two: Pa prop (Sd F^2 X) where Pa is the acoustical power into the room, Sd is the emissive area of the cone, F is the frequency being produced, and X is the peak excursion. Notice that this relation is completely consistent with the two examples I gave. Notice also that NOWHERE in this relation is there ANY term for cabinet volume. None, zilch. What I said was, and I will say it but one more time: box size DOES NOT AND CANNOT CHANGE REFERENCE EFFICIENCY. Having said that, I have to say that this is the end of my participation in this particular topic for the time being. I've explained it as well as I can, and I am at the point of repeating myself, and I feel I'm not doing you or anyone else any good on this subject. I don't know why you're having the trouble you are, and I don't want to be seen whipping this particular dead horse (this topic, that is) any further. If you want to continue on with other topics, fine. But, at this point, without the understanding as to the why, I have to say you're going to have to accept the principles here on faith: box size DOES NOT and CAN NOT change reference efficiency. I've tried to explain why, from several different tacts, this is so, atht, most importantly, from the standpoint of the fundamental physics alone, a speaker is operating in its mass-controlled region in the passband, NOT the stiffness controlled region. I'm sorry I can't do any better than laying out the facts in front of you. So, with no further adieu, I'll thank the collected readers for their extreme patience and hope that we can all move on. |
#267
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
To get more air moving, or SPL volume (as compared to cabinet size volume to differentiate between the two types of volumes), one must either have more speaker cone area or more power to the speaker - which would create a longer throw. This is what I was taught about speakers. ( Same ol' teachers - lol ) From what you are saying, the excursion of the throw parameters of the speaker have no effect on volume? Sorry, but it's difficult to parse this sentence, what with something about "excursion of throw parameters," but I absolutely did NOT say that excursion has not effect on volume. I made TWO statements: that the efficiency of the two example in the pass band of the system would have been the same for the same driver level AND the excursion would have been the same. Now, combine those two, you'll come to the conclusion that for a given sound pressure level at a given frequency, a specific excursion is required NO MATTER WHAT the cabinet size is. Logic seems to say that the more power, the further the voice coil would have to move in/out. More power to the coil should repel/attract the voice coil in the gap? I didn't say that, did I. Please read what I wrote: I SPECIFICALLY talked about THE SAME DRIVE LEVEL, didin't I? The only thing I could think of would be that the cone movement would be faster with a higher current going through it? But that would change the frequency Wrong. Absolutely wrong. There's a difference between the peak or average velocity of the cone and its frequency. If my reasoning is off on this (and I am sure it is), how does a speaker get more SPL volume? Well, I don't know what your reasoning is, but you're seriously misinterpreting what I said. Sound pressure level IS most definitely linked to excursion, I never said otherwise. Indeed, one finds that in the piston band, there is a direct link between the two: Pa prop (Sd F^2 X) where Pa is the acoustical power into the room, Sd is the emissive area of the cone, F is the frequency being produced, and X is the peak excursion. Notice that this relation is completely consistent with the two examples I gave. Notice also that NOWHERE in this relation is there ANY term for cabinet volume. None, zilch. What I said was, and I will say it but one more time: box size DOES NOT AND CANNOT CHANGE REFERENCE EFFICIENCY. Having said that, I have to say that this is the end of my participation in this particular topic for the time being. I've explained it as well as I can, and I am at the point of repeating myself, and I feel I'm not doing you or anyone else any good on this subject. I don't know why you're having the trouble you are, and I don't want to be seen whipping this particular dead horse (this topic, that is) any further. If you want to continue on with other topics, fine. But, at this point, without the understanding as to the why, I have to say you're going to have to accept the principles here on faith: box size DOES NOT and CAN NOT change reference efficiency. I've tried to explain why, from several different tacts, this is so, atht, most importantly, from the standpoint of the fundamental physics alone, a speaker is operating in its mass-controlled region in the passband, NOT the stiffness controlled region. I'm sorry I can't do any better than laying out the facts in front of you. So, with no further adieu, I'll thank the collected readers for their extreme patience and hope that we can all move on. |
#268
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Old speaker - See what you did Dick - lol
Sanders wrote in message ...
To get more air moving, or SPL volume (as compared to cabinet size volume to differentiate between the two types of volumes), one must either have more speaker cone area or more power to the speaker - which would create a longer throw. This is what I was taught about speakers. ( Same ol' teachers - lol ) From what you are saying, the excursion of the throw parameters of the speaker have no effect on volume? Sorry, but it's difficult to parse this sentence, what with something about "excursion of throw parameters," but I absolutely did NOT say that excursion has not effect on volume. I made TWO statements: that the efficiency of the two example in the pass band of the system would have been the same for the same driver level AND the excursion would have been the same. Now, combine those two, you'll come to the conclusion that for a given sound pressure level at a given frequency, a specific excursion is required NO MATTER WHAT the cabinet size is. Logic seems to say that the more power, the further the voice coil would have to move in/out. More power to the coil should repel/attract the voice coil in the gap? I didn't say that, did I. Please read what I wrote: I SPECIFICALLY talked about THE SAME DRIVE LEVEL, didin't I? The only thing I could think of would be that the cone movement would be faster with a higher current going through it? But that would change the frequency Wrong. Absolutely wrong. There's a difference between the peak or average velocity of the cone and its frequency. If my reasoning is off on this (and I am sure it is), how does a speaker get more SPL volume? Well, I don't know what your reasoning is, but you're seriously misinterpreting what I said. Sound pressure level IS most definitely linked to excursion, I never said otherwise. Indeed, one finds that in the piston band, there is a direct link between the two: Pa prop (Sd F^2 X) where Pa is the acoustical power into the room, Sd is the emissive area of the cone, F is the frequency being produced, and X is the peak excursion. Notice that this relation is completely consistent with the two examples I gave. Notice also that NOWHERE in this relation is there ANY term for cabinet volume. None, zilch. What I said was, and I will say it but one more time: box size DOES NOT AND CANNOT CHANGE REFERENCE EFFICIENCY. Having said that, I have to say that this is the end of my participation in this particular topic for the time being. I've explained it as well as I can, and I am at the point of repeating myself, and I feel I'm not doing you or anyone else any good on this subject. I don't know why you're having the trouble you are, and I don't want to be seen whipping this particular dead horse (this topic, that is) any further. If you want to continue on with other topics, fine. But, at this point, without the understanding as to the why, I have to say you're going to have to accept the principles here on faith: box size DOES NOT and CAN NOT change reference efficiency. I've tried to explain why, from several different tacts, this is so, atht, most importantly, from the standpoint of the fundamental physics alone, a speaker is operating in its mass-controlled region in the passband, NOT the stiffness controlled region. I'm sorry I can't do any better than laying out the facts in front of you. So, with no further adieu, I'll thank the collected readers for their extreme patience and hope that we can all move on. |
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Old speaker - See what you did Dick - lol
Having said that, I have to say that this is the end of my participation in this particular topic for the time being. I've explained it as well as I can, and I am at the point of repeating myself, and I feel I'm not doing you or anyone else any good on this subject. I don't know why you're having the trouble you are, and I don't want to be seen whipping this particular dead horse (this topic, that is) any further. Sorry ! |
#270
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Old speaker - See what you did Dick - lol
Having said that, I have to say that this is the end of my participation in this particular topic for the time being. I've explained it as well as I can, and I am at the point of repeating myself, and I feel I'm not doing you or anyone else any good on this subject. I don't know why you're having the trouble you are, and I don't want to be seen whipping this particular dead horse (this topic, that is) any further. Sorry ! |
#271
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Old speaker - See what you did Dick - lol
Having said that, I have to say that this is the end of my participation in this particular topic for the time being. I've explained it as well as I can, and I am at the point of repeating myself, and I feel I'm not doing you or anyone else any good on this subject. I don't know why you're having the trouble you are, and I don't want to be seen whipping this particular dead horse (this topic, that is) any further. Sorry ! |
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