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#321
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... My argument was with your statement that "voltage drop in a conductor is only dependent upon its resistance". I see that you have now understood that error. Yes, yes, yes, mea culpa. I did say that I should not have included the word "only" when I made that statement. I wasn't using the term to exclude current from the equation but rather to exclude the load from any bearing on the resistance of the conductor -- and therefore it's voltage drop characteristics. |
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#322
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. Strictly speaking, as I stated above, it *is* equal to the current times the conductor impedance, *regardless*. Do the math to see. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? Yes, simple enough - and the current will be half, and the voltage will still equal the current times the resistance. Thanks for re-iterating the point I made several posts ago. I think (hope) we're saying the same thing. When you said voltage drop is purely proportional to the resistance, you are implying a linear relationship, like V(drop)=kRc. I just showed to you that it is not a linear relationship. In fact, it is V(drop)= Vs*Rc/(Rc+Rload). That is not a linear relationship between V(drop) and Rc. In fact, wehn Rload=0, V(drop) is *independent* of Rc. Now are you saying the same thing? You say it is *independant*, I say it is still equal to the current times the resistance. Am I wrong? |
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#323
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. Strictly speaking, as I stated above, it *is* equal to the current times the conductor impedance, *regardless*. Do the math to see. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? Yes, simple enough - and the current will be half, and the voltage will still equal the current times the resistance. Thanks for re-iterating the point I made several posts ago. I think (hope) we're saying the same thing. When you said voltage drop is purely proportional to the resistance, you are implying a linear relationship, like V(drop)=kRc. I just showed to you that it is not a linear relationship. In fact, it is V(drop)= Vs*Rc/(Rc+Rload). That is not a linear relationship between V(drop) and Rc. In fact, wehn Rload=0, V(drop) is *independent* of Rc. Now are you saying the same thing? You say it is *independant*, I say it is still equal to the current times the resistance. Am I wrong? |
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#324
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. Strictly speaking, as I stated above, it *is* equal to the current times the conductor impedance, *regardless*. Do the math to see. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? Yes, simple enough - and the current will be half, and the voltage will still equal the current times the resistance. Thanks for re-iterating the point I made several posts ago. I think (hope) we're saying the same thing. When you said voltage drop is purely proportional to the resistance, you are implying a linear relationship, like V(drop)=kRc. I just showed to you that it is not a linear relationship. In fact, it is V(drop)= Vs*Rc/(Rc+Rload). That is not a linear relationship between V(drop) and Rc. In fact, wehn Rload=0, V(drop) is *independent* of Rc. Now are you saying the same thing? You say it is *independant*, I say it is still equal to the current times the resistance. Am I wrong? |
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#325
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. Strictly speaking, as I stated above, it *is* equal to the current times the conductor impedance, *regardless*. Do the math to see. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? Yes, simple enough - and the current will be half, and the voltage will still equal the current times the resistance. Thanks for re-iterating the point I made several posts ago. I think (hope) we're saying the same thing. When you said voltage drop is purely proportional to the resistance, you are implying a linear relationship, like V(drop)=kRc. I just showed to you that it is not a linear relationship. In fact, it is V(drop)= Vs*Rc/(Rc+Rload). That is not a linear relationship between V(drop) and Rc. In fact, wehn Rload=0, V(drop) is *independent* of Rc. Now are you saying the same thing? You say it is *independant*, I say it is still equal to the current times the resistance. Am I wrong? |
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#326
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. Strictly speaking, as I stated above, it *is* equal to the current times the conductor impedance, *regardless*. Do the math to see. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? Yes, simple enough - and the current will be half, and the voltage will still equal the current times the resistance. Thanks for re-iterating the point I made several posts ago. I think (hope) we're saying the same thing. When you said voltage drop is purely proportional to the resistance, you are implying a linear relationship, like V(drop)=kRc. I just showed to you that it is not a linear relationship. In fact, it is V(drop)= Vs*Rc/(Rc+Rload). That is not a linear relationship between V(drop) and Rc. In fact, wehn Rload=0, V(drop) is *independent* of Rc. Now are you saying the same thing? You say it is *independant*, I say it is still equal to the current times the resistance. Am I wrong? Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. |
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#327
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. Strictly speaking, as I stated above, it *is* equal to the current times the conductor impedance, *regardless*. Do the math to see. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? Yes, simple enough - and the current will be half, and the voltage will still equal the current times the resistance. Thanks for re-iterating the point I made several posts ago. I think (hope) we're saying the same thing. When you said voltage drop is purely proportional to the resistance, you are implying a linear relationship, like V(drop)=kRc. I just showed to you that it is not a linear relationship. In fact, it is V(drop)= Vs*Rc/(Rc+Rload). That is not a linear relationship between V(drop) and Rc. In fact, wehn Rload=0, V(drop) is *independent* of Rc. Now are you saying the same thing? You say it is *independant*, I say it is still equal to the current times the resistance. Am I wrong? Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. |
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#328
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. Strictly speaking, as I stated above, it *is* equal to the current times the conductor impedance, *regardless*. Do the math to see. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? Yes, simple enough - and the current will be half, and the voltage will still equal the current times the resistance. Thanks for re-iterating the point I made several posts ago. I think (hope) we're saying the same thing. When you said voltage drop is purely proportional to the resistance, you are implying a linear relationship, like V(drop)=kRc. I just showed to you that it is not a linear relationship. In fact, it is V(drop)= Vs*Rc/(Rc+Rload). That is not a linear relationship between V(drop) and Rc. In fact, wehn Rload=0, V(drop) is *independent* of Rc. Now are you saying the same thing? You say it is *independant*, I say it is still equal to the current times the resistance. Am I wrong? Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. |
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#329
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: "chung" wrote in message vers.com... afh3 wrote: Yes, you are right. What I was trying to say here is that the amount of resistance of the conductor doesn't change based on the load value, and therefore the voltage drop on the conductor will always be proportional to it's own resistance (and equal to the resistance times the current). Strictly speaking, that is not correct. For example, if the load is a short-circuit, then the voltage across the conductor is the same as the open-circuit votage of the source, *regardless* of the conductor impedance. Strictly speaking, as I stated above, it *is* equal to the current times the conductor impedance, *regardless*. Do the math to see. If you do the math, you will see that in the case of a short-circuit load, the drop across the conductor is *not* proportiional to its resistance as you were saying. Here is what you said, to refresh your memory: "therefore the voltage drop on the conductor will always be proportional to it's own resistance". If the short applied is a true (z=0) short, then the only remaining resistance in the circuit is the conductor (ideal voltage sources having zero impedance and all), right? The amount of current flowing will be equal to the source voltage divided by the conductor resistance, right? So the amount of voltage present across the conductor (necessarily equal to the source voltage) will be the current through the conductor times it's resistance, right? If that ain't proportional, I guess I don't the meaning of the word. Here a simple way to look at it. If you double the resistance of the wire, the voltage across it does not double, if the load is a short (or an open). Simple enough? Yes, simple enough - and the current will be half, and the voltage will still equal the current times the resistance. Thanks for re-iterating the point I made several posts ago. I think (hope) we're saying the same thing. When you said voltage drop is purely proportional to the resistance, you are implying a linear relationship, like V(drop)=kRc. I just showed to you that it is not a linear relationship. In fact, it is V(drop)= Vs*Rc/(Rc+Rload). That is not a linear relationship between V(drop) and Rc. In fact, wehn Rload=0, V(drop) is *independent* of Rc. Now are you saying the same thing? You say it is *independant*, I say it is still equal to the current times the resistance. Am I wrong? Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. |
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#330
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Bi-wiring - Hogwash?
On Thu, 04 Mar 2004 14:12:04 GMT, "afh3" wrote:
"Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Stewart, I've read your posting here before and I know you understand this, so I've got to believe that you either WAY misread what I wrote above, or something is fundamentally wrong with how I was communicating my point. It is simply a fact that (to paraphrase directly what I said previously): "The voltage at the end of one of [a] conductor pair will be exactly the same as the voltage at end of the [same] conductor pair, minus the difference in the voltage drop across [the] pair of conductors REGARDLESS of what passive or reactive component is connected to each end." That is *not* a paraphrase of what you said earlier, which referred to the voltage at the ends of *both* conductor pairs being exactly the same. They're not, which makes your whole intial premise fundamentally wrong. The reason for this is that biwiring is just one pair of conductors with a voltage supply in the middle and a load on each end. The only possible difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair. And those voltage drops will be different if the currents are different - which they are. -- Oh ferchristsakes, of fricking course they will be different -- who ever said they would not be??? You implied it in your *original* claim above, which seems to have been backtracked as you went along. Read the sentence. It says that the "difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair" You either agree or you don't. Which is it? I can't tell from your non-sequitor reply. My reply is both concise and accurate. The voltages at the speaker terminals will be different, because the currents in the two pairs of conductors will be different. Your storu seems to be changing as you go along, and you are fundamentally incorrect regarding the isolation effect. OTOH, as Arny says, it's too small to make any audible difference. It does however exist, and is typically 50-70dB below the fundamental. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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#331
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Bi-wiring - Hogwash?
On Thu, 04 Mar 2004 14:12:04 GMT, "afh3" wrote:
"Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Stewart, I've read your posting here before and I know you understand this, so I've got to believe that you either WAY misread what I wrote above, or something is fundamentally wrong with how I was communicating my point. It is simply a fact that (to paraphrase directly what I said previously): "The voltage at the end of one of [a] conductor pair will be exactly the same as the voltage at end of the [same] conductor pair, minus the difference in the voltage drop across [the] pair of conductors REGARDLESS of what passive or reactive component is connected to each end." That is *not* a paraphrase of what you said earlier, which referred to the voltage at the ends of *both* conductor pairs being exactly the same. They're not, which makes your whole intial premise fundamentally wrong. The reason for this is that biwiring is just one pair of conductors with a voltage supply in the middle and a load on each end. The only possible difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair. And those voltage drops will be different if the currents are different - which they are. -- Oh ferchristsakes, of fricking course they will be different -- who ever said they would not be??? You implied it in your *original* claim above, which seems to have been backtracked as you went along. Read the sentence. It says that the "difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair" You either agree or you don't. Which is it? I can't tell from your non-sequitor reply. My reply is both concise and accurate. The voltages at the speaker terminals will be different, because the currents in the two pairs of conductors will be different. Your storu seems to be changing as you go along, and you are fundamentally incorrect regarding the isolation effect. OTOH, as Arny says, it's too small to make any audible difference. It does however exist, and is typically 50-70dB below the fundamental. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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#332
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Bi-wiring - Hogwash?
On Thu, 04 Mar 2004 14:12:04 GMT, "afh3" wrote:
"Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Stewart, I've read your posting here before and I know you understand this, so I've got to believe that you either WAY misread what I wrote above, or something is fundamentally wrong with how I was communicating my point. It is simply a fact that (to paraphrase directly what I said previously): "The voltage at the end of one of [a] conductor pair will be exactly the same as the voltage at end of the [same] conductor pair, minus the difference in the voltage drop across [the] pair of conductors REGARDLESS of what passive or reactive component is connected to each end." That is *not* a paraphrase of what you said earlier, which referred to the voltage at the ends of *both* conductor pairs being exactly the same. They're not, which makes your whole intial premise fundamentally wrong. The reason for this is that biwiring is just one pair of conductors with a voltage supply in the middle and a load on each end. The only possible difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair. And those voltage drops will be different if the currents are different - which they are. -- Oh ferchristsakes, of fricking course they will be different -- who ever said they would not be??? You implied it in your *original* claim above, which seems to have been backtracked as you went along. Read the sentence. It says that the "difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair" You either agree or you don't. Which is it? I can't tell from your non-sequitor reply. My reply is both concise and accurate. The voltages at the speaker terminals will be different, because the currents in the two pairs of conductors will be different. Your storu seems to be changing as you go along, and you are fundamentally incorrect regarding the isolation effect. OTOH, as Arny says, it's too small to make any audible difference. It does however exist, and is typically 50-70dB below the fundamental. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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#333
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Bi-wiring - Hogwash?
On Thu, 04 Mar 2004 14:12:04 GMT, "afh3" wrote:
"Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Stewart, I've read your posting here before and I know you understand this, so I've got to believe that you either WAY misread what I wrote above, or something is fundamentally wrong with how I was communicating my point. It is simply a fact that (to paraphrase directly what I said previously): "The voltage at the end of one of [a] conductor pair will be exactly the same as the voltage at end of the [same] conductor pair, minus the difference in the voltage drop across [the] pair of conductors REGARDLESS of what passive or reactive component is connected to each end." That is *not* a paraphrase of what you said earlier, which referred to the voltage at the ends of *both* conductor pairs being exactly the same. They're not, which makes your whole intial premise fundamentally wrong. The reason for this is that biwiring is just one pair of conductors with a voltage supply in the middle and a load on each end. The only possible difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair. And those voltage drops will be different if the currents are different - which they are. -- Oh ferchristsakes, of fricking course they will be different -- who ever said they would not be??? You implied it in your *original* claim above, which seems to have been backtracked as you went along. Read the sentence. It says that the "difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair" You either agree or you don't. Which is it? I can't tell from your non-sequitor reply. My reply is both concise and accurate. The voltages at the speaker terminals will be different, because the currents in the two pairs of conductors will be different. Your storu seems to be changing as you go along, and you are fundamentally incorrect regarding the isolation effect. OTOH, as Arny says, it's too small to make any audible difference. It does however exist, and is typically 50-70dB below the fundamental. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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Bi-wiring - Hogwash?
On Thu, 04 Mar 2004 14:43:27 GMT, "afh3" wrote:
"Stewart Pinkerton" wrote in message .. . On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote: I'm no tube bigot, but there are certainly some very fine Class A SET amps out there that offer pretty great performance. No there aren't - not one single one. Opinion of course. I don't want one, but there are many others with Bel Canto designs (among others) who would disagree with your assessment. When you refer to 'performance', that is generally taken to be a technical claim. Technically, *all* SETs are ****. There are those who will have nothing else (negative feedback being completely evil and all) because of the softer clipping and numerous other reasons. Triodes have internal negative feedback, and soft clipping is a sign of inadequate linearity below the clipping point. You may *like* that sound, but it ain't high fidelity! I was of course, referring the to the use of negative feedback in the sense that is typically applied when speaking of amplifier design - EXTERNAL negative feedback - just like about 99% of people would assume was meant in this context. And 90% of all SET amps have *external* negative feedback in each stage - it used to be called 'regeneration'. What they often don't have is *global* negative feedback - but 100% of SET amps have negative feedback of one kind or another. It's just another example of the technical incompetence (or sheer crookedness) of SET designers. I don't care for tubes. Others do. So be it. Sure, preference is another matter, as I said before. I still prefer solid state, but even among those you have to spend quite a bit of money to get output impedance values below or approaching .1 ohms. No, you don't, a Yamaha AX-592 has plenty low output impedance. I could find no references to the output impedance or damping factor for this model, so I will assume your statement is accurate -- very much unlike the one you make below. I have measured its predecessor, the AX-570, at a couple of milliohms. Thankfully, the brain-dead 'damping factor' measure was dropped in the mid-80s, after it climbed above 1,000 for the cheapest and nastiest 'Jap crap' amps. It is of course irrelevant when compared with the 10-15 feet of 18AWG wire generally used with such amps........... While Krells, Meridians, CJs and other quality solid-state amps can handily beat even that value, Actually, modern Krells do *not* have particularly low output impedance. Krell KAV500i = 0.044 ohms Krell FPB-300c = 0.069 ohms Are these modern enough? Should I go on? Those are hardly ultra-low figures, given that many much cheaper amps have output impedances well below 0.01 ohms. OTOH, that is a pretty useless measure in any case, and tells you very little about the real ability of the amplifier when driving tough loads. Note for instance that the 'cheap' KAV amp has *lower* output impedance than the 'cutting edge' FPB which has vastly higher drive capability. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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Bi-wiring - Hogwash?
On Thu, 04 Mar 2004 14:43:27 GMT, "afh3" wrote:
"Stewart Pinkerton" wrote in message .. . On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote: I'm no tube bigot, but there are certainly some very fine Class A SET amps out there that offer pretty great performance. No there aren't - not one single one. Opinion of course. I don't want one, but there are many others with Bel Canto designs (among others) who would disagree with your assessment. When you refer to 'performance', that is generally taken to be a technical claim. Technically, *all* SETs are ****. There are those who will have nothing else (negative feedback being completely evil and all) because of the softer clipping and numerous other reasons. Triodes have internal negative feedback, and soft clipping is a sign of inadequate linearity below the clipping point. You may *like* that sound, but it ain't high fidelity! I was of course, referring the to the use of negative feedback in the sense that is typically applied when speaking of amplifier design - EXTERNAL negative feedback - just like about 99% of people would assume was meant in this context. And 90% of all SET amps have *external* negative feedback in each stage - it used to be called 'regeneration'. What they often don't have is *global* negative feedback - but 100% of SET amps have negative feedback of one kind or another. It's just another example of the technical incompetence (or sheer crookedness) of SET designers. I don't care for tubes. Others do. So be it. Sure, preference is another matter, as I said before. I still prefer solid state, but even among those you have to spend quite a bit of money to get output impedance values below or approaching .1 ohms. No, you don't, a Yamaha AX-592 has plenty low output impedance. I could find no references to the output impedance or damping factor for this model, so I will assume your statement is accurate -- very much unlike the one you make below. I have measured its predecessor, the AX-570, at a couple of milliohms. Thankfully, the brain-dead 'damping factor' measure was dropped in the mid-80s, after it climbed above 1,000 for the cheapest and nastiest 'Jap crap' amps. It is of course irrelevant when compared with the 10-15 feet of 18AWG wire generally used with such amps........... While Krells, Meridians, CJs and other quality solid-state amps can handily beat even that value, Actually, modern Krells do *not* have particularly low output impedance. Krell KAV500i = 0.044 ohms Krell FPB-300c = 0.069 ohms Are these modern enough? Should I go on? Those are hardly ultra-low figures, given that many much cheaper amps have output impedances well below 0.01 ohms. OTOH, that is a pretty useless measure in any case, and tells you very little about the real ability of the amplifier when driving tough loads. Note for instance that the 'cheap' KAV amp has *lower* output impedance than the 'cutting edge' FPB which has vastly higher drive capability. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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Bi-wiring - Hogwash?
On Thu, 04 Mar 2004 14:43:27 GMT, "afh3" wrote:
"Stewart Pinkerton" wrote in message .. . On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote: I'm no tube bigot, but there are certainly some very fine Class A SET amps out there that offer pretty great performance. No there aren't - not one single one. Opinion of course. I don't want one, but there are many others with Bel Canto designs (among others) who would disagree with your assessment. When you refer to 'performance', that is generally taken to be a technical claim. Technically, *all* SETs are ****. There are those who will have nothing else (negative feedback being completely evil and all) because of the softer clipping and numerous other reasons. Triodes have internal negative feedback, and soft clipping is a sign of inadequate linearity below the clipping point. You may *like* that sound, but it ain't high fidelity! I was of course, referring the to the use of negative feedback in the sense that is typically applied when speaking of amplifier design - EXTERNAL negative feedback - just like about 99% of people would assume was meant in this context. And 90% of all SET amps have *external* negative feedback in each stage - it used to be called 'regeneration'. What they often don't have is *global* negative feedback - but 100% of SET amps have negative feedback of one kind or another. It's just another example of the technical incompetence (or sheer crookedness) of SET designers. I don't care for tubes. Others do. So be it. Sure, preference is another matter, as I said before. I still prefer solid state, but even among those you have to spend quite a bit of money to get output impedance values below or approaching .1 ohms. No, you don't, a Yamaha AX-592 has plenty low output impedance. I could find no references to the output impedance or damping factor for this model, so I will assume your statement is accurate -- very much unlike the one you make below. I have measured its predecessor, the AX-570, at a couple of milliohms. Thankfully, the brain-dead 'damping factor' measure was dropped in the mid-80s, after it climbed above 1,000 for the cheapest and nastiest 'Jap crap' amps. It is of course irrelevant when compared with the 10-15 feet of 18AWG wire generally used with such amps........... While Krells, Meridians, CJs and other quality solid-state amps can handily beat even that value, Actually, modern Krells do *not* have particularly low output impedance. Krell KAV500i = 0.044 ohms Krell FPB-300c = 0.069 ohms Are these modern enough? Should I go on? Those are hardly ultra-low figures, given that many much cheaper amps have output impedances well below 0.01 ohms. OTOH, that is a pretty useless measure in any case, and tells you very little about the real ability of the amplifier when driving tough loads. Note for instance that the 'cheap' KAV amp has *lower* output impedance than the 'cutting edge' FPB which has vastly higher drive capability. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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Bi-wiring - Hogwash?
On Thu, 04 Mar 2004 14:43:27 GMT, "afh3" wrote:
"Stewart Pinkerton" wrote in message .. . On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote: I'm no tube bigot, but there are certainly some very fine Class A SET amps out there that offer pretty great performance. No there aren't - not one single one. Opinion of course. I don't want one, but there are many others with Bel Canto designs (among others) who would disagree with your assessment. When you refer to 'performance', that is generally taken to be a technical claim. Technically, *all* SETs are ****. There are those who will have nothing else (negative feedback being completely evil and all) because of the softer clipping and numerous other reasons. Triodes have internal negative feedback, and soft clipping is a sign of inadequate linearity below the clipping point. You may *like* that sound, but it ain't high fidelity! I was of course, referring the to the use of negative feedback in the sense that is typically applied when speaking of amplifier design - EXTERNAL negative feedback - just like about 99% of people would assume was meant in this context. And 90% of all SET amps have *external* negative feedback in each stage - it used to be called 'regeneration'. What they often don't have is *global* negative feedback - but 100% of SET amps have negative feedback of one kind or another. It's just another example of the technical incompetence (or sheer crookedness) of SET designers. I don't care for tubes. Others do. So be it. Sure, preference is another matter, as I said before. I still prefer solid state, but even among those you have to spend quite a bit of money to get output impedance values below or approaching .1 ohms. No, you don't, a Yamaha AX-592 has plenty low output impedance. I could find no references to the output impedance or damping factor for this model, so I will assume your statement is accurate -- very much unlike the one you make below. I have measured its predecessor, the AX-570, at a couple of milliohms. Thankfully, the brain-dead 'damping factor' measure was dropped in the mid-80s, after it climbed above 1,000 for the cheapest and nastiest 'Jap crap' amps. It is of course irrelevant when compared with the 10-15 feet of 18AWG wire generally used with such amps........... While Krells, Meridians, CJs and other quality solid-state amps can handily beat even that value, Actually, modern Krells do *not* have particularly low output impedance. Krell KAV500i = 0.044 ohms Krell FPB-300c = 0.069 ohms Are these modern enough? Should I go on? Those are hardly ultra-low figures, given that many much cheaper amps have output impedances well below 0.01 ohms. OTOH, that is a pretty useless measure in any case, and tells you very little about the real ability of the amplifier when driving tough loads. Note for instance that the 'cheap' KAV amp has *lower* output impedance than the 'cutting edge' FPB which has vastly higher drive capability. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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Bi-wiring - Hogwash?
On Thu, 04 Mar 2004 14:43:27 GMT, "afh3" wrote:
"Stewart Pinkerton" wrote in message .. . On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote: I'm no tube bigot, but there are certainly some very fine Class A SET amps out there that offer pretty great performance. No there aren't - not one single one. Opinion of course. I don't want one, but there are many others with Bel Canto designs (among others) who would disagree with your assessment. They may not disagree. BelCanto's SET models are discontinued and all of the current line are Tripath-based SS units. Kal |
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Bi-wiring - Hogwash?
On Thu, 04 Mar 2004 14:43:27 GMT, "afh3" wrote:
"Stewart Pinkerton" wrote in message .. . On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote: I'm no tube bigot, but there are certainly some very fine Class A SET amps out there that offer pretty great performance. No there aren't - not one single one. Opinion of course. I don't want one, but there are many others with Bel Canto designs (among others) who would disagree with your assessment. They may not disagree. BelCanto's SET models are discontinued and all of the current line are Tripath-based SS units. Kal |
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Bi-wiring - Hogwash?
On Thu, 04 Mar 2004 14:43:27 GMT, "afh3" wrote:
"Stewart Pinkerton" wrote in message .. . On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote: I'm no tube bigot, but there are certainly some very fine Class A SET amps out there that offer pretty great performance. No there aren't - not one single one. Opinion of course. I don't want one, but there are many others with Bel Canto designs (among others) who would disagree with your assessment. They may not disagree. BelCanto's SET models are discontinued and all of the current line are Tripath-based SS units. Kal |
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Bi-wiring - Hogwash?
On Thu, 04 Mar 2004 14:43:27 GMT, "afh3" wrote:
"Stewart Pinkerton" wrote in message .. . On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote: I'm no tube bigot, but there are certainly some very fine Class A SET amps out there that offer pretty great performance. No there aren't - not one single one. Opinion of course. I don't want one, but there are many others with Bel Canto designs (among others) who would disagree with your assessment. They may not disagree. BelCanto's SET models are discontinued and all of the current line are Tripath-based SS units. Kal |
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. Fine. Did I disagree? Why do you feel the need to keep reiterating this? If the Rload=0 then the *only* resistive element in the circuit is the conductor so *of course* it's going to have the source voltage across it. And, as I have stated, in reply each time, it will *still* always be equal the current flowing through the conductor multiplied by it's resistance. Why is it neccesary to discuss this over and over? Did I ever contend otherwise? Quotes please. This is what I said about 75 posts ago. Put a voltage source in the middle of two conductors and hang a load on each end. Each load will have exactly the same voltage across it -- minus any minute differences contributed by the conductor's voltage drop. Therefore, the same signal is being applied to each driver circuit. No story change here. Granted, if you have an exceedingly small output impedance, you may be able to isolate *some* of the back-emf created by the individual drivers -- thereby reducing the interaction error -- but we've already agreed it won't be audible. What is the problem here then? Man, let's just invoke Godwin and stick a fork in it. -afh3 |
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. Fine. Did I disagree? Why do you feel the need to keep reiterating this? If the Rload=0 then the *only* resistive element in the circuit is the conductor so *of course* it's going to have the source voltage across it. And, as I have stated, in reply each time, it will *still* always be equal the current flowing through the conductor multiplied by it's resistance. Why is it neccesary to discuss this over and over? Did I ever contend otherwise? Quotes please. This is what I said about 75 posts ago. Put a voltage source in the middle of two conductors and hang a load on each end. Each load will have exactly the same voltage across it -- minus any minute differences contributed by the conductor's voltage drop. Therefore, the same signal is being applied to each driver circuit. No story change here. Granted, if you have an exceedingly small output impedance, you may be able to isolate *some* of the back-emf created by the individual drivers -- thereby reducing the interaction error -- but we've already agreed it won't be audible. What is the problem here then? Man, let's just invoke Godwin and stick a fork in it. -afh3 |
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. Fine. Did I disagree? Why do you feel the need to keep reiterating this? If the Rload=0 then the *only* resistive element in the circuit is the conductor so *of course* it's going to have the source voltage across it. And, as I have stated, in reply each time, it will *still* always be equal the current flowing through the conductor multiplied by it's resistance. Why is it neccesary to discuss this over and over? Did I ever contend otherwise? Quotes please. This is what I said about 75 posts ago. Put a voltage source in the middle of two conductors and hang a load on each end. Each load will have exactly the same voltage across it -- minus any minute differences contributed by the conductor's voltage drop. Therefore, the same signal is being applied to each driver circuit. No story change here. Granted, if you have an exceedingly small output impedance, you may be able to isolate *some* of the back-emf created by the individual drivers -- thereby reducing the interaction error -- but we've already agreed it won't be audible. What is the problem here then? Man, let's just invoke Godwin and stick a fork in it. -afh3 |
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. Fine. Did I disagree? Why do you feel the need to keep reiterating this? If the Rload=0 then the *only* resistive element in the circuit is the conductor so *of course* it's going to have the source voltage across it. And, as I have stated, in reply each time, it will *still* always be equal the current flowing through the conductor multiplied by it's resistance. Why is it neccesary to discuss this over and over? Did I ever contend otherwise? Quotes please. This is what I said about 75 posts ago. Put a voltage source in the middle of two conductors and hang a load on each end. Each load will have exactly the same voltage across it -- minus any minute differences contributed by the conductor's voltage drop. Therefore, the same signal is being applied to each driver circuit. No story change here. Granted, if you have an exceedingly small output impedance, you may be able to isolate *some* of the back-emf created by the individual drivers -- thereby reducing the interaction error -- but we've already agreed it won't be audible. What is the problem here then? Man, let's just invoke Godwin and stick a fork in it. -afh3 |
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Bi-wiring - Hogwash?
"Stewart Pinkerton" wrote in message ... On Thu, 04 Mar 2004 14:12:04 GMT, "afh3" wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Stewart, I've read your posting here before and I know you understand this, so I've got to believe that you either WAY misread what I wrote above, or something is fundamentally wrong with how I was communicating my point. It is simply a fact that (to paraphrase directly what I said previously): "The voltage at the end of one of [a] conductor pair will be exactly the same as the voltage at end of the [same] conductor pair, minus the difference in the voltage drop across [the] pair of conductors REGARDLESS of what passive or reactive component is connected to each end." That is *not* a paraphrase of what you said earlier, which referred to the voltage at the ends of *both* conductor pairs being exactly the same. They're not, which makes your whole intial premise fundamentally wrong. What???? Please go back and give the direct quote that this isn't a direct paraphrase of. Nevermind, I'll do it for you. "The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end." What I did say, for about the 10th time thank you, is that they would be identical MINUS THE DIFFERENCE IN THE VOLTAGE DROPS OF THE TWO SECTIONS OF CONDUCTORS. The reason for this is that biwiring is just one pair of conductors with a voltage supply in the middle and a load on each end. The only possible difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair. And those voltage drops will be different if the currents are different - which they are. -- Oh ferchristsakes, of fricking course they will be different -- who ever said they would not be??? You implied it in your *original* claim above, which seems to have been backtracked as you went along. Read the sentence. It says that the "difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair" You either agree or you don't. Which is it? I can't tell from your non-sequitor reply. My reply is both concise and accurate. The voltages at the speaker terminals will be different, because the currents in the two pairs of conductors will be different. Your storu seems to be changing as you go along, and you are fundamentally incorrect regarding the isolation effect. OTOH, as Arny says, it's too small to make any audible difference. It does however exist, and is typically 50-70dB below the fundamental. Changing my story? Oh this is rich. Please, if you are under the impression that my original contention that the voltage across the two end loads on a single conductor pair with a voltage source is middle will *NOT* be identicle save for the difference in voltage drops of the two conductors, then let's just call me a Nazi, I'll swear some more and you can have the last word -- which of course mean you've "won". Amazing. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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Bi-wiring - Hogwash?
"Stewart Pinkerton" wrote in message ... On Thu, 04 Mar 2004 14:12:04 GMT, "afh3" wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Stewart, I've read your posting here before and I know you understand this, so I've got to believe that you either WAY misread what I wrote above, or something is fundamentally wrong with how I was communicating my point. It is simply a fact that (to paraphrase directly what I said previously): "The voltage at the end of one of [a] conductor pair will be exactly the same as the voltage at end of the [same] conductor pair, minus the difference in the voltage drop across [the] pair of conductors REGARDLESS of what passive or reactive component is connected to each end." That is *not* a paraphrase of what you said earlier, which referred to the voltage at the ends of *both* conductor pairs being exactly the same. They're not, which makes your whole intial premise fundamentally wrong. What???? Please go back and give the direct quote that this isn't a direct paraphrase of. Nevermind, I'll do it for you. "The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end." What I did say, for about the 10th time thank you, is that they would be identical MINUS THE DIFFERENCE IN THE VOLTAGE DROPS OF THE TWO SECTIONS OF CONDUCTORS. The reason for this is that biwiring is just one pair of conductors with a voltage supply in the middle and a load on each end. The only possible difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair. And those voltage drops will be different if the currents are different - which they are. -- Oh ferchristsakes, of fricking course they will be different -- who ever said they would not be??? You implied it in your *original* claim above, which seems to have been backtracked as you went along. Read the sentence. It says that the "difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair" You either agree or you don't. Which is it? I can't tell from your non-sequitor reply. My reply is both concise and accurate. The voltages at the speaker terminals will be different, because the currents in the two pairs of conductors will be different. Your storu seems to be changing as you go along, and you are fundamentally incorrect regarding the isolation effect. OTOH, as Arny says, it's too small to make any audible difference. It does however exist, and is typically 50-70dB below the fundamental. Changing my story? Oh this is rich. Please, if you are under the impression that my original contention that the voltage across the two end loads on a single conductor pair with a voltage source is middle will *NOT* be identicle save for the difference in voltage drops of the two conductors, then let's just call me a Nazi, I'll swear some more and you can have the last word -- which of course mean you've "won". Amazing. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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Bi-wiring - Hogwash?
"Stewart Pinkerton" wrote in message ... On Thu, 04 Mar 2004 14:12:04 GMT, "afh3" wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Stewart, I've read your posting here before and I know you understand this, so I've got to believe that you either WAY misread what I wrote above, or something is fundamentally wrong with how I was communicating my point. It is simply a fact that (to paraphrase directly what I said previously): "The voltage at the end of one of [a] conductor pair will be exactly the same as the voltage at end of the [same] conductor pair, minus the difference in the voltage drop across [the] pair of conductors REGARDLESS of what passive or reactive component is connected to each end." That is *not* a paraphrase of what you said earlier, which referred to the voltage at the ends of *both* conductor pairs being exactly the same. They're not, which makes your whole intial premise fundamentally wrong. What???? Please go back and give the direct quote that this isn't a direct paraphrase of. Nevermind, I'll do it for you. "The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end." What I did say, for about the 10th time thank you, is that they would be identical MINUS THE DIFFERENCE IN THE VOLTAGE DROPS OF THE TWO SECTIONS OF CONDUCTORS. The reason for this is that biwiring is just one pair of conductors with a voltage supply in the middle and a load on each end. The only possible difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair. And those voltage drops will be different if the currents are different - which they are. -- Oh ferchristsakes, of fricking course they will be different -- who ever said they would not be??? You implied it in your *original* claim above, which seems to have been backtracked as you went along. Read the sentence. It says that the "difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair" You either agree or you don't. Which is it? I can't tell from your non-sequitor reply. My reply is both concise and accurate. The voltages at the speaker terminals will be different, because the currents in the two pairs of conductors will be different. Your storu seems to be changing as you go along, and you are fundamentally incorrect regarding the isolation effect. OTOH, as Arny says, it's too small to make any audible difference. It does however exist, and is typically 50-70dB below the fundamental. Changing my story? Oh this is rich. Please, if you are under the impression that my original contention that the voltage across the two end loads on a single conductor pair with a voltage source is middle will *NOT* be identicle save for the difference in voltage drops of the two conductors, then let's just call me a Nazi, I'll swear some more and you can have the last word -- which of course mean you've "won". Amazing. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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Bi-wiring - Hogwash?
"Stewart Pinkerton" wrote in message ... On Thu, 04 Mar 2004 14:12:04 GMT, "afh3" wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote: The statement I made was an assertion that the voltage drop across the conductor is proportional to the resistance of the conductor. The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end. Read Kirchoff's laws if this is unclear. No, it won't, because the currents in the two legs will be quite different. Read Ohm's Law if this is unclear. Stewart, I've read your posting here before and I know you understand this, so I've got to believe that you either WAY misread what I wrote above, or something is fundamentally wrong with how I was communicating my point. It is simply a fact that (to paraphrase directly what I said previously): "The voltage at the end of one of [a] conductor pair will be exactly the same as the voltage at end of the [same] conductor pair, minus the difference in the voltage drop across [the] pair of conductors REGARDLESS of what passive or reactive component is connected to each end." That is *not* a paraphrase of what you said earlier, which referred to the voltage at the ends of *both* conductor pairs being exactly the same. They're not, which makes your whole intial premise fundamentally wrong. What???? Please go back and give the direct quote that this isn't a direct paraphrase of. Nevermind, I'll do it for you. "The voltage at the end of one of the conductor pairs will be exactly the same as the voltage at end of the other conductor pair, minus the difference in the voltage drop across each pair of conductors REGARDLESS of what passive or reactive component is connected to each end." What I did say, for about the 10th time thank you, is that they would be identical MINUS THE DIFFERENCE IN THE VOLTAGE DROPS OF THE TWO SECTIONS OF CONDUCTORS. The reason for this is that biwiring is just one pair of conductors with a voltage supply in the middle and a load on each end. The only possible difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair. And those voltage drops will be different if the currents are different - which they are. -- Oh ferchristsakes, of fricking course they will be different -- who ever said they would not be??? You implied it in your *original* claim above, which seems to have been backtracked as you went along. Read the sentence. It says that the "difference in voltage at either end of the conductor pair is due to the voltage drop of that section of conductor pair minus the drop of the other section of conductor pair" You either agree or you don't. Which is it? I can't tell from your non-sequitor reply. My reply is both concise and accurate. The voltages at the speaker terminals will be different, because the currents in the two pairs of conductors will be different. Your storu seems to be changing as you go along, and you are fundamentally incorrect regarding the isolation effect. OTOH, as Arny says, it's too small to make any audible difference. It does however exist, and is typically 50-70dB below the fundamental. Changing my story? Oh this is rich. Please, if you are under the impression that my original contention that the voltage across the two end loads on a single conductor pair with a voltage source is middle will *NOT* be identicle save for the difference in voltage drops of the two conductors, then let's just call me a Nazi, I'll swear some more and you can have the last word -- which of course mean you've "won". Amazing. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. Fine. Did I disagree? Why do you feel the need to keep reiterating this? Because you quoted me wrong? And you seem to have trouble with what I meant by "independent"? If the Rload=0 then the *only* resistive element in the circuit is the conductor so *of course* it's going to have the source voltage across it. And, as I have stated, in reply each time, it will *still* always be equal the current flowing through the conductor multiplied by it's resistance. Why is it neccesary to discuss this over and over? Did I ever contend otherwise? Quotes please. See above. The objection is, of course, your saying that the drop is proportional to the resistance of the conductor. And since you have not admitted that mistake, we can go on  .Man, let's just invoke Godwin and stick a fork in it. Then don't ask any more questions. -afh3 |
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#351
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. Fine. Did I disagree? Why do you feel the need to keep reiterating this? Because you quoted me wrong? And you seem to have trouble with what I meant by "independent"? If the Rload=0 then the *only* resistive element in the circuit is the conductor so *of course* it's going to have the source voltage across it. And, as I have stated, in reply each time, it will *still* always be equal the current flowing through the conductor multiplied by it's resistance. Why is it neccesary to discuss this over and over? Did I ever contend otherwise? Quotes please. See above. The objection is, of course, your saying that the drop is proportional to the resistance of the conductor. And since you have not admitted that mistake, we can go on .Man, let's just invoke Godwin and stick a fork in it. Then don't ask any more questions. -afh3 |
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#352
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. Fine. Did I disagree? Why do you feel the need to keep reiterating this? Because you quoted me wrong? And you seem to have trouble with what I meant by "independent"? If the Rload=0 then the *only* resistive element in the circuit is the conductor so *of course* it's going to have the source voltage across it. And, as I have stated, in reply each time, it will *still* always be equal the current flowing through the conductor multiplied by it's resistance. Why is it neccesary to discuss this over and over? Did I ever contend otherwise? Quotes please. See above. The objection is, of course, your saying that the drop is proportional to the resistance of the conductor. And since you have not admitted that mistake, we can go on .Man, let's just invoke Godwin and stick a fork in it. Then don't ask any more questions. -afh3 |
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#353
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Bi-wiring - Hogwash?
afh3 wrote:
"chung" wrote in message rvers.com... Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. Fine. Did I disagree? Why do you feel the need to keep reiterating this? Because you quoted me wrong? And you seem to have trouble with what I meant by "independent"? If the Rload=0 then the *only* resistive element in the circuit is the conductor so *of course* it's going to have the source voltage across it. And, as I have stated, in reply each time, it will *still* always be equal the current flowing through the conductor multiplied by it's resistance. Why is it neccesary to discuss this over and over? Did I ever contend otherwise? Quotes please. See above. The objection is, of course, your saying that the drop is proportional to the resistance of the conductor. And since you have not admitted that mistake, we can go on .Man, let's just invoke Godwin and stick a fork in it. Then don't ask any more questions. -afh3 |
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#354
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... afh3 wrote: "chung" wrote in message rvers.com... Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. Fine. Did I disagree? Why do you feel the need to keep reiterating this? Because you quoted me wrong? And you seem to have trouble with what I meant by "independent"? No more so, it would appear, than you have with what I meant by "equal". If the Rload=0 then the *only* resistive element in the circuit is the conductor so *of course* it's going to have the source voltage across it. And, as I have stated, in reply each time, it will *still* always be equal the current flowing through the conductor multiplied by it's resistance. Why is it necessary to discuss this over and over? Did I ever contend otherwise? Quotes please. See above. The objection is, of course, your saying that the drop is proportional to the resistance of the conductor. And since you have not admitted that mistake, we can go on .I believe what you are attempting to say here, is that I made a mistake when I qualified the proportion with the word "only" about 10 posts back -- and then stated that it was a mistake, for about the fourth time now. Voltage drop most assuredly *is* proportional to the resistance of the conductor. The fact that it is also proportional to the current flowing through the conductor certainly does not invalidate the previous statement relating it to it's resistance. -afh3 |
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#355
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... afh3 wrote: "chung" wrote in message rvers.com... Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. Fine. Did I disagree? Why do you feel the need to keep reiterating this? Because you quoted me wrong? And you seem to have trouble with what I meant by "independent"? No more so, it would appear, than you have with what I meant by "equal". If the Rload=0 then the *only* resistive element in the circuit is the conductor so *of course* it's going to have the source voltage across it. And, as I have stated, in reply each time, it will *still* always be equal the current flowing through the conductor multiplied by it's resistance. Why is it necessary to discuss this over and over? Did I ever contend otherwise? Quotes please. See above. The objection is, of course, your saying that the drop is proportional to the resistance of the conductor. And since you have not admitted that mistake, we can go on .I believe what you are attempting to say here, is that I made a mistake when I qualified the proportion with the word "only" about 10 posts back -- and then stated that it was a mistake, for about the fourth time now. Voltage drop most assuredly *is* proportional to the resistance of the conductor. The fact that it is also proportional to the current flowing through the conductor certainly does not invalidate the previous statement relating it to it's resistance. -afh3 |
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#356
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... afh3 wrote: "chung" wrote in message rvers.com... Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. Fine. Did I disagree? Why do you feel the need to keep reiterating this? Because you quoted me wrong? And you seem to have trouble with what I meant by "independent"? No more so, it would appear, than you have with what I meant by "equal". If the Rload=0 then the *only* resistive element in the circuit is the conductor so *of course* it's going to have the source voltage across it. And, as I have stated, in reply each time, it will *still* always be equal the current flowing through the conductor multiplied by it's resistance. Why is it necessary to discuss this over and over? Did I ever contend otherwise? Quotes please. See above. The objection is, of course, your saying that the drop is proportional to the resistance of the conductor. And since you have not admitted that mistake, we can go on .I believe what you are attempting to say here, is that I made a mistake when I qualified the proportion with the word "only" about 10 posts back -- and then stated that it was a mistake, for about the fourth time now. Voltage drop most assuredly *is* proportional to the resistance of the conductor. The fact that it is also proportional to the current flowing through the conductor certainly does not invalidate the previous statement relating it to it's resistance. -afh3 |
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#357
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... afh3 wrote: "chung" wrote in message rvers.com... Please be careful quoting me. I said it is *independent* in the case of a *shorted* load. Of course, the drop is equal to current times resistance; it's Ohm's Law, and that was what I said several posts ago. Fine. Did I disagree? Why do you feel the need to keep reiterating this? Because you quoted me wrong? And you seem to have trouble with what I meant by "independent"? No more so, it would appear, than you have with what I meant by "equal". If the Rload=0 then the *only* resistive element in the circuit is the conductor so *of course* it's going to have the source voltage across it. And, as I have stated, in reply each time, it will *still* always be equal the current flowing through the conductor multiplied by it's resistance. Why is it necessary to discuss this over and over? Did I ever contend otherwise? Quotes please. See above. The objection is, of course, your saying that the drop is proportional to the resistance of the conductor. And since you have not admitted that mistake, we can go on .I believe what you are attempting to say here, is that I made a mistake when I qualified the proportion with the word "only" about 10 posts back -- and then stated that it was a mistake, for about the fourth time now. Voltage drop most assuredly *is* proportional to the resistance of the conductor. The fact that it is also proportional to the current flowing through the conductor certainly does not invalidate the previous statement relating it to it's resistance. -afh3 |
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#358
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... See above. The objection is, of course, your saying that the drop is proportional to the resistance of the conductor. And since you have not admitted that mistake, we can go on .Look, the only physical property that a wire has with relation to current and voltage, is it's resistance. Voltage is not a property of wire. Current is not a property of wire. To say the voltage drop on a wire is proportional to it's resistance is just plain not wrong. It's a simple case of a analysis with a dependant variable and an dependent variable. You tell me the current, I'll apply the proportion of the resistance and calculate the voltage drop. This is exactly like an analogy of the length of a building's shadow. It's a function of it's height and the sun's angle. Nothing I do to the building can impact the sun angle. The only parameter of the building I can adjust is it's height. Therefore, the length of a building's shadow is indeed proportional to it's height. When the sun angle changes, so does the shadow length, but sun angle is not a property of the building. It is perfectly valid syntax to state that the length of a building's shadow is proportional to it's height. In much the same way it's perfectly valid to state that the voltage drop on a conductor is proportional to it's resistance -- that being the only physical property the wire possesses that can change the voltage drop. -afh3 I'll try not to mention, over and over, as has been your style so far, that in your last sentence above you claimed I have "not" admitted a mistake. Turns out, I didn't have to. |
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#359
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... See above. The objection is, of course, your saying that the drop is proportional to the resistance of the conductor. And since you have not admitted that mistake, we can go on .Look, the only physical property that a wire has with relation to current and voltage, is it's resistance. Voltage is not a property of wire. Current is not a property of wire. To say the voltage drop on a wire is proportional to it's resistance is just plain not wrong. It's a simple case of a analysis with a dependant variable and an dependent variable. You tell me the current, I'll apply the proportion of the resistance and calculate the voltage drop. This is exactly like an analogy of the length of a building's shadow. It's a function of it's height and the sun's angle. Nothing I do to the building can impact the sun angle. The only parameter of the building I can adjust is it's height. Therefore, the length of a building's shadow is indeed proportional to it's height. When the sun angle changes, so does the shadow length, but sun angle is not a property of the building. It is perfectly valid syntax to state that the length of a building's shadow is proportional to it's height. In much the same way it's perfectly valid to state that the voltage drop on a conductor is proportional to it's resistance -- that being the only physical property the wire possesses that can change the voltage drop. -afh3 I'll try not to mention, over and over, as has been your style so far, that in your last sentence above you claimed I have "not" admitted a mistake. Turns out, I didn't have to. |
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#360
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Bi-wiring - Hogwash?
"chung" wrote in message rvers.com... See above. The objection is, of course, your saying that the drop is proportional to the resistance of the conductor. And since you have not admitted that mistake, we can go on .Look, the only physical property that a wire has with relation to current and voltage, is it's resistance. Voltage is not a property of wire. Current is not a property of wire. To say the voltage drop on a wire is proportional to it's resistance is just plain not wrong. It's a simple case of a analysis with a dependant variable and an dependent variable. You tell me the current, I'll apply the proportion of the resistance and calculate the voltage drop. This is exactly like an analogy of the length of a building's shadow. It's a function of it's height and the sun's angle. Nothing I do to the building can impact the sun angle. The only parameter of the building I can adjust is it's height. Therefore, the length of a building's shadow is indeed proportional to it's height. When the sun angle changes, so does the shadow length, but sun angle is not a property of the building. It is perfectly valid syntax to state that the length of a building's shadow is proportional to it's height. In much the same way it's perfectly valid to state that the voltage drop on a conductor is proportional to it's resistance -- that being the only physical property the wire possesses that can change the voltage drop. -afh3 I'll try not to mention, over and over, as has been your style so far, that in your last sentence above you claimed I have "not" admitted a mistake. Turns out, I didn't have to. |
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