Rower,sorry to respond before you have a chance to see my last post..
These two paragraphs tell me everything I need to know about you.
"quote"
So let's say you arrive at your "complex" impedance by adding the vector sum of the real and imaginary (capacitance and inductance) parts. That would be HUGELY capacitive to get to an 8-ohms value with such low real component resistance and inductance. Most of the magnitude is a CAPACITOR! Why on earth would you want to load the circuit with all that capacitance when POWER or VOLTAGE is NOT dropped across imaginary values but only the resistive one? Capacitors and Inductors store voltage and current, only to release it later on (minus their internal resistance, anyway). Add a bunch of imaginary capacitive component to your speaker leads and you create a messy situation even at RF. Talk about phase shift and imaging issues, there is no transfer of energy, just storage and release of energy at in opportune times. It doesn't sit around forever. The higher the capacitance, the worse it gets. We aren't storing nuts for the winter, we want to eat them as the come down the line.
The skin effect calculation is "wrong"? Well, All I see is you have a different opinion right now. Multiple credible sites use the most common methods and all arrive at about 18-mils at 20 KHz. Where is your documentation on your method? I agree that "approximations" can boil stuff down too far. Saying so is one thing, showing us is another. We're all tired of sayings.
"end of quote"
It is quite clear that you have not taken any courses in E/M theory.
I would be happy to recommend some texts for you. Don't worry, I won't recommend Jackson nor Becker. Both are far too involved for you (actually, for all humans). I was thinking Rojansky or Shadowitz perhaps.
Or ask.
As a start, several equations along your path of learning.
E = 1/2 L I squared
E = 1/2 C V squared
Z =sqr(L/C)
Using these equations, you should be able to determine the relationship between the energy stored within the cable as a consequence of capacitance, and the energy stored within the cable as a consequence of inductance. You may even note that they are equal when the signal travelling down the wire has the voltage to current relationship which matches the characteristic impedance of the cable.
You may or may not understand that the only signal that can travel down a wire pair is a signal which has the voltage to current relationship consistent with the cable impedance. It does.
You may or may not realize that is is impossible for a cable to carry a signal at the base propagation velocity if that signal has a V to I ration which is not that of the cable.
Therefore, when an amp says "100 volts" into a cable which has a characteristic impedance of 100 ohms, that 1 amp signal will travel to the load at the line impedance. NOT THE LOAD IMPEDANCE. If the load impedance is 100 ohms, the event is over once the signal arrives, typically 2 nSec per foot.
If the load is 10, one transit will NOT produce 10 amps into the load. It will take quite a few.
Very important point... YOU ARE CLAIMING SUPERLUMINAL SIGNAL VELOCITIES.
Doesn't happen.
AND, what I am saying has been measured. Get a reflection bridge.
you said:
""Information is power...get some.""
pssst. I have news for you..
Be nice, get nice. Be arrogant, get same.
John
These two paragraphs tell me everything I need to know about you.
"quote"
So let's say you arrive at your "complex" impedance by adding the vector sum of the real and imaginary (capacitance and inductance) parts. That would be HUGELY capacitive to get to an 8-ohms value with such low real component resistance and inductance. Most of the magnitude is a CAPACITOR! Why on earth would you want to load the circuit with all that capacitance when POWER or VOLTAGE is NOT dropped across imaginary values but only the resistive one? Capacitors and Inductors store voltage and current, only to release it later on (minus their internal resistance, anyway). Add a bunch of imaginary capacitive component to your speaker leads and you create a messy situation even at RF. Talk about phase shift and imaging issues, there is no transfer of energy, just storage and release of energy at in opportune times. It doesn't sit around forever. The higher the capacitance, the worse it gets. We aren't storing nuts for the winter, we want to eat them as the come down the line.
The skin effect calculation is "wrong"? Well, All I see is you have a different opinion right now. Multiple credible sites use the most common methods and all arrive at about 18-mils at 20 KHz. Where is your documentation on your method? I agree that "approximations" can boil stuff down too far. Saying so is one thing, showing us is another. We're all tired of sayings.
"end of quote"
It is quite clear that you have not taken any courses in E/M theory.
I would be happy to recommend some texts for you. Don't worry, I won't recommend Jackson nor Becker. Both are far too involved for you (actually, for all humans). I was thinking Rojansky or Shadowitz perhaps.
Or ask.
As a start, several equations along your path of learning.
E = 1/2 L I squared
E = 1/2 C V squared
Z =sqr(L/C)
Using these equations, you should be able to determine the relationship between the energy stored within the cable as a consequence of capacitance, and the energy stored within the cable as a consequence of inductance. You may even note that they are equal when the signal travelling down the wire has the voltage to current relationship which matches the characteristic impedance of the cable.
You may or may not understand that the only signal that can travel down a wire pair is a signal which has the voltage to current relationship consistent with the cable impedance. It does.
You may or may not realize that is is impossible for a cable to carry a signal at the base propagation velocity if that signal has a V to I ration which is not that of the cable.
Therefore, when an amp says "100 volts" into a cable which has a characteristic impedance of 100 ohms, that 1 amp signal will travel to the load at the line impedance. NOT THE LOAD IMPEDANCE. If the load impedance is 100 ohms, the event is over once the signal arrives, typically 2 nSec per foot.
If the load is 10, one transit will NOT produce 10 amps into the load. It will take quite a few.
Very important point... YOU ARE CLAIMING SUPERLUMINAL SIGNAL VELOCITIES.
Doesn't happen.
AND, what I am saying has been measured. Get a reflection bridge.
you said:
""Information is power...get some.""
pssst. I have news for you..
Be nice, get nice. Be arrogant, get same.
John