Update..
Jim e-mailed me..I splained my issues with the whole shebang...not a problem.
I sent him this diagram with a (better) explanation..perhaps it will float Rod's, Bruno's, Bruce's , and Gene's
boats...
Gene/Clint...I believe that it is allowed for me to post the contents of MY e-mail to him. If this is in bad form, feel free to spank me and remove this post..
start of transmission.
The crux of my argument pertains to the current node between the feed wire from the amp, to the reactive branches of the crossover.
I've attached a diagram for you, it is of one of the nodes feeding the speaker. For ease of discussion, current A is hf sine, and current B is lf sine. The crossover components, the amp, and the frequencies are unimportant and independent from the currents within the wires, I'm sure you agree with that..Note that I choose "simple sines" as the mathematics of analysis using Vivaldi's "Spring" would be, shall we say, daunting?
The top is a generic diagram, showing the resistive dissipation on each of the wires. The middle is a monowire case, note the dissipation of the feed line, which of course contains the 2AB component. The bottom case is the biwiring case, where the losses of the feedlines are simply the squared currents without the 2AB component.
The 2AB component as a standalone entity is of course, nonsense. It has equal area above zero as well as below. It integrates to zero long term, and cannot exist without the current square terms. Since it is indeed integral zero, it in no way alters the RMS dissipation. But the 2AB term is not consistent with either load dissipation profiles when the loads branch, and the 2AB is precisely the difference between mono and biwire.
Audibility, I do not argue. I am not concerned with that.
The argument you put forth was that of one load, one wire. As such, your analysis is spot on. However, as you can see, it doesn't address the difference entity, the 2AB.
Thanks for responding..I hope you will entertain further discussion.."""
end of transmission..(beep)
edit:. Kirchhoffs current law states that the sum of all currents entering a node is zero. So at the node depicted, if you have a hf sine at 1 ampere and a lf sine at one ampere, the currents must add up in the feedwire.
the power that is dissipated within a resistor is P=I<sup>2</sup>R
And from simple algebra, (A+B)<sup>2</sup> = A<sup>2</sup> + B<sup>2</sup> + 2AB
Cheers, John