Don Pearce wrote:

OK - you snipped the substance, as you had no way of dealing with it.

I think the phrase "couldn't be bothered" springs to mind. But since you
insist...

So I will repeat it again:

yawn

First lets assume that your experiment is applicable. Now complete
it. Use you second case - with the two wires run separately right
back to the source ( for maximum brightness). Now take your short
piece of wire and join the two bulbs together. If biwiring made a
difference, there should be a change in brightness as you do this.
There is no change. All you have shown is that it is usually better
to have thicker wire.

My experiment was a very simple one, designed to demonstrate the effect
of a long run of cable acting like a series resistor. By strapping the
bulbs together when "bi-wired", you're then achieving "C = R / 2", where
C is the series resistance of the cable(s) between the bonding strap and
the amplifier (or power source in this case), and R is the series
resistance of each individual length of cable. Of course this assumes
that both runs of cable are identical.

Now, this experiment uses DC, so only demonstrates resistance. A DC
signal is not affected by inductance (or indeed capacitance between the
cores - series capacitance will block DC but parallel won't). So it
serves only to demonstrate one of several mechanisms in play that means
bi-wiring will give an improvement.

Go back and read my earlier posting where I stated that eventually
you'll reach a thickness of cable where the series resistance becomes so
low that this effect no longer occurs. But a cable this thick will
either be very unwieldy or (comparatively) very expensive, therefore
it's more convenient (or cost effective, as the case may be) to use two
runs of thinner/cheaper cable.

The second point is that the tweeter and woofer are not in parallel.
Does that surprise you? This because we are dealing with signals in
defined frequency bands, and we have a crossover, which presents a
high impedance to the cable in the stopband of each driver. [snip]

Yes, except - there's an equivalent series resistor in the feed to the
"high impedance load". So a voltage drop across the LF *will* affect the HF.

By bi-wiring, you split the signal before this "series resistor",
therefore a voltage drop across the LF cable won't affect the HF.

Right, enough time wasted on this for now. Let's see if your limited
understanding can make sense of the above...

--
Glenn Richards Tel: (01453) 845735
Squirrel Solutions http://www.squirrelsolutions.co.uk/

IT consultancy, hardware and software support, broadband installation