On Sun, 18 Jun 2006 12:55:14 +0100, Glenn Richards
wrote:

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.

Really, is that all it was? Then why did you claim it illustrated the
beneficial effects of biwiring?

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.

C=R/2 whether they are strapped together or not. Do you really not see
this? (incidentally, please don't use C for a resistance, it is used
for capacitance).

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.

But it clearly shows that biwiring (ie not strapping together at the
bulb end) has no effect. What does have an effect is using a
sufficiently thick cable. If you are going to use a thought experiment
- even as poor a one as this - to prove a point, at least draw the
correct conclusion.

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.

Fine, but the effect is the reverse of what you claim. If you have a
cable so thin that it is harming the sound, then you don't improve it
by biwiring - the same cable resistance is still feeding each driver,
so the loss is the same. What you need to do is parallel the cables so
they are both connected together at both ends. This will have a good
effect because you have now halved the resistance of the cable feeding
each unit.

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.


No it won't. Go read some theory on diplexing.

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, you are living in a muddle. Please go and study.

d

--
Pearce Consulting
http://www.pearce.uk.com