In article , Glenn
Richards wrote:
Serge Auckland wrote:

You've obviously convinced yourself that it makes a difference. I've
never managed to hear any myself, and when I do the sums, I'm not a
bit surprised. By the way, if you are indeed sure it does make a
difference, have you tried to analyse why and how? What mechanism can
be acting to make the sound better (or even different)?

[snip]

Once you take the bridging straps off however, something else is
occurring. And what happens makes an audible difference. Too tired to do
the maths atm, but I'm sure a few ASCII art diagrams will help.

Actually, if people visit the webpages I've been mentioning they can see
the relevant diagrams displayed as gif images on the pages. Also the
relevant equiations modelling the arrangements, and the results for a
simple example. :-)

[snip]


Treat the run of cable between the amp and speaker as if it were a
resistor and it makes it easier to understand what's happening. And
remember that the LF driver can take a hefty current when you're trying
to move a lot of air - and that current is effectively being drawn
through a series resistor (ie the run of speaker cable). By bi-wiring,
you're no longer drawing that high current required by the LF driver
through the same series resistor as the HF driver.

That is correct. but you are drawing through another series resistor.

If the bi-wire cables are similar to the one used for conventional
wiring, then the HF signal will pass through the same series resistance
as before, and endure a similar Ohmic drop.

The result - much more linearity from the HF driver, as it's no longer
suffering from current drain via a series resistor when the LF driver
draws current.

That does not seem to follow from the above, and seem either incorrect or
confused. You gave changed one resistor (linear device) with another.
None of the above gives any reason for the result to be "much more linear".

If you change from using a common cable of series resistance, R, to one of
R feeding the HF driver, it will see a cable resistance of R in both cases.

The resistance of the cable will essentially be a linear element in both
cases.

Thus there is no need to assume any significant result in terms of "much
more linearity from the HF driver".

Therefore no current sag to the HF driver, resulting in a cleaner and
more dynamic HF response.

Not clear what 'current sag' you are assuming, nor why...

Slainte,

Jim

--
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