One also needs to bypass the passive crossovers and opening the cabinet is needed for this.
You're right, I should not have omitted this point.
In a buy-wireable speaker, the crossover is already split behind those terminals inside the speaker. The strap will not cause it to split or not. Look at it as if you are just extending that strap to the amp. Or, if you replace that strap with a longer wires, same length as the speaker wire, then just moving it to the speaker, nothing has changed as far as the crossovers are concerned.
Not exactly true, although the crossover is already split behind those terminals, the straps are causing the "split" crossover to be seen by the circuit that starts from the amp's speaker output terminals to the binding posts as one network, that is, not split. So the amp terminals will send the signals down this path, no matter how many pair of wires, according to the combined/equivalent impedance of both of the LF and HF XO circuits.
Once you remove the straps, each pair of the wires (bi-wire) will be able to see the individual LF & HF crossover circuits separately. You are right about "nothing has changed as far as the crossovers are concerned", but to the wires, it is different when the straps are removed. Think about it, if the straps are not removed, electrically the situation would be just as you described, i.e. like extending the straps to the amp's terminals, and I may add, it would also be like extending the straps to the HF & LF XO's terminals (i.e. effectively still joining the two XO's. With the straps removed, and using two pair of wires, the amp's output terminals now clearly see two separate paths, one goes from the amp to the LF driver's XO (impedance A), and the other goes from the amp's same terminals, but to the HF's XO (impedance B). With no linkes in between! Now, impedance A (LF drivers) has the characteristics of a low pass filter, so it would tend to block HF signals. Likewise, impedance B (HF drivers) would block LF signals. So in bi-wiring, each pair of wires will carry its own intended band of frequencies, as opposed to the full spectrum of frequencies flowing down the same pair (or pairs regardless of how many) of wires. Whether that would improve the sound we are able to hear is a separate argument. I am only trying to back up the claim that there is a difference electrically.
I have read about many claims that there is no electrical difference but none offer any valid argument that is based on electrical theory. I wish people would sit down, draw out a simple electrical circuit diagram, and then apply the theory. All it takes is Ohm's law, Kirchoff's law, basic series/parallel circuit theory and maths. Some understanding of the impedance characteristics of XO/filters will help. That said, I did read about one article (linked by an article named "the ten audio myth....or something like that) where someone attempted to prove bi-wiring makes no difference, and yes he included a diagram and some forumla, but it's presentation was so confusing that I have no idea what he was doing. Unfortunately articles like that ended up misleading people.
therefore each part of the crossover will present a different frequency/impedance characteristics to each pair of the wires.
All you are doing here is reducing the wire resistance by 1/2. Not an issue.
This is not correct. See explanation above.
One last point, apparently the jury is still out for the benefit of even true amping (with active XO), as people are still arguing about the pros and cons.
The experts are not arguing.
You may be right about this one, it depends on who we consider as "experts", I can post a link or two of those arguments if you wish.
With an active crossover, you are not at the mercy of a passive crossover and those issues; you can design circuits that have much better, sharper crossover characteristics, eliminating two different drivers from reproducing the same sounds, etc.
