Polk432 replied but I see that content in my mailbox, not here on this forum. Polk432, thanks. Any reply helps. So does yours, by confirming my story for instance, and confirming what I thought to know already. But my post doesn't seem to be complete enough.
I know how to replace the resistors. WITT v1s have no access hatch like model II. The heavy-weight resin block with the x-over concealed in it has the resistors' soldering contacts on top of it, so it needs to come out completely, together with the the heavy-weight bakelite bottom panel on which it is mounted. The fragile wires running to the speakers are too short to comfortly handle the x-over, and are VERY in the way when putting stuff together again. It's a two-man job for sure to get to the soldering contacts, and I'd like to fix the problem once and for all.
I bought my WITTs used too. To get a complete set of new resistors from Wilson to my place in the Netherlands would cost about 400 dollars/euros. That would be resistors which are common to be blown, from a manufacturer who says they can't handle compressed audio, in a set that is sold elsewhere for about 50 bucks. So I'd like to use different resistors, heavier duty anyway. From the same Caddock product line I can assemble a heavy duty set for abt 160 bucks.
In similar configurations audio enthusiasts seem to use inductive-free wirewound resistors to great satisfactory though. Something like bifilar nickel chromium wire, Ayrton-Perry-style wound over a ceramic-aluminium composite core, by manufacturer Mills, but I can only find 12w rated resistors in that product line. Something like 16w AchrOhmiC resistors would be abt the same story. So how could I replace the bass resistors with wirewounds?
Even if my WITTs were prefectly serviced for free, at my house, guaranteed to never blow again, I'd still like to learn how and why the resistors are wired, and why it's common to blow them. BTW, due to different mid & low speakers, WITT II resistors have different values: 7.9Ω(treble) / 5.6Ω(mid) / 14.2Ω(bass) / 4.2Ω(phase). I got all resistor values from the Wilson website.
Please don't get me wrong. I'm not grumpy over this problem. To me it's an opportunity to learn. Meanwhile I happily listen to another set of speakers.