Dielectrics and such...??

Hi Ben,

I'm not sure if you are saying that the four cables would be laid out between two parallel strips of wide tape, so that most of what is between adjacent cables is air, or if the tape would be very thickly wound around each cable, so that most of what is between them is tape.

In either case, though, I don't think that the dielectric properties of the tape would have much if any relevance. Although of course I would be more certain of that in the first of those two situations.

Given that each of the cables has its + and - conductors closely and extensively intertwined, and considering the one inch spacing between cables, I would expect any interaction between adjacent cables to be insignificant compared to what occurs between the intertwined conductors, and the dielectric effects of their individual insulation.

So I would expect the net effects of what you are proposing to simply be a factor of 4 reduction of the resistance and inductance that is specified for an individual cable, and an increase in the specified capacitance by a factor of 4. None of those changes strike me as unreasonable in any way, with this particular cable, but your guess is as good as mine as to what the resulting sonic effects will be.

P.S: Just saw Kijanki's response. Although the four cables would pick up noise unequally, wouldn't the noise picked up by any one of them be cancelled just as well as if it were the only cable that was present?

Best regards,
-- Al

Al, Yes part of the noise would be cancelled. Twisting just offers better cancellation.

Reduction of inductance might be more important than reduction in resistance since xover inductor in series with the woofer has most likely around 0.08ohm limiting DF to 100. Assuming he can cut inductance by 2 it will be 200nH per ft less, making two runs of 10' equal 4uH equivalent to 0.5ohm reactance reduction at 20kHz.

Hi Kijanki,

I suspect that you are basing those calculations on typical numbers for two conductors that are not twisted. See the link in my previous post, though, which shows a photo and the specs of the specific cable. And note that the inductance number is for 2.5 meters.

So due to the braided and twisted construction of the cable, it appears that the inductance is only about 87 nH per foot-run, for a single cable. A double-shotgun configuration (four two-conductor cables in parallel for each speaker) would reduce that to about 22 nH per foot-run, which of course is extremely low.

Best regards,
-- Al

Al, I didn't notice the link you provided. I agree that 22nH per foot would be extremely low and we can take it out of the equation. So what else is there (other than resistance, inductance or dielectric) to consider?

So what else is there (other than resistance, inductance or dielectric) to consider?

Nothing else that I can think of, beyond what we've already said. I don't think that dielectric absorption will be an issue. The electric field between adjacent paralleled cables will be minimal, partly because of the one-inch distance, and partly because of the cancellation resulting from the twisted and braided construction of the two conductors in each of the four paralleled cables. Also, depending on which of the two physical configurations I asked about is used, the dielectric between paralleled cables may be mostly air, which as I understand it has essentially no susceptibility to dielectric absorption.

Total capacitance (between + and -) for a 2.5 meter run (about 8.2 feet) would be 1448 pf, somewhat on the high side but I suspect still low enough to not have audible consequences with the great majority of amplifiers.

Best regards,
-- Al