Sub woofer cable help please...

Glad to help. Although I doubt you will have to deal with the preamp loading issue that I had, just in case, the formula is:

L1 x L2/L1 + L2 + CI (Combined Impedance)

L1 = Main 1 load impedance
L2 = Main 2 Load impedance

So, for example, if L1 is 300K Ohms and L2 is 20K Ohms, CI = 300K x 20K/320K, or 18.75K Ohms. In my post above, I mentioned that the Ref 3 pre should NOT be presented with a load that presents LESS than 20K Ohms. Therefore, in my case, because the recommended load should not be less than 20K Ohms, the Reg 3 would be overloaded because the CI is only 18.75K Ohms. Overloading my Ref 3 could result in denigrated sonics and could prematurely age the tubes. FWIW

Thank you again I really appreciate it!

Good comments above.

I would add, though, that IMO even if the two sets of preamp outputs are separately buffered it would still not be good practice to sum the two channels together for the sub with a y-adapter. In saying so, I recognize that many people do exactly that, with reasonable results.

To the extent that the signals on the two channels differ, the output stages that drive the two channels will fight each other, conceivably with adverse sonic effects. That would be particularly likely to occur if an instrument producing deep bass tones were off to one side (i.e., essentially in just one channel).

A more technical way of putting it is that the output stage of each channel will have to drive the stereo component of the signal into a load impedance equal to (actually, slightly less than) the output impedance of the stage driving the other channel, which will be very low compared to the impedances it is designed to drive.

Regards,
-- Al

Thank you Al and all the rest who have posted. I appreciate the help as it turns out there was more to this than I anticipated.

Correction: In my post above, I wrote the formula to compute Combined Impedance is:

L1 x L2/L1 + L2 + CI (Combined Impedance). I should have used an equal sign in the formula, sorry. As corrected:

(L1 x L2)/(L1 + L2) = CI (Combined Impedance)