A conductor and a shield don't have significant capacitance by themselves. The capacitance exists BETWEEN conductors or BETWEEN a conductor and a shield.
I don't know exactly what the cable configuration is, but I'll hazard a guess that it provides two conductors for each channel, within a shield, and the cable terminates in rca connectors at the destination end, and both the shield and one of the two conductors for each channel are grounded to the rca ground sleeve. And that the 15pf/ft represents the capacitance between conductors, and the 37pf/ft represents the capacitance from the ungrounded conductor to the shield.
If that is correct, then the capacitance seen by the cartridge would correspond to the sum of the two capacitance figures you cited.
I would say that a very rough guess as to headshell and tonearm wiring capacitance, given that the total distance involved is probably a little more than a foot, would be 40pf.
So the total capacitance, based on those assumptions, is 40 + ((15 + 37) x (the number of feet in 1.2 meters)), which works out to 245 pf for each channel.
Regards,
-- Al
I don't know exactly what the cable configuration is, but I'll hazard a guess that it provides two conductors for each channel, within a shield, and the cable terminates in rca connectors at the destination end, and both the shield and one of the two conductors for each channel are grounded to the rca ground sleeve. And that the 15pf/ft represents the capacitance between conductors, and the 37pf/ft represents the capacitance from the ungrounded conductor to the shield.
If that is correct, then the capacitance seen by the cartridge would correspond to the sum of the two capacitance figures you cited.
I would say that a very rough guess as to headshell and tonearm wiring capacitance, given that the total distance involved is probably a little more than a foot, would be 40pf.
So the total capacitance, based on those assumptions, is 40 + ((15 + 37) x (the number of feet in 1.2 meters)), which works out to 245 pf for each channel.
... times the total number of feet right and left cable ?The capacitance on one channel has essentially no effect on the other channel, so the two cable lengths are not added together.
Regards,
-- Al