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I took a B50K pot (measured 50K5 Ω) and a C50K pot (measured 47K8) and measured them at the following positions:

Position     “B” Value     “C” Value

  • Full Left         0 Ω              0 Ω
  •  8:00            3K1 Ω          0K2 Ω
  •  9:00            8K2 Ω         11K1 Ω
  • 10:00         14K8 Ω         31K3 Ω
  • 11:00         20K8 Ω         39K6 Ω
  • 12:00          25K Ω          41K1 Ω
  •  1:00          30K9 Ω         43K4 Ω
  •  2:00          36K8 Ω         44K8 Ω
  •  3:00          43K8 Ω         46K6 Ω
  •  4:00          48K7 Ω         47K2 Ω
  • Full Rt        50K5 Ω         47K8 Ω

Then I tied 51K (2 x 25K5) Ω across their lugs #1 & #3 respectively

The “B” then measured @ 25K2 Ω and the “C” @ 24K8 Ω across lugs #1 & #3  . . . .

Position     “B” Value     “C” Value

  • Full Left         0 Ω               0 Ω
  •  8:00            3K3 Ω          0K6 Ω
  •  9:00            9K2 Ω         11K7 Ω
  • 10:00         15K1 Ω         20K1 Ω
  • 11:00         17K2 Ω         23K8 Ω
  • 12:00         19K4 Ω         24K1 Ω
  •  1:00          21K9 Ω         24K4 Ω
  •  2:00          23K5 Ω         24K8 Ω
  •  3:00          24K7 Ω         24K6 Ω
  •  4:00          25K4 Ω         24K2 Ω
  • Full Rt        25K2 Ω         24K8 Ω

Sorry for sloppy list format, I couldn’t find my graph paper to do a proper graph. Still, it tends to support the concept that adding a resistor across lugs #1 & #3 of a pot does change its taper.

And after doing all of this, it occurs to me that the “sweep” behavior (taper) of a Tone pot is not as critical as Volume, Gain or Drive.

So, I shall go with the B50K dual-gang concentric pot and strap a 50K resistor across lugs #1 & #3 for the Tone pot. I’ll just live with the tiny bit of weirdness toward the extreme full-right position.

Thanks for the feedback Wilkie. Your initial counter-point prompted me to do the manual measurements and re-acquaint myself with the “mathematic visual” aspect of pot tapes. Good link too, by the way. Thanks for that. I gotta new bookmark!