Ok I got it!!! Sorry for all the math but this should help anyone stumbling into this.

Left side of the circuit is the high and low pass based on the Baxendall circuit. Frequency cutoffs are

1/2Pi*RC

so.. for the high side its 1/2Pi*5600*8.2E-10 = 35K Hz (wow this doesn't look right?? should be 12.5K)

low side is 1/2Pi*55000*4.7E-8= 62Hz (close 70Hz)

Right side is a sweepable mid based on the Wien Bridge

1/2Pi*sqrt(r46*c27*r47*c28) when the pot is at zero ohm. (high sweep)

So its 1/2Pi* sqrt(3300*1.2E-8*3300*4.7E-9) = 6.4k Hz (spot on)

1/2Pi*sqrt(r46+vr4*c27*r47+vr4*c28) when the pot is at 55K ohm. (low sweep)

So… 1/2Pi*sqrt(103300*1.2E-8*103300*4.7E-9) = 205Hz (spot on)

If I want to lower the 205Hz I can double C27 and C28 but it also reduces the high sweep to 3.25k

1/2Pi* sqrt(3300*2.2E-8*3300*1E-8) = 3.25k Hz

1/2Pi*sqrt(103300*2.2E-8*103300*1 E-8) = 104 Hz

Or I can double only C27 and get 150 hz and 4.74, sound like a nice compromise?

1/2Pi* sqrt(3300*2.2E-8*3300*4.7E-9) = 4.74k Hz

1/2Pi*sqrt(103300*2.2E-8*103300*4.7E-9) = 151 Hz

Either way they are all getting replaced with NPO or mylar. thanks guys.

sources

http://www.headwize.com/projects/equal_prj.htm fig 1b

http://freecircuitdiagram.com/2008/09/15/wien-bridge-oscillator-the-circuit-schematic-diagram-and-formula/