I've built a guitar pedal using the daisy seed and have embedded a pd patch onto it, the only problem is that in the finishing stages i've realised i want some of the controls to on a linear scale but i have log potentiometers. Does anyone know how to convert log to linear? Everything i see is about converting lin to log.

Just figured I ask before I have to buy new pots.

Cheers

@callummalcolm In my experience, "audio taper" pots and faders aren't mathematically logarithmic like Pd's are, which is why you can't make Pd's log sliders go to 0 without extra effort. So here's a solution that works in Pd, but what I'm saying is that it probably won't linearize a physical audio taper pot well. I made a table-driven transfer function to linearize the CC messages coming from the faders on my Yamaha 01V96i and the line is somewhat arbitrary, especially at the bottom of the travel. Let me know if you want to try that route and I'll tell you more. Capturing the physical curve is kinda messy.
log2lin.pd

@callummalcolm You can get a decent linearization of log pots with a resistor. I would assume these pots are voltage dividers? if so just place a resistor between +V and the center lug of the pot. Calculating the value of the resistor is simple, just a standard voltage divider with parallel resistances in the top resistor. I generally measure the resistance between those two pins at half a dozen settings and use that for the calculations since pots tend not to be great on the tolerances.

@oid said:

Calculating the value of the resistor is simple

Yes, but you neglected to give the OP the formula, which is a little harder to figure out . It's approx 0.13 of the value of the potentiometer. So for a 10k audio taper pot you want a 1.3k resistor. And for the record, it increases the load that the pot presents to the power rails by a factor of 10 when it's fully counter-clockwise.

To me, both solutions are just linear-ish, You can find a discussion of it here: https://www.modwiggler.com/forum/viewtopic.php?t=56318. You can see that they discuss 0.12 so 0.13 is not exact, even more so since the audio taper pot isn't exactly log. I'm not claiming that linear pots are perfect, but if it were me I'd just go buy the right pots and chalk up the extra expense as tuition. Unless there was a show breathing down my neck.

@jameslo I named the equations needed to calculate it based off those measurements I mentioned, voltage divider and parallel resistance. 0.13 assumes the taper is accurate but for most pots the taper is approximate and can be quite off. I was a bit lazy in my explanation and on reread I was more lazy than I thought. The actual taper of the log/audio taper pot is two linear curves overlapping at the halfway point, first half of rotation covers a straight line from 0 to ~10% of the max value, then we transfer to the second curve to get the ~10% to 100% in the final half of rotation. "Not exactly log" is a bit of an understatement but it does the job most of the time. This goofy curve combined with the poor tolerances of pots and difficulties involved in matching pots for curve is why high end studio gear often goes to rotary switches when matched controls are needed.