There are two types of volume slider I've encountered thus far, "too logarithmic", and "not logarithmic enough".

embedding-shape 8 hours ago | parent | next [-]

That's because one ear is logarithmic-based and the other is exponential-based. Which one differs per person.

Do you have a source, that seems unlikely at face value to me, though I've never gone and looked for perception studies myself.

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graypegg 8 hours ago | parent | next [-]

I think they're joking, this is on a thread about silly volume control UX

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KolmogorovComp 8 hours ago | parent | prev [-]

It’s satire.

	▲	JulianWasTaken 8 hours ago \| parent [-]
		Always hard to tell.

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ErroneousBosh 5 hours ago | parent | prev [-]

It's actually possible to turn a linear pot into an approximation of a log pot by wiring a resistor in parallel with the wiper and one end. The volume pot is a voltage divider so the amount it "scales" by is given by Scale = Rbottom / (Rtop + Rbottom).

But, if you put a resistor in parallel, you need to work out that:

R = 1 / ((1 / R1) + (1 / R2)) or Rbottom = 1 / ((1 / Rbottom) + (1 / Rfixed)) where Rfixed is the amount you're "bending" it by.

So you could make the amount of "logness" be adjustable by having another (linear) control to vary Rfixed.

You'd work out, for a pot rotation Vol from 0 to 1:

Rbottom = 1 / ((1 / Vol) + (1 / Rfixed)) Rtop = 1 - Vol Scale = Rbottom / (Rbottom + Rtop)

Now for those better at arithmetic than me, how can you reverse this? Imagine you've got a pot in a piece of equipment with a resistor between the wiper and ground giving a log curve, and you've got to read that with an ADC and turn it back into the linear position of the wiper.

It ought to be possible but I've always sucked at arithmetic.