3 ms·
There are two types of volume slider I've encountered thus far, "too logarithmic", and "not logarithmic enough".
by Retr0id 8mo ago
There are two types of volume slider I've encountered thus far, "too logarithmic", and "not logarithmic enough".
- embedding-shape 8mo agoThat's because one ear is logarithmic-based and the other is exponential-based. Which one differs per person.
- JulianWasTaken 8mo agoDo you have a source, that seems unlikely at face value to me, though I've never gone and looked for perception studies myself.
- KolmogorovComp 8mo agoIt’s satire.
- JulianWasTaken 8mo agoAlways hard to tell.
- graypegg 8mo agoI think they're joking, this is on a thread about silly volume control UX
- ErroneousBosh 8mo agoIt'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.