Il giorno ven 24 feb 2023 alle ore 00:59 Jeanette C. <jul...@mail.upb.de> ha scritto: > > Feb 23 2023, Fons Adriaensen has written: > ... > > So if the start and end values are A and B, you would make a linear > > function from log (A) to log (B), and then use exp () on that to > > find the tempo at any point. > ... > Thanks! This is fascinating. Starting with an exponential curve i.e. > first get exp(a) and exp(b),, get the linear function and calculate the > values by log(a + inc), it makes a huge change in the curve shape when I > scale in input arguments a and b. So starting with exp(120) to exp(150) > is a very steep curve, whereas exp(1.2) to exp(1.5) and a rescaling post > all other calculations gives a much gentler slope. Whereas the same > exercise beginning with log makes no difference going from log(1.2) to > log(1.5) and scaling up the final values to 120 to 150 is no different, > at least to three or four places after the decimal point, to starting > with log(120) to log(150).
Given any three points (x1, y1), (x2, (y1+y3)/2), (x3, y3) where y1>y2>y3 or y1<y2<y3 you can find an exponential function passing through them as explained here: https://www.orastron.com/blog/potentiometers-parameter-mapping-part-1 (+ "output scaling"). In other words, you can choose a midpoint of your liking and build an exponential function passing through it and the extremes. -- Stefano D'Angelo https://www.orastron.com/ - https://dangelo.audio/ _______________________________________________ Linux-audio-dev mailing list -- linux-audio-dev@lists.linuxaudio.org To unsubscribe send an email to linux-audio-dev-le...@lists.linuxaudio.org
