Hi Fons, I don't understand, exactly what you mean. So here's an example. I record some music in a DAW. I have a beat based timeline (bars, beats, ticks and a tempo). Then I want to gradually change the tempo. AFAIK several DAWs offer predefinede curves to do such changes. You may have a linear change, log curve, exp curve or some other type. Such predefined automation curves might also be applied to volume changes or other whatever. In my example: I want to change the tempo from 120BPM to 150BPM over four bars of equal length. I click on logarithmic curve. Which part of the logarithmic function is used to generate the logarithmic curve? Is there something that is commonly used as a formula? My assumption you choose one segment of the log function, one range of x coordinates, x1 to x2. You find the number of clock ticks (for a MIDI clock or similar) inside the four bar interval (say 50 ticks per bar, so 200 ticks for four bars. Scale = abs(tempo1 - tempo2) / (log(x2) - log(x1)) so: Scale = abs(120 - 150) / abs(log(x1) - log(x2)) Increment = (x2 - x1) / number_of ticks so: Increment = (x2 - x1) / 200 ; 200 ticks in four bars Then incrementally in some kind of loop or whatever: for (cur_tick = 0;cur_tick<200);cur_tic++) { cur_tempo = (log(x1 + (cur_tick * Increment)) - log(x1)) * Scale + 120 } OK, too simplistic, only allowing for going up in tempo. But hopefully the basic idea is clearer now. Question is: are there default values or segments that have proven to give desireable results? If a DAW will allow curves of different steepness, what do the change? Will they choose x1 and x2 farther apart? Will they leave the absolute interval between x1 and x2 but move it along the x-axis? Best wishes, Jeanette Feb 23 2023, Fons Adriaensen has written: -- * Website: http://juliencoder.de - for summer is a state of sound * Youtube: https://www.youtube.com/channel/UCMS4rfGrTwz8W7jhC1Jnv7g * Audiobombs: https://www.audiobombs.com/users/jeanette_c * GitHub: https://github.com/jeanette-c Baby, take the time to realize I'm not the kind to sacrifice <3 (Britney Spears)

Feb 23 2023, Fons Adriaensen has written: ... ... 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). Best wishes, Jeanette -- * Website: http://juliencoder.de - for summer is a state of sound * Youtube: https://www.youtube.com/channel/UCMS4rfGrTwz8W7jhC1Jnv7g * Audiobombs: https://www.audiobombs.com/users/jeanette_c * GitHub: https://github.com/jeanette-c Baby, take the time to realize I'm not the kind to sacrifice <3 (Britney Spears)

Il giorno ven 24 feb 2023 alle ore 12:12 Stefano D'Angelo &lt;zanga.mail(a)gmail.com&gt; ha scritto: Sorry, typo: I meant points (x1,y1), ((x1+x3)/2,y2), (x3,y3). -- Stefano D'Angelo https://www.orastron.com/ - https://dangelo.audio/

Il giorno ven 24 feb 2023 alle ore 13:21 Fons Adriaensen &lt;fons(a)linuxaudio.org&gt; ha scritto: Of course you need a third degree of freedom. An offset term is sufficient for the purpose: F(x)=A*exp(B*x)+C IIRC, imposing the midpoint value leads to a nonlinear yet solvable system. -- Stefano D'Angelo https://www.orastron.com/ - https://dangelo.audio/