24 Feb
2023
24 Feb
'23
12:14 p.m.
Il giorno ven 24 feb 2023 alle ore 12:12 Stefano D'Angelo
<zanga.mail(a)gmail.com> ha scritto:
Il giorno ven 24 feb 2023 alle ore 00:59 Jeanette C.
<julien(a)mail.upb.de> ha scritto:
Sorry, typo: I meant points (x1,y1), ((x1+x3)/2,y2), (x3,y3).
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Stefano D'Angelo
https://www.orastron.com/ - https://dangelo.audio/
Feb 23 2023, Fons Adriaensen has written:
...
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"). 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).