schlam

I use mc.function and its curves to interpolate presets with pattrstorage.

It's working good when the mc.function curves are "lines", if they have curves, I need to translate those informations...It's surely so easy ...but sooo difficult for my math skills..

There is some math operations that I am not able to understand.

May a little math guru will have a look at this little problem..

I use a slider that "scrubs" the m.function lines to interpolate pattrstorage.

if line goes from 1 to 6, pattrstorage will interpolate from 1 to 6 but "directly', it will not interpolate with all the other presets (2 3 4 and 5), which is the normal behaviour of mc.function.

The problem happens when the lines are curves.

-> mc .function indicates each curve with a float from -1. to 1. (0 is a straight line)

-> but the scale object I have to use to translate this curve to pattrstorage, calculates its exponential curve with a float from 0. to 100. (even not sure of the maximum...) and here 1.06 is a straight line.

so...how to translate those 2 ways to calculate a same curve ??

I copied below an extract of the scale reference and I really understand nothing to the formulas....

You can see a snapshot that shows the scale exponential reference and some values of some curves in mc.function

Please :open each red object to see where the scale object is...

thank you so much if someone can have a look !

EDIT : maybe [scale] is not the good object for that purpose and [expr] is more appropriate ?

image.png

mc.function2pattr.interpo.zip

Hello.
I use mc.function and its curves to interpolate presets with pattrstorage.
It's working good when the mc.function curves are "lines", if they have curves, I need to translate those informations...It's surely so easy ...but sooo difficult for my math skills..

I use a slider that "scrubs" the m.function lines to interpolate pattrstorage.

if line goes from 1 to 6, pattrstorage will interpolate from 1 to 6 but "directly', it will not interpolate with all the other presets (2 3 4 and 5), which is the normal behaviour of mc.function.

-> mc .function indicates each curve with a float from -1. to 1. (0 is a straight line)
-> but the scale object I have to use to translate this curve to pattrstorage, calculates its exponential curve with a float from 0. to 100. (even not sure of the maximum...) and here 1.06 is a straight line.


EDIT : maybe [scale] is not the good object for that purpose and [expr] is more appropriate ?


Hello.
I use mc.function and its curves to interpolate presets with pattrstorage.
It's working good when the mc.function curves are "lines",

math-question-to-format-exponential-curve-from-mc-function-to-scale

functions such as log(), exp() or pow() end up changing the input range to something else. but that factor can be found by trial an error and then you can change the range accordingly. it is always the same value ;)

expr seems most appropiate because then you can combine scaling and distortion, just as the scale object does, but better and with more "user control".

mapping a range is expr ( ( ($f1- $f2) / ( $f3- $f2) ) * ( $f5- $f4) ) + $f4

i strongly recommend to use a [function] range of 0.-1. and scale it up later.


functions such as log(), exp() or pow() end up changing the input range to something else. but that factor can be found by trial an error and then you can change the range accordingly. it is always the same value ;)

expr seems most appropiate because then you can combine scaling and distortion, just as the scale object does, but better and with more "user control".

mapping a range is  expr ( ( ($f1- $f2) / ( $f3- $f2) ) * ( $f5- $f4) ) +  $f4

i strongly recommend to use a [function] range of 0.-1. and scale it up later.




..but I don't really understand what are the 5 floats you put in expr...

the only float I have is the one between -1 and 1 delivered by mc.function.

I understand that I have to scale it to 0. 1. before [expr] but where do the other float come from ?

Thank you for your advice.
..but I don't really understand what are the 5 floats you put in expr...
the only float I have is the one between -1 and 1 delivered by mc.function.
I understand that I have to scale it to 0. 1. before [expr] but where do the other float come from ?
can you explain me ?


that 0-1 thing was actually an alternative to scaling.

. 1 is the input and 2-5 are the arguments - same order as [scale] or [zmap] uses it.

would be [expr log(($f1)* ($f2-1)+1)/log($f2)]

 would be [expr (exp($f1*log($f2))-1)/($f2-1)]

to reverse the direction of the curve you could simply invert the input range.


that 0-1 thing was actually an alternative to scaling.

that expression above is scaling. 1 is the input and 2-5 are the arguments - same order as [scale] or [zmap] uses it.

log for 0-1 would be [expr log(($f1)* ($f2-1)+1)/log($f2)]

exp for 0-1 would be [expr (exp($f1*log($f2))-1)/($f2-1)]

to reverse the direction of the curve you could simply invert the input range.


What I dont understand is that, in my problem, my only information is one float outputted from mc.function that "describes" a curve (this float is between -1. and 1.)

-imagine a curve going in x from 0. to 1. and in y from 0 to 10,

-mc.function gives me 0.85 to describe this curve

-I have an horizontal slider elsewhere going from 0 to 1 too and I want that this slider reflects the curve in its progression.

Here I don't have any min or max value to scale from.

if the 5th inlet of scale admitted a curve value calculated like mc.function calculates it, it will be so easy, 

But the scale object calculate curves very differently. I event don't understand the min and max values for that parameter...

So I am looking for a way to "feed" the exponential inlet of [scale] with a translated version of the "0.85" that gave me [mc.function], or to use [expr]...

thank you again for your time.
What I dont understand is that, in my problem,  my only information is one float outputted from mc.function that "describes" a curve (this float is between -1. and 1.)


-imagine a curve going in x from 0. to 1. and in y from 0 to 10,
-mc.function gives me 0.85 to describe this curve
-I have an  horizontal slider elsewhere going from 0 to 1 too and I want that this slider reflects the curve in its progression.

Here I don't have any min or max value to scale from.

if the 5th inlet of scale admitted a curve value calculated like mc.function calculates it, it will be so easy, 
But the scale object calculate curves very differently. I event don't understand the min and max values for that parameter...

So I am looking for a way to "feed" the exponential inlet of [scale] with a translated version of the "0.85" that gave me [mc.function], or to use [expr]...

this is probably only scaled (yet again) to that magic number of the log/exp distortion.

like -1 is log 100, 0. is dont use distortion, + 0.5 is log 25 or so ... i think if you try to use sqrt() you should come close. (i dont use modern version of function object)


this is probably only scaled (yet again) to that magic number of the log/exp distortion. 

when i think about it, i have no idea what woud happen between log 1.01 and no distortion.

grafik.png


when i think about it, i have no idea what woud happen between log 1.01 and no distortion.

but pow 2 works okay for that parameter.



I am really so noooob with these maths...:/

math question to format exponential curve from mc.function to [scale]