dovemouse

not reeeally a max related question here but i'm having difficulties.

I want to have a transposition dial in my app that will "speed" the sound file up by a multiple of semitones.

i figured that 1 is normal speed. 2 is double speed and therefore an octave up.

So the number of the dial * 0.833 then + 1 so that 0 on the dial will play at normal speed.

It just doesn't work and i can't think why. It's probably something really obvious that i've missed but any helped would be great.

Okay,
not reeeally a max related question here but i'm having difficulties.
I want to have a transposition dial in my app that will "speed" the sound file up by a multiple of semitones.
i figured that 1 is normal speed. 2 is double speed and therefore an octave up.
And octave is 12 semitones.
1/12 = 0.833333
So the number of the dial * 0.833 then + 1 so that 0 on the dial will play at normal speed.

It just doesn't work and i can't think why. It's probably something really obvious that i've missed but any helped would be great.
Thanks.


changing-speed-by-a-semitone

This is leftover from the code I use for keyboard transposing in my own patches, but it should work fine....

A simple [expr pow(2.,($f1/12))] will do the trick.

Ch, your one has a few errors. For example: if you put in 12 it should come out with 2 but its just a smidgen under.

Gregory, that's a nice solution but not as elegant as the final one posted.

Thanks guys.
Ch, your one has a few errors. For example: if you put in 12 it should come out with 2 but its just a smidgen under.

thereishopeforus@hotmail.com wrote on Wed, 24 June 2009 14:00

where the 60 is the key number (middle C)

thereishopeforus@hotmail.com wrote on Wed, 24 June 2009 14:00A simple [expr pow(2.,($f1/12))] will do the trick.

The [expr pow(2.,($f1/12))] solution is preferable because expr will perform the internal arithmetic in double-precision whereas values such as 0.057762 in an object or message box are initially parsed in single-precision. It is precisely the single-precision issue that leads to the minuscule (and absolutely inaudible) discrepancy noted in Ch's example.

Of course, the easiest thing to do is to use [mtof]. That's what it's there for.

how would you do a more acccurate conversion for note

to rate? when i made mine last century i tested it against

and that gives the same results like using exp()

btw, isnt [mtof] doing the less exact calculation, too?

how would you do a more acccurate conversion for note
to rate? when i made mine last century i tested it against

Roman Thilenius wrote on Wed, 24 June 2009 22:25

of course i couldnt notice the difference when trying 

only a few keys upwards. from 3 octaves on i do.

Roman Thilenius wrote on Wed, 24 June 2009 22:25
pow(2.,(($f1-$f2)/12.))

wow. uh oh. 
of course i couldnt notice the difference when trying 
only a few keys upwards. from 3 octaves on i do.

[mtof] is (almost certainly) using an internal double-precision representation of pow(2, 1/12).

When you write 1.059463 (or any other float) in an object box, the object can only get a single-precision value. The content of an object box is passed to the object code in the form of atoms, and these can only contain 32-bit values.

[mtof] is (almost certainly) using an internal double-precision representation of pow(2, 1/12). 

Could anyone help me out with the equation going the other way? I want to work out how much I alter the pitch of a file when i play it back over a shorter or longer time frame. For example the sample is 2000 msecs long, but i play i back over 700 msecs. How many semitones have I shifted the pitch?

Take the ratio of original length to new length and feed it into [expr (log($f1)/0.693147)*12]. For 2000/700, the answer is 18.17.

Changing speed by a "semitone"