Forums > MaxMSP

Nonlinear scaling of parameter values?

January 10, 2012 | 7:44 pm

Hi All,

My dabblings in Max have improved greatly, and I am almost starting to feel comfortable about using Max, and a big thanks for that goes out to all of you! Without this forum I’d still be stuck with trying to get my head around the environment :)

But yeah, have another question to ask y’all (tried searching but no help) – I need to scale a parameter value (using 0. – 1. range for the input) to the range of 0.4 – 10.0, which itself is easy enough.. but the thing is, I want the value of 1. to be exactly at 50% of the parameter range (ie. 0.5 in the controlling input param)!!

Is there an easy way of getting such a range, or do I have to resort to building some kind of param eval logic to the knob? Something like "if value less than 1 -> scale like this / else scale like this"? Sounds like a hack to me.. there has to be a more elegant solution?

TIA & Cheers


January 10, 2012 | 7:52 pm

http://cycling74.com/forums/topic.php?id=36275

try the knee thing (or try all of them)

note that it also would be possible to do this knee thing only with only one [expr] object.


January 10, 2012 | 7:58 pm

Hi Roman,
Thanks for this!

Cheers

EDIT: Alas, your kneepatches went over my head.. I’ll get a mate who’s a C programming whiz to explain a bit about expr stuff.. ATM I am using a function object with interpolation to do the scaling for me (as found in one of the patches in that brilliant thread) :p


January 10, 2012 | 10:12 pm

depends on how you want the function to ‘look’. For a smooth function you could try something of the form

y = a + b*x^c

we get a and b from the constraints directly, a = 0.4, b = 9.6, then solving for the case when y=1, x=0.5, we get

1 = 0.4 + 9.6*0.5^c
0.6/9.8 = 0.5^c

c = ln(0.6/9.6)/ln(0.5)

which just happens to be c = 4

so your mapping is y = 0.4 + 9.6*x^4

which you can patch directly.

EDIT

Here’s a little sample patch for you

– Pasted Max Patch, click to expand. –

January 10, 2012 | 10:40 pm

Or in one object: [expr pow($f1,4)*9.6+0.4]


January 10, 2012 | 10:45 pm

You’ve got me there Luke :)


January 11, 2012 | 9:56 am

Wow! Much obliged for your assistance. You guys are the best!


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