If the formula from the reference is:

y = b e-a log c ex log c

Where:

y = result

x = number to scale

a,b,c = the “three typed in arguments”

e = base of the natural log

My question:

There are 4 mandatory arguments, 5 with exponential scaling (which is why I’m interested in the formula). Given only three (a,b,c) – how does this work?

Also, what are the operators in the formula? I’m guessing:

y = b * (e-a) * log(c) * ln(x) * log(c)

Anyone dug in deep enough into their own scaling object?

]]>c + (d-c)*((v-a)/(b-a))

I don’t know about the scaling exponent, whether it’s applied after all this, or is worked into the formula somehow. And I’m not sure this is even the right formula for [scale], but it works in the patch below:

– Pasted Max Patch, click to expand. –

]]>
the basic expr here is:

expr (exp((($f1-$f2)/($f3-$f2)-1)*$f6)-1)/(exp(-$f6)-1)*($f4-$f5)+$f5

which works like scale but has an easy logarithmic scaling, where 0 is linear. ]]>

Those are really great, and likely useful for my students as well.

Cheers!

]]>well ll.og is just one of many objects, that are included in the ppooll distribution. there is more useful stuff in the whole package ;)

cheers ]]>