## math behind scale?

May 17 2010 | 7:30 am
Can someone give me some details about the math behind the scale object? Thanks!

• May 17 2010 | 10:21 am
I don't have access to the scale source code, but I can make a guess.
The linear part is easy:
- Subtract the low input value from the number, then divide the result by the difference between high input value and low input value. This gives you a number between 0 and 1.
- Then multiply this again with the difference between low output and high output value, and finally add the low output value.
As an expression, this would look like this:
expr "((\$f1-\$f2)/(\$f3-\$f2)) * (\$f5-\$f4) + \$f4" \$f1 being your number input, \$f1 the low input value, \$f2 the high input value, \$f3 the low output value, \$f4 the high output value.
This still ignores the exponential part, but I don't think this should be too hard to figure out as well.
EDIT:
eh, I'm silly. Should have looked into the documentation of [scale] first where the formula is written down. So that's what you should do as well!
• May 17 2010 | 11:26 am
here's the formula from the docs:
y = b*e^(-a*log(c)) * e^(x*log(c))
it says "a,b and c are the three typed-in arguments". So c is the exponential factor and e is the base of the natural logarithm. But what's a and b ??
• May 17 2010 | 8:22 pm
Woops, I didn't now that the formula is inside the docs.
Thanks a lot guys!
• Jun 28 2010 | 10:39 am
I think that what is written in the docs is for exponential mode calculations. I was interested as well to use a jit.expr expression that simulates the scale object.
I managed to do something similar to [scale 0 255 -1. 1.] with the following expresion
[expr (in[0]/255-0.5)*2]
but I don't have a clue how to do [scale 0 255 1. -1.]
Does anyone know? I'm not good at maths either
• Jun 30 2010 | 10:07 pm
Please, if anyone knows how to reverse a scale in jit.expr. I am trying to reverse a short matrix with 2d xy coordinates for a 640x480 pixels to have 3d xy coordinates which are -1. 1. for both x and y (with the center as 0.).
I did the first part for the x coordinate where 0 becomes -1. and 640 becomes 1. But for the y value, 0 is 1. and 480 is -1.
So I would need an expresion for jit.expr to turn a (-1. 1) matrix the other way around (1. -1.)
• Jun 30 2010 | 10:34 pm
Could you just multiply by -1. after the scale?
[jit.op @op * @val -1.]
• Jun 30 2010 | 11:04 pm
I guess it's mid school mathematics! I'm loosing my mind!!
Thank you so much, Jesse!!