### math behind scale?

Can someone give me some details about the math behind the scale object?

Thanks!

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!

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 ??

Woops, I didn’t now that the formula is inside the docs.

Thanks a lot guys!

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

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 640×480 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.)

**all**of the following text. Then, in Max, select

*New From Clipboard*.

```
----------begin_max5_patcher----------
528.3ocwV9saaBCEF+Z3ovxWsMQQ1l+TXWs8bTEU4BtotBrQfiFaU8ce1GBK
YILRYohdii3bL1mue93OxK9d3Gz8hNL5qn6Pddu364AgbA71+rGtl2WTw6fo
gKz00BkAGLjyH5MPbk1f1p0kiIZ3lhmjps22JJLCaPFMNjDfnI4teRRciLRH
AsY+67nVY5j+R3lNkYSODVsqVppDFn.n6CJKgsU+vy2PYw3iVAEuFVA72ak7
J7gEQuyLtJDWvW88cCAWoxunpAgRycBBEG89o5n0P0JwOr61YhtqfWIPjPTb
lc3FZHhFNCDtkDczQOkjeFEFJSyOaDCuA9wJM291KgRoSQI1RoD88iROKMgh
9lVz2fwOIU2Q1D1XG9hU6jHJK4yN18mDzgDwzzTWhwErRpDE5cJXUi92XNMG
.aDk.XNB57hylCy1R79ZtoU1iCP3kwa1D7lr3tR1+AuqEcc7shy.dTbdXhsw
hDlLa2HbWj4HS.51AWoYozUyk7O51vgyXDCA2qhXHJhNKhn.hxFfCK4xWXul
NoI72xhWiFo2j8FEb3lkVwG4wuh1aQoevsUMsxCeRbFGobv+ONc0aitJ6HXy
A22S9ORfRcw+ap0o20VLJjwlXzAwVJ5LRE2H0pilD3YdzrdRVVJfILJxZYYi
1B58UAZyjGhu0h5z8a5pZcqIm+3EKoSn4xpI6Cu5+aPQ0fnU
-----------end_max5_patcher-----------
```

Could you just multiply by -1. after the scale?

[jit.op @op * @val -1.]

I guess it’s mid school mathematics! I’m loosing my mind!!

Thank you so much, Jesse!!