Hallo, CK hat gesagt: // CK wrote:

I read:

...

i need to use the scale object from maxlib and the one from gem in the same patch... opening pd twice with different libs loaded is not really the most elegant way to solve that problem i think...

I think so far your only option is to recompile one of them and rename one object to sth. else (uppercase, oder mxl_scale / gem_scale or sth. like that)

Yes, true. But at least scale from maxlib is very easy to emulate as an abstraction. I'M quite sure I posted at least two abstraction emulations of scale without exp-interpolation here some weeks ago (or on pd-dev) and there's also rrad.scale.pd in CVS.

Ciao

Frank Barknecht _ ______footils.org__

Hallo, CK hat gesagt: // CK wrote:

right I was about to whip one up but then I felt a bit stupid cloning an abstraction by reading c code of an external ;)

It's not difficult math, it's easy to understand if visualised. Please turn on a non-proportional font:

^ | y1.....+......+ | . / . | . / . | . / . | . / . | ./ . y0.....+......+ Figure 1. | . . | . . +-----x0-----x1--->

The slope "k" of the line connecting (x0,y0) and (x1,y1) is given as:

k = (y0-y1)/(x0-x1)

(In rrad.scale.pd this is a sender called "$0-k")

Now if x0 = y0 = 0 the formula to find y out of x would be very easy:

y = f(x) = k*x

Normally x0 and y0 are not 0. Assuming y0 not equals 0, but x0 = 0:

^ | y1......+ | / . | / . | / . | / . | / . y0......+ Figure 2. | . | . +------x1---> x0=0

You see, that now the curve simply is found by adding y0 to every value found by the simplest formula and thus shifting the line upwards (or downwards, if y0 < 0):

y = k*x + y0

To handle the x-shift similarily, lets look again at figure 1:

^ | y1.....+......+ | . / . | . / . | . / . | . / . | ./ . y0.....+......+ | . . | . . +-----x0-----x1--->

Compared with the previous figure, note that every x in the range from x0 to x1 actually results in the same y values from figure 2, only the x values are bigger by x0. We can counter this by first substracting x0 from our input x and basically transforming Figure 1 into Figure 2 on the way, so we can then use the formula from above again:

y = k*x + y0 using x -> x-x0 ==> y = k*(x-x0) + y0

Or in linear function lingo using:

k = m b = y0 - k*x0

it would be:

y = mx + b

Voila. [scale] done in math. Scaling exponential is a bit harder, but not much different. You can always look at scale.c ;)

Ciao

Frank Barknecht _ ______footils.org__