Forum: Poser - OFFICIAL


Subject: Nodes for Dummies

RobynsVeil opened this issue on Jan 24, 2009 ยท 490 posts


bagginsbill posted Fri, 19 November 2010 at 3:56 PM

Quote - I'll do it again, but I'm pretty sure that I used exact the same colour to subtract in the Poser shader as I did in the SM shader (0, 128, 0).

But but but I can see the color of the green is different. How can you say you're pretty sure they are the same when just looking at your two screen shots the color values are clearly different shades of green.

Yes the outcome between the Poser and the SM versions should be the same, and they are. The final result looks the same as the input color map. I thought we were talking about how the intermediate value is different. The intermediate value is not going to be the same because you're using a different color.

f(x) = m(x - k) + k

Where you used m = 1.

The result is

f(x) = (x - k) + k = x

Clearly the output of f(x) is just x, regardless of what value of k you use, but only when m is 1.

Anyway, you have m=1.

So the middle node is (x - k).

That is NOT the same unless k is the same.

k is your reference color green. You used two different colors. So the middle value is two different outcomes. The final value is the same in both.

However, if the multiplier, m, had not been equal to 1, but equal to something else, then the outcome would be different as well. The output would no longer be exactly the same as the input.

The only time the f(x) = x is when m=1.

Given f(x) = m(x - k) + k

We can examine under what conditions f(x) = x, i.e. when does f do nothing.

f(x) = x

m(x - k) + k = x

mx - mk + k = x

mx - x = mk - k

(m-1)x = (m-1)k

There are two ways this can happen.

(m-1) = 0

i.e. m = 1.

The other way is:

x = k

Which is to say when the input texture is entirely made of the one color k - then m doesn't matter.


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