What does "Output Levels" really do?
If I start with an image that has information all across the histogram, then I set Black Output Levels at say..30, and I set White Output levels at 245...can someone go into detail as to what is exactly happening? Basically, I *LOVE* this look...starting with Output levels where I described, then using Contrast or Curves to add punch BACK into the image. To my eyes, it gives this really filmy look...but I'm curious as to what is occuring from a technical standpoint. Could it be said that I am increasing detail? (By taking the same information and squeezing it into a smaller space in the histogram, is it similar to taking a certain number of pixels and squeezing them into a smaller space? i.e. increased resolution?)
Or does it just look more detailed because the shadows are in a more visible area of the histogram?
Also, how come if I do the same thing in Photoshop, I get a straight "brick-wall" sort of output clipping in the histogram, whereas C1 seems to allow some color to slip past the Output level limits? (as seen in the C1 histogram)
Thoughts?
Or does it just look more detailed because the shadows are in a more visible area of the histogram?
Also, how come if I do the same thing in Photoshop, I get a straight "brick-wall" sort of output clipping in the histogram, whereas C1 seems to allow some color to slip past the Output level limits? (as seen in the C1 histogram)
Thoughts?
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For a start, you might want to go through this:
viewtopic.php?p=25945#p259450 -
Thanks Paul, good read:) I guess there's not that much too it, I'm just amazed at how cool playing with Print Output levels is and how other people are playing with it. I too use it to recover highlights and it works great! Changing Output levels doesn't work quite the same in PS (or I guess any non-RAW file), there it just hard limits the output. C1 is certainly taking the RAW into account, and it has a much more natural look to it (and you can see this in the histogram where the cutoffs have slight graduated tails on them). Anyway, thanks again for the reply Paul! 0
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