Datacolor User to User Support Group.

This is very similar to my experience.
Clearly there is some point at which science gives way to art.
I am assuming that the hue representation is accurate and all we are looking at is tonal range (If that means grading from light to dark)

David T is right about blacks. I have included a picture of the excellent spydercube under my album "color cast".
The black trap drops away to 0,0,0 and the specular highlight is 255,255,255.
The virtual gretag card's black of 50,50,50 is very weak by comparison.
where possible a color rgb value of the image in the computer would be the same as the value measured by the spectro when printed on the paper.
On reflection this may not be the case, with deliberate distortions introduced to "take advantage" of the sensitivities of the paper/ink combinations at different parts of the tonal range. (To bring out detail in that range)

And I presume these distortions change with rendering intent!

Dave Sainsbury





> David,
> I spent quite a bit of time with SpyderProof and my base measurement file and incrementally tweaked and built profiles that I then Spyderproofed until it looked pretty good.  Increasing the Brightness to +4 gave the best view in SpyderProof.  But when I printed out one of the "Matrix" images (the little girl with the flowers) I had used to tweak the profile, it was still too dark.  I increased the Brightness to +10 and printed and it looked pretty good.  The contrast seems a bit lacking.  I tried playing with the Contrast amount when tweaking the profile but didn't like the effect it had on light to dark transitions.
> I tried, as you suggested, printing a "Matrix" image both from SpyderProof and from CS4 with the new profile and they both appeared to be the same.  So I am printing via CS4 OK apparently.
> So, other than the contrast issue I think I have created a better profile than the one that Epson provided.
> I am going to try the 729 patch target next to see if that improves it a bit more.
> Thanks for your help.
> Rick
>