Re: Hue Angle for Black
Re: Hue Angle for Black
- Subject: Re: Hue Angle for Black
- From: "dpascale" <email@hidden>
- Date: Wed, 26 Aug 2009 13:07:44 -0400
Hello Bill,
Pure black, and all neutrals, can have any hue angle. In practice there is
always a small "tint" and a definite hue angle can be computed. However, for
very low Chroma (low C) colors, even large changes in angle (Delta-h*)
correspond to a small DeltaE*. In other words, you could have a 180 degree
Delta-h* difference for an almost zero DeltaE*. If you are concerned about
hue error, use Delta-H* (with a capital H), which gives you the contribution
of the hue error to the total DeltaE* error (and is NOT an angular error).
Also computing the hue angle can be tricky since you have to test for angles
smaller or larger than 90, divides by zeto (as you noticed) and make sure
you use radians instead of degrees.
You can consult the Help manual of BabelColor CT&A which gives all the
formulas for the various flavors of DeltaE.
http://www.babelcolor.com/download/BabelColor_CT&A_Help.pdf
Look in the Technical data/Definitions and theory/DeltaE section.
Danny Pascale
www.babelcolor.com
----- Original Message -----
From: <email@hidden>
To: <email@hidden>
Sent: Wednesday, August 26, 2009 12:16 PM
Subject: Hue Angle for Black
Greetings,
(FYI, I posted a similar question on the Print Planet list - for those
that are concerned about cross posting. Thanks, as
always, to Mike E for your reply)
Can black have a hue angle?
To help me in my G7 qualification runs on our flexo presses, I measure 3
consecutive press sheets and average 6 readings
(3 from each side of the press). I have created an Excel file that enters
these 6 Lab measurements as I measure them with
my 528 (using a MAC) and then calculates chroma, hue, dE (under CMC 2:1);
and plots L*a*b* to tolerances, and shows an
approximated hue angle view from the average of these measurements.
My challenge is creating the dE for the black. I use the formulas below my
name that I received from a colleague. These
formulas work for CMY when compared to the results achieved on Bruce
Lindbloom's web page.
However, I am not able to compute a hue angle for black. Since my math
education ended with Geometry, I am a little
challenged with these formulas, but I did notice that with black's a* and
b* being 0, the formula for dH*ab returns a 0
divided by 0, and I know dividing by 0 is undefinable. My guess is that
this is what is "breaking" the formula to compute
hue for black. But once again can black have a hue angle?
I am able to get a tolerance when comparing two blacks using Bruce's site,
so I am hoping some one can share with me
how that is computed.
Thanks to anyone that takes a little time to help me and if seeing my
Excel file would help, I will gladly send it.
-Bill-
Since as I mentioned, I received these formulas from a colleague (in an
Excel file), I had to translate the cell in his file to its
appropriate LabCH values so I could then use them in my file format, which
differs from his. The second line in each
formula below is that translation.
SL =IF(B3>16,(0.04097*B3)/(1+0.01765*B3),0.511)
=IF(L#1>16,(0.04097*L#1)/(1+0.01765*L#1),0.511)
SC =((0.0638*E3)/(1+0.0131*E3))+0.638
=((0.0638*C#1)/(1+0.0131*C#1))+0.638
F =((E3^4)/(E3^4+1900))^0.5
=((C#1^4)/(C#1^4+1900))^0.5
T
=IF(F3>164,IF(F3<345,0.56+ABS(0.2*COS(RADIANS(F3+168))),0.38+ABS(0.4*COS(RADIANS(F3+35)))),0.38+ABS(0.4*COS(R
ADIANS(F3+35))))
=IF(h>164,IF(h<345,0.56+ABS(0.2*COS(RADIANS(h+168))),0.38+ABS(0.4*COS(RADIANS(h+35)))),0.38+ABS(0.4*COS(RADI
ANS(h+35))))
SH =+AE3*((AG3*AF3)+1-AF3)
=+SC*((T*F)+1-F)
dC*ab =((H3^2+I3^2)^0.5)-((C3^2+D3^2)^0.5)
=((a#2^2+b#2^2)^0.5)-((a#1^2+b#1^2)^0.5)
dH*ab =((H3*D3)-(C3*I3))/((0.5*((J3*E3)+(H3*C3)+(I3*D3)))^0.5)
=((a#2*b#1)-(a#1*b#2))/((0.5*((c#2*c#1)+(a#2*a#1)+(b#2*b#1)))^0.5)
dL Term =+((B3-G3)/(2*AD3))^2
=+((L#1-L#2)/(2*SL))^2
dC Term =(AI3/AE3)^2
=(deltaC/SC)^2
dH Term =+((AJ3)/AH3)^2
=+((deltaH)/SH)^2
dEcmc2:1 =+(AK3+AL3+AM3)^0.5
=+(L term+C term+H term)^0.5
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