There is a little bit of a hole in the most recent version of
the ISO spec for flexo printing[1].
I will explain the hole, discuss my analysis, and provide a patch for this hole,
but first a bit of background.

There is a family of standards called ISO 12647, which all
cover printing. In particular, they provide some target color values, along
with acceptance tolerances around those colors. In almost all case, those
tolerances are in terms of ΔE, with the newer revisions of the standards
shifting from ΔE

_{ab}to ΔE_{00}.
Part 6 of this standard is devoted to standards for
flexographic printing. Flexo printing is used on such a wide variety of
substrates that it is often not possible to nail a particular color. As a
result, tolerances based on a ΔE-type color difference are not possible. One
has to be content with getting the correct hue.

ISO 12747-6:2012 states target hue angles for C, M, and Y, and
provides a ±6

^{o}tolerance. Spot colors are also given a tolerance in terms of hue angle, but it’s ±8^{o}. Black, however, does not have a hue angle tolerance. This makes a great deal of sense, since black is neutral. The hue angle is undefined.
But, “black” is not the only black. My Pantone book lists six
shades of black. Presumably, the target a* and b* values for these inks are
also 0, so the hue angle is undefined. This is the question that led to the
realization of the hole in the standard: “how can analysis software decide if
the ink name ‘midnight’ or ‘inkwell’ is equivalent to black?”[2]

Worse yet, what if the target for a spot color is not quite
exactly neutral, but

*near*neutral? If the chroma[3] of a spot color is small, then a tolerance of 8^{o}might be impossibly tight. For example, in my Pantone book, the ink “Warm gray 5” has an L*a*b* value of {69.61, 2.58, 1.00}. Let’s say the hue of this ink is rotated by 8° to {69.61, 2.70, 1.63} so as to be at the very edge of acceptance in terms of hue. The color difference is a meager 0.38 ΔE_{00}. Holding a printer to a 8° tolerance is pretty tight.
For highly saturated colors, a tolerance of a specified number
of degrees around a target hue is perfectly reasonable. But as a color gets
closer to a near neutral, it would be preferable to shift over to a ΔE

_{00}tolerance. At what chroma value do you need to shift between a ΔE_{00 }tolerance to a hue angle tolerance?**First naïve approach**

I call this the naïve approach, because I am going to make a
mistake. It turns out that the mistake is not huge, and it gives some intuitive
understanding, so I will leave it in on this first pass. Watch for the mistake.

The figure below shows two colors r

_{1}and r_{2}, in the a*b* plane. Both have a chroma of*c*, but are separated by 8^{o}. If r_{1}is the target hue, then r_{2}is at the edge of the 12647-6 tolerance window.
The arc
length between the two colors is fairly trivial to compute. I could use the
normal trig stuff, but since the arc is so small, I will just estimate the ΔE
(which I a straight-line distance) with the arc length.

Thus, if we assume a tolerance of 1.5 ΔE, the crossover is close
to a chroma of 10. Below that point, a hue tolerance is overly restrictive.

**Second, not so naïve approach**

Did anyone catch the mistake I made? Yes, you in the back?
Ahhhh… the old accidentally used ΔE

_{ab}instead of ΔE_{00}mistake! The formula for ΔE_{00}is just a tiny bit more complicated than the formula for ΔE_{ab}.[4] So, I need a slightly less naïve way to look at this problem.
Suppose I picked a color at random, and then rotated it by 8°
- that is to say, shifted the hue by 8° without changing L* or the chroma, C*

_{ab}. What is the color difference (in ΔE_{00}) between the original color and the hue-shifted color? Then let’s say you computed that for a whole big bunch of colors just to see how it all played out. You would expect that, the larger the chroma, the larger the color difference, right?
Well, I just happened to have a large collection of colors
(11,488 of them to be exact) just laying around from a blog of mine on counting colors. Below is a plot of the color
difference caused by an 8° hue shift as a function of chroma.

*Color difference caused by an 8° hue shift*

Is this cool?

To answer the rhetorical question, yes, it is cool. But the
data is a bit sparse down there in the lower left-hand corner around 1.5 ΔE

_{00}where we are looking for our answer. This is not a surprise, since my color database only includes colors on a grid with spacing of 5.
So, I took a little different approach. For my next plot, I
looked at points in color space with a chroma of 10. (This is the estimate we
came up with by analyzing ΔE

_{ab}.) These are all shades of gray, essentially, with a moderate amount of a color cast to them. I looked at how much color difference there was for each of them when they shifted in hue by 8°. The plot below shows this color change as a function of original hue angle.*The color difference caused by an 8° change in hue, as a function of hue angle*

We see that
the amount of color change oscillates between 1 and 2 ΔE

_{00}, so the average must be somewhere around 1.5 ΔE_{00}. The true average is 1.45 ΔE_{00}. Going back to Table 4 from ISO 12647-6, this is really close to the variation tolerance for spot colors.
For an ink with a target chroma of 10, a tolerance of 8° is
roughly equivalent to a tolerance of 1.5 ΔE

_{00}.[5] Based on that, I have a simple recommendation for spot colors:*If the target chroma is less than 10, then the appropriate tolerance is 1.5 ΔE*_{00}. For target chroma of 10 or greater, then the hue angle tolerance of 8° should be used.

Just in case someone wants to repeat this analysis for angles
other than 8° or color differences other than 1.5 ΔE

_{00}, I provide the useful nomograph below. The t-shirt version should be out just in time for Christmas.*Nomograph for determining the crossover chroma*

for various combinations of color difference tolerance and hue tolerance

for various combinations of color difference tolerance and hue tolerance

The red line on the nomograph shows the determination of the
crossover chroma between tolerances of 1.5 ΔE

_{00}and 8° of hue. The graph shows a crossover at C* = 10.5. Just between friends, let’s call it 10.
[1] I
am not blaming any of the diligent folks on TC 130 who reviewed this document.
I was one of them!

[2]
Thanks to Bruce Bachmann and Mike Sisco for this realization.

[4]
Well, to be fair, the formula for ΔE

_{00}is just a tiny bit more complicated than the formula to get a balanced budget.
[5]
One caveat: I have made the assumption that the only change in color is in hue
angle.

I assume that the Star Trek "Neutral Zone" would fall into these equation somehow given the appropriate space.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteJohn,

ReplyDeleteGood stuff. The question is: cuold magenta (for example) have a chroma less than 10. It wouldn't be much of a magenta, but I don't think it fair to discriminate. Maybe you could take this a step further and just use this and specify: process variance X<10, X>10, spot variance X<10, X>10, process deviation X, spot deviation X.

John,

ReplyDeleteYour timing is impeccable. Was recently debating this topic with a mutual associate of ours in that the behavior of the hue angle of near-neutrals "freaks out" the closer they are to the neutral axis. The moral of the story we were trying to tell is that as chroma decreases, so does the hue angle's ability to describe what is truly at fault with a color. So much so that if you have a sufficient dE tolerance for a near-neutral, and whilst holding L* and C* constant, you effectively find that h* can rotate in any direction and still yield an in-tolerance result. All the more reason I believe the triviality of dE76 completely fails to provide useful tolerances. Given how hard the industry works to address the neutrality of neutrals (e.g. G7) it seems that there has to be a breakpoint as to where one set of tolerances are used -- if not exactly the point you make here for your C* < 10 example.

My spot color tolerance is based on the hue angle (h') of the Farnsworth-Munsell color acuity test. All 84 chips have about the same L* = 58 and C* = 27 but vary in hue angle betwee 0-360'. The average dh' between chips is 4.2 which corresponds to an averge dE'00 = 1.8.

ReplyDeleteSteve Suffoletto SSuffoletto@BuffNews.com