Color: the 1931 CIE color-matching functions and chromaticity chart

Introduction

The CIE chromaticy chart is one of the things we are headed for. Here is a black-and-white drawing of its boundary. I’ll show you later how I got it.

You should be able to find color versions of the CIE chromaticity chart (1931) all over the Internet. Here’s one that shows some curves of constant hue. I’ll point out now that these curves are not straight lines. I will also use that image to decide that something is a greenish-yellow or yellow-green.

Wikipedia seems to have a lot of information, such as this, for example.

What my drawing shows is that I can reproduce the analytical content of it. Black and white, but no color.

How did I do that?

the color-matching functions and chromaticity coordinates

I have four tables of data, from Wyszecki and Stiles, for which see the bibliography; and I will abbreviate them “w&s” in this post.

“rgb” are chromaticity coordinates (row sums = 1), “rgb bar” are color-matching functions. The full heading on the combined “rgb” and “rgb bar” tables is “Chromaticity Coordinates and Color-matching Functions of the CIE 1931 Standard Colorimetric System, with respect to real primary stimuli:
R at nm;
G at nm;
B at nm”

We also have “xyz” chromaticity coordinates (row sums = 1), and “xyz bar” color-matching functions.

All 4 tables are at 5 nm intervals from 380 to 780.

Well, these 4 tables come in a variety of sets. I chose to use the 1931 tables at 5 nm intervals.

They have been interpolated to 1 nm intervals. They have been corrected by several authors in different ways. They were re-done in 1964 by the CIE to use a 10° field instead of the 2° field of the 1931 standard.

My choice of the uncorrected 5 nm 1931 standard is not universal — but it is an extremely common choice, and an awful lot of the work done since 1931 has used it.

At my level of usage, it probably doesn’t matter a whole lot. In fact, a whole lot of other things will turn out to be more important.

Since I’ve had more than a little time to get used to the 4 different tables, and you may not have, let me list them:

• rgb bar
• rgb
• xyz bar
• xyz

My source for the tables, as I said, was w&s. I am not going to reproduce them — as far as I can see the CIE charges for some of them — but I wouldn’t be surprised if you can find them on the Internet.

In principle, three of the tables are secondary. Their source is the table of rgb bar. First, let me show you what they look like.

Here is a picture of the rbar, gbar, bbar data. The x-axis runs through wavelengths from 380-780, the points are 5 nm apart.

It is clear that rbar is significantly negative over part of its range (440-545 nm); it is true, but not clear, that gbar is very slightly negative over 380-435, and bbar is very slightly negative over 550-655 — and it’s shown as 0 for longer wavelengths.

In fact, at any given wavelength, exactly one of the three is negative; nevertheless, only the red rbar gets seriously negative.

What do these negative numbers mean?

I suspect that the simple description of the color matching experiment is too simple experimentally, but I also believe it suffices in principle. Imagine that you are looking at a split screen, at two halves of a colored disk. The left half of the disk is a monochromatic color. (Think of a tunable laser — but of course they didn’t have lasers in the 1920s when the experiments were done!) The right half of the disk is controlled by the viewer, who has three dials and may choose mixtures of red, green, and blue in order to make the two halves of the disk match.

It turns out that we can’t do that. Over most of the range, we have to change the color of the left half of the disk as well as the right half. We can’t match a wavelength of 380 nm (the violet end of the visble spectrum) with red, green, and blue. But if we add a tiny of green to the violet, we can match that combination with red and blue.

We record a negative number for green at that wavelength.

Until the blue goes to zero thru negative values, exactly one of the 3 numbers is negative. At long wavelengths, when blue is zero, we just use red and green.

So much for the negative values.

Oh, having said that I will not reproduce the table, I will present — so you can do these calculations yourself — one quarter of it. Here is the rgb bar table at 20-nm intervals.

The second table, rgb, is derived from the first in principle — but not in practice!

In principle, take the rgb bar table and recsale it by setting every row sum to 1.

In parctice, we don’t have enough digits to do that. They provide a table. It is called the rgb (chromaticity coordinates).

Here is what the rgb table looks like at 5 nm intervals.

And here is a subset of the rgb table, at 20-nm intervals.

Let me show you an example of why we need their table, instead of trying to generate it ourselves. At 380 nm, the rgb bar table has values…

which sum to 0.00119 .

If I divide that row by its sum, and round, I get…

but the corresponding row of the rgb table is…

.

Close, but not the same.

That’s two of the four tables. As it happens, they’re the two we may be able to dispense with in practice — but in principle, rgb bar is the starting point.

Let me emphasize that these values depend on a specific choice of red, green, and blue.

We can fix that.

The CIE defined a specific change-of-basis. The tri-stimulus coordinates (rbar, gbar, bbar) or (R,G,B) are a vector in R^3; they defined a vector denoted either (xbar, ybar, zbar) or (X,Y,Z).

Oh, although the table is denoted rgb bar, specific coordinates wrt it are usually denoted RGB; similarly, specific coordinates wrt the xyz bar table are usually denoted upper-case XYZ.

Bear in mind that these RGB are not the same as coordinates in RGB-color space. (The latter is bounded by 0 and 1; the former is not!) We’ll see more about this down the road.

Let me show it to you:

And let me show you the cut-down data, every 20-nm; this is a subset of the “xyz bar” table:

Here is the change-of-basis matrix:

They tell us we can check it, for example, by applying their matrix T to RGB coordinates, such as for example at 500 nm…

Apply T, and round off, getting

… and compare that to the XYZ coordinates at 500 nm:

.

The same.

Still… raise your hands and step away from the keyboard.

What we just did was apply their matrix T to RGB components to get XYZ components. Perfectly natural. But we know that the transition matrix delivers old components when it is applied to new ones. If we want to call that given matrix T the transition matrix, then we must think of XYZ as the old components.

In practice, this really just means that we better know which way the transformation goes. They gave us the transition matrix for old components XYZ and new components RGB — even though they derived the XYZ basis from the RGB basis.

OK? If not, please be patient.

Now just where did that change-of-basis transformation come from? Malacara has a readable discussion. The most important point, to my mind, is that the Y column vector, the second column, is chosen to be the “phototopic response” (1924) of the human eye, that is, daylight vision, that is, color vision. There’s not much point in my digging out that data and plotting it: it is exactly the green curve in the drawing of the xyz bar table.

Second, all the values X,Y,Z are non-negative. Third, they also set the column sums almost equal to the value for the Y column. There was another criterion, but I’ll show it to you after I show you the fourth table.

Table 4 comes from table 3 as table 2 comes from table 1: get table 4 by setting the row sums of table 3 to 1, just as we got table 2 by setting row sums of table 1 to 1. All in principle, of course, or having more digits available than were printed.

Here is what the xyz table looks like:

and here is the cut-down data:

Just for the fun of it, check the column sums of XYZ. I said they were almost equal. I get

.

I do not know if that is deliberate, or whether its round-off and the real tables have more digits. (I don’t really care. But I couldn’t very well say they were equal….)

The CIE chromaticity chart

Ok, here’s the next major idea. Since x+y+z = 1, this table is essentially 2D, and we can plot y vs x.

That is exactly the picture I started this post with, the boundary of the chromaticity chart.

That’s all it is: take the first two columns of the xyz table, read each row as (x,y) coordinates, and plot each row.

The fourth criterion was to make the line x+y = 1 tangent to the curve. (In fact, the earlier criterion about X,Y,Z being non-negative was implemented by setting the y-axis tangent to the boundary.

Let me show you:

So, 4 tables, one 2D chart. We could also draw a 2D chart of g vs r, or b vs r, from the rgb table, since its rows each sum to 1.

What we have lost, in that chart, is the intensity. To recover XYZ from xy, we would need, for example, xyY; x,y and any one of X, Y, or Z would suffice.

Let me start using that chart right now. We have RGB basis vectors (1,0,0), (0,1,0), and (0,0,1}. What are their (x,y) coordinates?

There are a few ways to do this. Given the scale at which we are working, the easiest is to interpolate the x,y coordinates from the xyz table at wavelengths 435.8, 546.1, and 700 (that last one is a table value).

I get x,y coordinates of

R = {0.7347,0.2653}

G = {0.2737,0.717411}

B = {0.166541,0.00892768}

Let’ s plot them on the chromaticity chart… and let’s draw lines connecting them.

Think back to simplices. All of the points inside (and on) that triangle can be written as linear combinations of the coordinates of the 3 vertices; most importantly, the coefficients of those linear combinations are between 0 and 1, inclusive. That is, when the coefficients are in the closed interval [0,1], the combination is inside or on the triangle.

That restriction to [0,1] is what restricts us to the inside (and on) the triangle. That triangle is said to describe the gamut of these three R,G,B vertices.

The points outside the triangle have either negative coefficients (or perhaps coefficients larger than 1; I’m not sure). This is why we have negative values inside the primary table, rgb bar: all the points on the curved boundary are outside, or just on, the rgb bar triangle.

That triangle is said to to show the gamut of those choices for R,G,B primaries. One of the most common uses of the chromaticity chart is to compare gamuts.

From Foley, van Dam et al., I get the following x,y coordinates for the “standard NTSC RGB phosphor” (the color TV standard):

R = {0.67,0.33},

G = {0.21,0.71},

B = {0.14,0.08}

Let’s add them to the CIE, triangle in red:

A TV with these phosphors can reproduce colors in the red triangle.

Another set of coordinates from Foley & van Dam et al., for “short-persistence phosphors” are:

R = {0.61,0.35},

G = {0.29,0.59},

B = {0.15,0.063}

Let’s add them to the drawing, with their triangle in green:

Whatever these are, they can only produce the comparatively limited set of colors bounded by the green triangle.

To reproduce the entire CIE chart, within a triangle — i.e. with linear combinations having components in [0,1], we would need to have B at x,y = (0,0), R at x,y = (1,0), and G at x,y = (1,0).

Easy enough to draw, but don’t ask me how to get such sources physically.

A different form of the same principle — linear combinations of vectors — leads to the idea of mixing two colors. That is, instead of looking at all that can be produced by a particular set of RGB phosphors, we take two points and consider the line joining them.

A particularly useful choice is when one of the points is defined as “white”. Here are the x,y coordinates of a white known as D65, a standard form of “daylight”; here also are x,y coordinates of a form of green (from Andrew Glassner’s “Principles of Digital Image Synthesis”, volume 2, Appendix G; Morgan kaufmann, 1995), known as “foliage” on the MacBeth Color Checker.

p65 = {0.31271,0.32902}

pf = {0.4002,0.3504}

I want the equation of the line thru those points, and considering triangles, I’m going to write the equation as

line[t_] := t pf + (1 – t) p65.

Like our discussion of triangles, if t is restricted to [0,1], then we get the line segment bounded by the two points. Only if we let t range outside [0,1] do we get line segments that extend beyond the two points.

For what it’s worth, the numerical equation of the line is

.

Here are the two points and the line segment joining them:

But I’d rather do something else: extend the line to the boundary, and consider the green point to be a combination of the white point and the boundary point.

According to my poster of the 1931 CIE, a wavelength of 570 is on the border between greenish-yellow and yellowish-green. Go to the first link in this post. So our “foliage” at the red point is a mixture of white and a wavelength very close to 570 nm. (Or, Malacara has a version of this drawing.)

(I meant to show the two points as black and red in both drawings, but I forgot. I didn’t want to imply that the dot was the “right” color.)

One final point. We could go from HSB to RGB (for some standard set of RGB) and compute curves of constant hue. They are not straight lines. Rather than prove this by computation, let me just point you at the very first link again: the boundaries which it shows are not straight lines, and those boundaries certainly ought to be curves of constant hue, namely the boundary hue. (It ought to be easy enough to show this, but not now.)

It is crucial that the XYZ values are independent of any particular RGB values. Of course, they originally came from the spectrallly pure ones — but once we have the XYZ basis, we can go to any other RGB basis we choose.

Having seen other gamuts, we should expect that people get XYZ values from a spectrum. I’ll show you that next.