Why Is It So .....

[Yves W]

On Fuji, I don't have the BAYER CFA, I have the X-Trans with different RGB but I have the same results with pure white with PTFE.
In nature there is little true white, Calystegia is not pure white.
 

[lukaszgryglicki]

Funny thing is that my Fuji GFX 50R is also said to have X-Trans Bayer: https://en.wikipedia.org/wiki/Fujifilm_GFX_50R -> https://en.wikipedia.org/wiki/Fujifilm_X-Trans_sensor but from what I've research it just have a standard bayer sensor.

Maxmax is debayering one for me so I will know for sure.

 

[Andrea B.]

Poor bun! Yeah, I mean, white isn't a spectral color, but lots of colors (pink, brown, etc.)

are not spectral colors, and that wasn't what I was getting at.

"Yes, understood", she said. 

Sometimes it is SO DIFFICULT to discuss tech online. In real life conversations, corrections and restatements happen so easily if the two parties are willing.

---

 

Of course, I got to wondering about pink and brown. In the English language we are sloppy about what "pink" means. Is it desaturated red or is it desaturated magenta? And how would one characterize a pink reflection? Is a pink reflection a mix of reflected white and reflected red? I'm not sure how that could happen. Or is a pink reflection  simply a weak red reflection? If so, what is meant by "weak"?

I need to go look up whether and/or how a desaturated color can be characterized in terms of reflected wavelengths.

There is always so much to learn.

[Andrea B.]

Oh geez. A whole nother Can of Worms.

 

Wikipedia says:

• Colorfulness, chroma and saturation are attribute of perceived color relating to chromatic intensity.

Later it says:

• As colorfulness, chroma, and saturation are defined as attributes of perception, they cannot be physically measured as such,
  but they can be quantified in relation to psychometric scales intended to be perceptually even.

(Italics are mine.)

OK, I'll go with that for a minute or two while trying to figure out what characterizes reflected "pink" (i.e., desaturated red)in terms of reflected wavelengths.

 

[[I feel like I have looked this up before and then forgotten it!]]

[Kai]

@ Andrea,
here is a thought on the fluorescence of Bayer dyes:
If fluorescence were to occur under any circumstances, the image contrast would have to decrease dramatically due to light diffusion, right?
Has anyone ever observed this?

[Andrea B.]

THESE ARE NOTES FROM MY ONLINE READING. THEY ARE PROBABLY NOT ALL ACCURATE.

 

Exploring Pink, a non-spectral color, as a Desaturated Red.

Warning:  This is a digression and could be rather boring.

 

Spectral Color = evoked in a typical human by a single wavelength of light in the visible spectrum.

May be extended to include a very narrow band of wavelengths.

 

 

Color Appearance Model versus Color Model

 

A Color Model defines a coordinate space to describe colors. Example:  RGB triplets.

• So a typical pink would be something like (255, 200, 200).

 

A Color Appearance Model is a mathematical model which describes the perceptual aspects of human color vision.

hue (color designation by wavelength, dominant wavelength, color wheel location, whatever)

For my pink square, the hue is red=0°, which is somewhere in the wavelength range of about 645-696 nm.

BTW, for brown, the hue is specified as orange which is somewhere around 615 nm.

Note that hue alone is not enough to describe a color perception.

brightness (perceived strength of color signal)

Seems OK to say the pink square is bright. The HSB color model gives it a 100% brightness.

colorfulness (perceived quantity of color)

Yes, the pink square has color.

chroma (perceived color designation)

Perceptually speaking, that is a pink square. It is certainly not perceived by me as being red even though that is the hue designation.

lightness (perceived strength of signal as compared to a white object)

The pink square is light, but not as light as the white screen it is on.

saturation (perceived quantity of color as compared to a white object)

The pink square has more color than white, but not by much. That is, I see the pink square as somewhat desaturated.

 

A CAM does assume many viewing conditions (background, illumination type, position of viewer and more)

A CAM is an attempt to map spectral stuff to perceptual stuff.

 

So as a CAM pink my square above would be described as hue=0°/695nm, lightness=yes pink seems luminous next to white, brightness=yes in good light, colorfulness=yes pink objects have color, chroma=?, saturation=pink appears desaturated to me.

 

The following two Color Model descriptions do not agree on the saturation concept. Interesting.

In my perception the pink square is not 100% bright as in HSB and it is not 100% saturated as in HSL.

HSB = HSV = (0°, 21.6%, 100%) hue, saturation, brightness

HSL=(0°, 100%, 89.2%) hue, saturation, lightness

 

I think I'd describe the pink square as (0°, 21.6% sat, 89.2% bright). So there!

 

---

 

 

Saturation in terms of wavelengths:

Saturation is determined by a combination of light intensity and distribution across different wavelengths.

The most saturated color = one wavelength at a high intensity.

That is a bit vague. Hmm......

 

We perceive a high intensity wavelength as more saturated than a low intensity wavelength?

Then that would indicate that my pink square of hue=0° is a spectral color.

Now I'm confused.

 

I do get that wavelength intensity is a function of wave amplitude.

But how does the human eye perceive differences in wavelength amplitude?

 

 

I have to leave it all there for now. There are other tasks to attend to.

 

 

LATER:  oh, duh! ding, ding, ding...

Higher wavelength amplitude of a given hue corresponds to a perception that the color is bright.

Lower wavelength amplitude of a given hue corresponds to a perception that the color is not as bright.

As an example, here is the pink square made less bright. RGB = (230, 200, 200).

 

 

 

 

 

---

 

The set of all visible wavelengths forms an infinite dimensional Hilbert space H_color.

Think of a function mapping wavelength to its intensity.

{(wavelength, amplitude of wavelength)}, RxR, where w in real interval [400, 700] and a in [0,?].

 

A visible wavelength stimulates the 3 human photoreceptor sets (cones),

so a humanly perceived color can be modeled by a number triplet representing

how much stimulation occurs in each cone set.

 

Thus human color perception can be modeled by a Euclidean space R³_color.

Added 03 Sept 2022: This is not correct.

 

There are physical limitations in the human eye/brain perception of color

which prevent the mapping from the Hilbert sp. to the Euclidean sp. from being 1-1.

 

Well, that explains a lot of things.

[Andrea B.]

Kai wrote:  If fluorescence [of the Bayer dyes] were to occur under any circumstances, the image contrast would have to decrease dramatically due to light diffusion, right? Has anyone ever observed this?

 

It seems hard to know whether dye fluorescence, if any, would cause a "bloom". Seems like it would depend on the intensity of the emitted fluorescence and the physical grid containing the bayer dyes and its position relative to the "pixel wells". (Not sure of the terminology there.) 

 

Given that we see increased fine surface details in UV photos due to the shorter wavelengths, I don't think any of us can say that we've routinely observed decreased contrast or light "smear" or anything like that. (I'm excepting generic flare in that statement.)

 

 

[colinbm]

Brown is orange in this video ....

 

[Andy Perrin]

7 hours ago, Andrea B. said:

A visible wavelength stimulates the 3 human photoreceptor sets (cones),

so a humanly perceived color can be modeled by a number triplet representing

how much stimulation occurs in each cone set.

Thus human color perception can be modeled by a Euclidean space R³_color.

 

 

 

Well, that explains a lot of things.

Not sure where you got all that, Andrea, but it's incorrect, or at most partly correct. You can't accurately model human color perception as a Euclidean vector space. You can't even model it as a curved space either, like they use in General Relativity, according to that article which Doug linked earlier this month. The problem is, just being able to assign a set of three numbers to each photoreceptor isn't enough to give it the *structure* of a vector space (in other words, how each color relates to adjacent colors is different in human vision than the way vectors in a Euclidean vector space are related). 

 

That doesn't stop people from trying (and failing...) to treat it like a vector space, though. Even Stefano tried just this past month to write a simulation where he tracked the transformation of each spectral color in the visible range into RGB and tried averaging them — he discovered for himself that you don't get white if you do this! And it's directly because colors are not truly vectors in a vector space. You can't add two colors that are represented as RGB triplets and have it behave the same way as shining two corresponding colors of light on top of each other and then converting to RGB afterwards. 

[Wayne Harridge]

On 9/2/2022 at 5:30 PM, colinbm said:

I have not seen the Bayer CFA fluoresce, but that is through the cover glass that should allow the UVA light in & the fluorescent light back out ?
I would like to see a graph of the Bayer CFA that reaches below 300nm like it would have with a fused silica window.
Here is one that I have found ....
https://www.flir.com.au/support-center/iis/machine-vision/application-note/understanding-color-interpolation/

Looking at this sensor response curve raises a question about the interpretation of a specific RGB value and its subsequent display on a screen or print.  If you look at the red curve for example and put a horizontal line through it at a low value of QE (e.g QE=3) there are up to 6 different wavelengths that could generate the same response at the sensor.  I presume when this value is displayed or printed the assumption is that the wavelength of the peak response is displayed but with a relatively low QE value, i.e. the wavelength of the red dotted vertical line.  I think this has some implications for white balance.

 

[Andy Perrin]

Yeah, that’s the same as the metamer issue — in a visible light context, if you have a yellow pixel, is it because the light was spectral yellow, or because you had equal amounts of spectral red and green light? This is why you can’t infer wavelengths from pixel colors (except in a very vague sense) under most circumstances. 
 

The implications for white balance are that multiple spectrums can trick our cameras into seeing “white” besides the standard D65. This is why you can have multiple lightbulbs with high CRI rating (CRI measures closeness of the spectrum to D65 unless the source has color temp below 5000K) and they look different!

[Stefano]

12 hours ago, Andy Perrin said:

That doesn't stop people from trying (and failing...) to treat it like a vector space, though. Even Stefano tried just this past month to write a simulation where he tracked the transformation of each spectral color in the visible range into RGB and tried averaging them — he discovered for himself that you don't get white if you do this! And it's directly because colors are not truly vectors in a vector space. You can't add two colors that are represented as RGB triplets and have it behave the same way as shining two corresponding colors of light on top of each other and then converting to RGB afterwards. 

In case anyone is curious...

 

Spectrum from 380 to 825 nm:

 

Sum of the above colors (using a finer interval, but the result is the same):

 

(255, 195, 172, which became 255, 195, 171 after the conversion to .jpg):

 

Using the 1931 CIE color matching functions modified by Judd (1951) and Vos (1978), plus conversion to sRGB and gamma (2.4).

[Andrea B.]

Andy thanks.  I was making reading notes up there which I have not really thought about!

I'll try to find the link claiming vector spaces.

 

For now I crossed out the offending statement!

[Andy Perrin]

I mean, people do try to APPROXIMATE them with vector spaces, just doesn't work very satisfactorily.

[Stefano]

How do you correctly sum colors?

[Andy Perrin]

Here's a direct answer:

https://en.wikipedia.org/wiki/CIE_1931_color_space#Mixing_colors_specified_with_the_CIE_xy_chromaticity_diagram

 

Obviously to use that you have to have the x and y values, which are in the same wiki page (above the formula a bit).

[Stefano]

Thanks. I was expecting it to be more complicated actually.

[Andrea B.]

Regarding the reference to modeling human vision as a Euclidean R³ space --

See Wikipedia Color Vision. Scroll to the section Mathematics of color perception.

However, in looking through that article's references, I do not find anything (yet) which discusses the Math used in the Wiki article.

 

What is being modeled? It seems to be the outputs from the human Short (blue), Medium (green) and Long (red) cones.

 

HOWEVER, see the next topic.

 

The 4 outputs from the S,M,L cones and from the rods -- combine in a certain way to trigger 3 brain responses: one in the red-green opponent channel, one in the blue-yellow opponent channel and one in the lightness-darkness (i.e., white-black) opponent channel. The combined brain responses determine a color perception. 

 

[Some think that the opponent channels are better thought of as cyan-magenta and lime-violet.]

[Calling the Long cones "red" is a bit of a misnomer. The peak is actually in a green-yellow region.]

 

 

[Andrea B.]

Here ya go!! As noted by Andy, perceptual color space is NOT Euclidean. In this paper it is shown that perceptual color space is also NOT Riemannian.

So who would like to correct Wikipedia's claim?

 

The non-Riemannian nature of perceptual color space

 

Abstract

The scientific community generally agrees on the theory, introduced by Riemann and furthered by Helmholtz and Schrödinger, that perceived color space is not Euclidean but rather, a three-dimensional Riemannian space. We show that the principle of diminishing returns applies to human color perception. This means that large color differences cannot be derived by adding a series of small steps, and therefore, perceptual color space cannot be described by a Riemannian geometry. This finding is inconsistent with the current approaches to modeling perceptual color space. Therefore, the assumed shape of color space requires a paradigm shift. Consequences of this apply to color metrics that are currently used in image and video processing, color mapping, and the paint and textile industries. These metrics are valid only for small differences. Rethinking them outside of a Riemannian setting could provide a path to extending them to large differences. This finding further hints at the existence of a second-order Weber–Fechner law describing perceived differences.

---

 

At this point the principle of diminishing returns also applies to my reading about color vision and color theory. 

Enough! Time to go do other things.

[Wayne Harridge]

4 hours ago, Andy Perrin said:

Here's a direct answer:

https://en.wikipedia.org/wiki/CIE_1931_color_space#Mixing_colors_specified_with_the_CIE_xy_chromaticity_diagram

 

Obviously to use that you have to have the x and y values, which are in the same wiki page (above the formula a bit).

Thanks for that link Andy, a very interesting read.

 

[Andy Perrin]

A

53 minutes ago, Andrea B. said:

Here ya go!! As noted by Andy, perceptual color space is NOT Euclidean. In this paper it is shown that perceptual color space is also NOT Riemannian.

So who would like to correct Wikipedia's claim?

 

The non-Riemannian nature of perceptual color space

Hah, I would be surprised if someone doesn't edit Wiki soon, if that article turns out to be correct. But let's get real here: all science is about modeling the real world making certain simplifying assumptions that work within some range of validity. For example, our actual space is not Euclidean either, and Einstein showed with General Relativity that it is Riemannian, but that doesn't mean that we always need to use General Relativity for every single calculation. If you want to find the path a baseball takes if you throw it, plain old Newtonian physics works fine, despite assuming a Euclidean space. The same thing goes for color theories. These more general theories might have something to offer in certain circumstances, but approximating things in a Euclidean way or a Riemannian way will work a lot of the time.

[photoni]

I understood 20% of the last things written.

Thinking of the old world of photographic slide films (I worked and loved the purity of Ektachrome 64)

white is not objectionable if you use a professional viewer with 6500 ° K neon tubes and IRC98%

Now digital allows us to very easily balance the three color channels in the visible

we use well calibrated monitors, with a wide gamut 99% AdobeRGB - P3 ???

but anything beyond the visible is a subjective ... and questionable interpretation

even using a Teflon / satin aluminum ... as a reference sample we will always have a bayer filter that falsely interprets the colors.

I think XD