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Imagine perfect monitor that can display pure blacks and is perfectly calibrated.

What would the contrast be between its second darkest shade of black, that is 1,1,1 values for blue green red sub pixel and brightest white with 256,256,256 values for rgb?

Please note that I mean second darkest shade of black, that means the pixel is active and light shines from it albeit very weakly. I specifically dont want people to mistake this with the darkest shade of black ( rgb 0,0,0 ) which means the pixels are completly turned off and no light exits them.

This is becose with monitors with perfect black, no matter the peak brightness value, the contrast is always infinite. So that is why I ask about second darkest black, not the total black.

I know SDR monitors have 2.2 gamma curve target and have 256 possible values of becose they are 8bit. I know 8 bit has 256-1 contrast ratio assuming the brightness change is the same between all the bit values but since the encoding of the digital 8 bit format is not linear like RAW files, I have no idea what contrast/dynamic range 8bit is supposed to have becose I dont know how much are the individual bit values spaced apart.

If this theoretical perfect SDR 8 bit monitor had peak brightness of 100 cd/m2, how bright would the second darkest shade of black be?

wav scientist
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    I'm voting to close this question as off-topic because theoretical contrast calculations of monitor values are not photography. – flolilo Aug 17 '19 at 21:50
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    There is no such thing as perfect black. There is projected black, which is the darkest value a display can show, vs. off-black, which is simply projecting nothing because it is off. Thus, the contrast ratio between (255, 255, 255) (and not 256, 256, 256) is measurable and not infinite. – scottbb Aug 17 '19 at 21:56
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    Monitors are essential part of digital photography, without them you will not see any of your digital photos. A monitor is just as important as camera when it comes to photography. – wav scientist Aug 17 '19 at 21:58
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    I generally agree with @flolilo that this is off-topic as written, but I feel that there is something here that can be of topical interest to photography. Yes, monitors are used in the editing and viewing process, but that doesn't mean any question pertaining to monitors is prima facie on-topic for photography. How can this question be edited to make it pertain more to photographic interests? – scottbb Aug 17 '19 at 22:00
  • The only way to make this question appear more about photography is ask this from point of view of how many stops of dynamic range my camera needs to have to be able to produce perfect 8 bit SDR Rec 709 photos where even the darkest shades are noise free. – wav scientist Aug 17 '19 at 22:19
  • @wavscientist The answer to that would be : "∞" – Michael C Aug 18 '19 at 01:21
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    however in the real life (and even in the definitions of dynamic range) we use certain criteria for acceptable noise, less demanding than "noise free", so there must be some answer more usable than "∞" ;-) and what is easier to calculate and maybe even more practical is the minimum number of raw bits in a camera if we want to be sure that the faithful reproduction is limited by the sRGB display and not the camera - and the answer is 12. – szulat Aug 18 '19 at 12:05

3 Answers3

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The contrast between the brightest and the second darkest shade in a 8-bit monitor can be as high as you want it to be, depending on your personal configuration and the intended color space. However, if you want to stick to sRGB then the contrast is 3294.6:1 because that's how the sRGB gamma is defined:

linear_brightness(1/255) = (1/255) / 12.92 = 1 / (255*12.92) = 1 / 3294.6

szulat
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    I applaud your approach, but the math is wrong. The formula for sRGB can't be applied to the ratio, it must be applied to the individual colors before dividing. – Mark Ransom Aug 18 '19 at 20:33
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    @MarkRansom nobody is applying the formula to the ratio, this is the linear brightness of the second darkest shade (having the sRGB value = 1/255). the brightest shade is 1 (you could use the formula but it's the max value so we know it's 1 without calculating) and then dividing these two brightness values 1/(1/3294.6) gives the contrast of 3294.6 : 1 – szulat Aug 19 '19 at 00:53
  • My apologies, since you weren't explicit about the final divide I assumed the 1/255 was a ratio and not a color. – Mark Ransom Aug 30 '19 at 23:54
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DEFINITIONS:

  1. CONTRAST: At the most basic, contrast is a defined relationship between two stimuli. But WHAT relationship and HOW derived is a subject that can take a couple chapters in a textbook to describe. If you want more, Google Weber Contrast, Michelson Contrast, and CIELAB Difference.
  2. Brightness is a perception, not a measure.
  3. Human perception is nonlinear relative to light, and also to spectral intensities.
  4. Luminance is a LINEAR measure of LIGHT, spectrally weighted. It is either relative Y (0-1 as in Y from CIEXYZ) or absolute L, as in 100 cd/m2.
  5. Perceptual Lightness is a non-linear measure of light, but is intended as "perceptually uniform" i.e. linear in regards to perception. L* (Lstar, not to be confused with L for Linear Luminance) from CIELAB (Lab*) is one such measure of perceptual lightness.
  6. Gamma: A curve used to encode image information to take best advantage of a given data space, taking advantage of the non linear aspect of human perception.
  7. Perfect black: there is no such thing, especially not in a monitor that is turned on. But if you want to discuss "theoretical black" then okay.

PAINT IT BLACK

In a real world environment, the actual "black" level (#000) including modest flare will typically be around 1 cd/m2, at least that's a good approximation for most purposes.

But if your question is, what is the luminance of #010101 if #FFF = 100 cd/m2?

In that case, #010101 = 0.0303526983548837 cd/m^2. If you use the piecewise sRGB linearization, as discussed in this answer I posted on StackOverflow.

However, monitors more typically use a more "pure" gamma function. And it's not necessarily 2.2 - it's how you have ADJUSTED it. But we'll go with 2.2 for maths sake. In the case of a simple 2.2 gamma, and the imaginary "perfect black" then: #010101 = 0.000507705190066176 cd/m^2

The Contrasts:

Using the sRGB piecewise maths: YLo=(1/255)/12.92

As this is normalized for 0-1, multiply by 100, i.e. YLo=((1/255)/12.92)*100

For YHi = 100.0 & YLo = 0.0303526983548837 Then:

  • Simple Ratio (YHi / YLo): 3294.6:1
  • Weber Contrast ((YHi - YLo) / YHi): 99.9696473016451%
  • Michelson Contrast ((YHi - YLo) / (YHi + YLo)): 99.9393130234252%

Using the simple 2.2 gamma maths: YLo=(1/255)2.2

As this is normalized for 0-1, multiply by 100, i.e. YLo=((1/255)2.2)*100

For YHi = 100.0 & YLo = 0.000507705190066176 Then:

  • Simple Ratio (YHi / YLo): 196964.699:1
  • Weber Contrast ((YHi - YLo) / YHi): 99.9994922948099%
  • Michelson Contrast ((YHi - YLo) / (YHi + YLo)): 99.9989845947751%

(There's a potential rounding error difference of ~0.000000000030352698380 depending on if you use the rounded coefficients).

Myndex
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The perceived performance can be much better than 8-bits because a whole lot of dithering may be going on. All plasma panels and DLPs (digital micromirrors) are 1-bit devices, but thanks to pulse code modulation and forms of dithering a.k.a. noise shaping a.k.a. error diffusion they can produce a fairly decent greyscale (to our slow and unsharp human eyes). It's easy to make a 6-bits image look as good as 10-bits. Sometimes the noise on the original image is already enough for this. 8-bits video was specified for the amount of analog noise of that time, more bits would only reproduce more precise noise.

The silver grains of "analog" photography are binary (black on white) too, but given enough grains and a random distribution they can make a pretty analog photo too.

Jeroen van Duyn
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