# Tag Info

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Because your monitor is not properly calibrated. On my screen at home the top and bottom parts have the same hue. At my office though, the top part tends to looks a bit yellow compared to the bottom part that looks more red. The difference: my screen at home was from a series that was decently calibrated out of the factory, and on top of that I did ...

6

Simple answer: because it costs. :) Some higher-end monitor models are indeed calibrated at the factory (and come with a calibration report sheet) but it would not be cost-effective to do so for all models in the product line. Color reproduction varies even between batches of the exact same panel model, so it isn't possible to just use a set calibration for ...

6

Without replicating the bits the LSBs will be 0, so for the maximum value of 0x1f (max for 5 bits) it would expand to 0xf8 when converted to 8 bit. What you want is 0xff so the range of 0x00->0x1f will be mapped to 0x00->0xff instead of 0x00->0xf8. Without merging the LSB you would not be able to convert 0x1f,0x1f,0x1f to white (0xff,0xff,0xff). Incidentally ...

6

There's actually a reasonably good mathematical reason for doing bit replication: First note that the n-bit string, $N$, actually represents the value $\frac{N}{2^n-1}$ and we want to produce the m-bit string, $M$, where $n<m$ and $$\frac{N}{2^n-1}\approx\frac{M}{2^m-1}$$ We first scale numerator and denominator with \frac{N.(2^n+1)}{(2^n-1)(2^n+1)}\...

4

The physical imaging elements of your monitor define your gamut. So things like how bright the elements can be as well as the color spectra of the color elements define the gamut. But, thats quite the simplification, its also affected by such things as the screen surface material and background lighting conditions. Can the gamut be changed? Sure, if you ...

4

In order to replace a color with another color, you need some sort of distance metric between the colors and a function for calculating the blending based on that distance. Finding a distance between 2 colors can be tricky. RGB is a bad color space to measure the distance between colors for perceptual uses. HSV or YCbCr are better. Once you have the ...

3

There are two aspects to this: The traditional chromaticity graph is actually a two-dimensional slice of a three-dimensional volume. The slice usually used is one that matches up (more or less) to the fully-saturated primaries of an RGB-based colorspace; the full RGB colorspace is a tetrahedron. CMYK is a four-primary colorspace, with primaries that are ...

2

It turns out that the chromaticity chart is harder to read than I anticipated. The CMYK slice is actually triangular of sorts its just that the chromaticity chart is not really linear. The chromaticity chart is made with the assumption that light is a spectra. The curved arc is the pure spectral color of the rainbow. Everything in between is interpolated ...

2

I realized that littlecms, suggested by joojaa in a comment above is very complete in dealing with the ICC profiles. I did a few tests in their color translator program with different destination color spaces and the differences are great as you can see: Original sRGB-like adobeRGB1998 (it seems it's the original color space) appleRGB And the most ...

2

Assuming you are looking to transform a particular component (r, g or b) of any color, this can be easily done by simple vector/matrix multiplication. The introduction on this page (Chapter 6.1) - from the most excellent immersivemath.com site - illustrates this perfectly: Chapter 6: The Matrix (Immersive Linear Algebra) | immersivemath.com Think of it as ...

2

Was interested in this myself and did some digging, I was wondering if they where spectral, but it appears not (although some interesting work on spectral BRDF is here http://web.cs.wpi.edu/~emmanuel/courses/cs563/write_ups/cliffl/cliffl_spectral_brdf.html ) This chat seems to indicate they are camera native linear, probably with sRGB primaries. https://...

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There has been quite some research into this using Barten contrast sensitivity function. It is the current formula behind the Dolby Perceptual Quantizer as featured in SMPTE 2084 and HDR10. This, coupled with a colour appearance model such as the work behind Dr. Mark Fairchild's CIECAM02, can result in very accurate predictions of quantization depth.

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A nearest-neighbor resize will do what you ask but I am not certain it will give you much smaller tiles. It is quite likely that the tiles made from the original image are smaller simply because they have less high frequency content and thus compress far better. You can look at it the other way around: there are more terrain features in the tiles of the ...

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Think about it this way. To calculate YCbCr of (1, 0, 0), you'll end up with Y = 0.299, Cb = -0.16873, Cr = 0.5. If you take the magnitude of the chroma channels (sqrt(Cb*Cb + Cr*Cr)), you get ~0.53 for the "saturation". The YCbCr color space has a different shape than RGB, so it may represent values differently enough that you don't get the exact same ...

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Someone points me that this article describes the spectral response of the camera the MERL paper used, and even provides colour transform matrices at the end. Unfortunately, using those transforms result in even worse results than assuming sRGB primaries.

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