You probably want model paint then, it comes in lots of bright colors, albeit in small doses
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Replying to @0xabad1dea
Hmm, thanks for the suggestion… I don’t know much about paint, but it seems to be hard to find actual color values online in a lot of cases
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Replying to @comex @0xabad1dea
Colorimetry is hard. The mapping depends on way too many factors (e.g. what light you use), even if you assume sRGB on the RGB side. An unqualified RGB tuple is almost meaningless.
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Replying to @marcan42 @0xabad1dea
Colorimetry is hard, sure, but the difference I’m talking about is large enough that it doesn’t really matter.
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Replying to @comex @0xabad1dea
The reason why there's no paint in pure sRGB corner colors is probably that it'd be too dark in normal lighting. If you only reflect a very narrow band of wavelengths as required, you're throwing away most of the spectrum. Things are always duller in reflective media.
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There's probably a way to do the math too. Assuming a D65 illuminant, there should be an optimal reflective response that visually matches an sRGB blue primary chromaticity, and then you can calculate its brightness relative to 100% white. I'm sure it's less than
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Replying to @marcan42 @0xabad1dea
More like a curve of visual closeness vs. brightness, no?
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Replying to @comex @0xabad1dea
Well, you could compute a 3D space of the optimal spectra representing a chromaticity from 100% sRGB blue to white, and then pick the one that matches the brightness of 100%B in sRGB (which will be washed out)
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Replying to @marcan42
I’m pretty sure sRGB (or any other gamut) doesn’t specify brightness.
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If your illuminant were highly narrowband, like 3 lasers covering a gamut wider than sRGB then you *could* have a reflector/paint that produces optimal brightness sRGB blue relative to white. But in the real world we don't usually get ceiling lights made of RGB lasers. Usually
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