RGB color conversion

MSX color component = 24bit color component / 32

No formula - you just drop (discard) the least significant bits that don't fit in the less-accurate end result.

Say for example you have a 24 bit value containing 8 bits that represent red, and you want to convert to an RGB value that has only 3 bits representing red. Then you take the 3 most significant bits (out of 8), and put 'em in that 3-bit space. The remaining 5 (less significant) bits you throw away.

Yes that means losing image quality, but that's the whole point of such a conversion: represent the same pixels using fewer bits. One could try various dithering and/or rounding methods to improve results (or perhaps an integer divide with a divisor other than a power of 2), but ultimately you will be discarding part of the image information.

Red / Green / Blue channels can be treated as independent, applying the same method to each. Or not. Again: whatever works, but one way or the other you'll be discarding those least significant bits.

From 8-bit colour component values to 3-bit values: divide by 255, multiply by 7, and round the result. Other way around: divide by 7, multiply by 255, round the result.

The 7 colour component values: 0, 36, 73, 109, 146, 182, 219, 255

When drawing graphics for screen 5, I try to pick only colours with these component values.

Why is this better than dropping the lowest 5 bits (/32)? Because in a 3-bit palette value 0 is fully off, 7 is fully on, and on a 8-bit palette value 0 is fully off and 255 is fully on. You can convert these to a scale from 0 to 1, by either dividing by 7 or dividing by 255.

Assuming all values are linearly distanced (disregarding things like gamma), you want to pick the 3-bit colour that is closest to the original 8-bit colour. For e.g. 8-bit value 224 (0.88), this should be 3-bit value 6 (0.85), not 7 (1.00). This is why you should divide, multiply and round, rather than just dropping bits with an integer division, as it minimises the error.

For this reason when you convert a 3-bit colour component to 24-bit, you also don’t want to pad the binary values with five 0’s, but rather nc = c << 5 + c << 2 + c >> 1 (integer logic) or nc = round(c / 7 * 255) (floating point logic).

@Grauw If you want to take that approach then (VALUE+16)/32 is better... The problem is that you think that 255 should become 7 while actually you should think that 256 should become 8.