Hi,
here's another attempt to explain the least significant bit.
Imagine you have 6 bits for each color.
G5 G4 G3 G2 G1 G0 R5 R4 R3 R2 R1 R0 B5 B4 B3 B2 B1 B0
Now the only difference between real 6 bits per color and the Q40/Q60
highcolor mode is, that you are forced to give G0, R0 and B0 the same value.
G5 G4 G3 G2 G1 G0 R5 R4 R3 R2 R1 R0 B5 B4 B3 B2 B1 B0
| | |
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use same value!
When you are forced to use the same value for G0, R0 and B0, it is enough
to use the same variable to describe all three, lets call it "RGB0".
G5 G4 G3 G2 G1 RGB0 R5 R4 R3 R2 R1 RGB0 B5 B4 B3 B2 B1 RGB0
And if you use the same variable, you don't need to hold it at 3 different
positions in memory, so you can skip two of them. The result is:
G5 G4 G3 G2 G1 R5 R4 R3 R2 R1 B5 B4 B3 B2 B1 RGB0
(The Q40/Q60 highcolor bit arrangement)
By the way, the 3 digital to analogue converters (DACs) which create the
color output signals, all have 6 bit digital inputs!
So it is indeed a true 6th bit for all three colors. The restriction is
that the 6th bit is tied together.
Has this helped or caused more confusion? :-)
All the best
Peter
