[net.physics] Newer pigments are outside the spanned vector space?

kpmartin@watmath.UUCP (Kevin Martin) (08/12/84)

The difference between a 3-D basis for a space in algebra class and a basis
for colours is that you can't apply negative co-ordinates in colour mixing
(for additive mixing, anyway), that is, you can't take a somewhat washed-out
green and red, mix them (to get a washed-out yellow), then take out some blue
to increase the saturation of the colour.

By projecting the colour to a constant-intensity plane (so we can ignore
the brightness), and by using two super-tints (for lack of a better name)
as a basis, all visible tints can be plotted. One choice of basis gives
the CIE colour diagram, in which the spectrum forms a horseshoe-shaped line,
with the 'white' spot being somewhere near the centre of the horseshoe.

Combining two colours A and B involves picking the colour X on the
line AB such that AX:XB = intensity ratio. This is due to the sum
being a vector sum which must be projected (from the 3-space) back onto the
constant-intensity plane.

This construction indicates that it is impossible to get a colour mixture
to match a spectral colour exactly. In fact, the triangle enclosed by the
three primary colours used to television is only about half of the total
possible tints. At the time the NTSC colour TV standard was made up, this
triangle enclosed almost every dye tint known, so it was considered
satisfactory.

If you want a look at such a diagram, practically every book on
colour TV technology or the NTSC standard seems to have one. They also
sometimes show eye sensitivity diagrams (and thus explain why NTSC
picked two wierd tints with different maximum bandwidths for their tint
basis, rather than just blue and red...)

As for the subject line...
Newer dyes which seem to reflect tints which are outside the space spanned
by a basis (with positive co-ordinates only) colour set such as the
three TV dyes are not impossible. In fact, the only ways to have
a colour basis be all-encompassing is to have either every spectral tint
in the basis (that is, and infinite number of basis tints), or to make the
basis from super-colours, which are outside the spectral curve (ha!).
If you just want to get close enough to every colour that your eye doesn't
notice the difference, A few (6? 20?) spectral colours suffice.
      -Kevin Martin, UoW Software Development Group.