Stephen Williams wrote:
>
> s57uuu@hamradio.si said:
> > (Yes, it's 3 - the luminance sensors (the 'rods') have the same
> > sensitivity as the 'green' channel 'cones')
>
> Not quite, they respond to a wider spectrum then the green sensors.
> Though they are highly sensitive to green, they respond to most all
> of the visible color range. I think.
>
They do have the same curve (they also use the same sensitive compuond).
None of the eye (or film/CCD) sensors is really a 'rectangular' band
pass. They all have relatively wide responses, with the skirts extending
all over the visual range.
The eye curves are well measured and documented, you can find them
tabulated and graphed in any serious book about imaging.
The film curves can be found in manufacturer's brochures. (the same
goes for CCD's)
The linear approximation for the output of such a channel sensor is
(let the S represent the integral sign and l lambda)
R = S r(l)i(l) dl
one such equation for each channel,
where r(l) is the sensitivity curve of the sensor, and i(l) is the
spectral intensity of the scene. Integration limits are over the
full range where r(l) is nonzero.
The eye and film can be highly non-linear, but CCD's are for practical
purposes perfectly linear (= ideal image sensors, they caused a
revolution in astronomical and other scientific imaging)
> But do the spectra of those three channels cause the RGB sensors in
> your scanner to respond similarly? That's a question for a photography
The consequence of the above equation is that you have crosstalk
between the channels, which depends on the shapes of all the curves
involved, and these can differ between various films, photo-papers,
printer inks, CRT phosphors and whatewer. Trouble guaranted! (and this
was only the linear model ;-) There is a 3x3 matrix linking two such
linear tricolor systems. Ideally, it would be a identity matrix, but as
Murphy designed it, it never is.
Even if the artificial sensor's curves would match the eye curves
perfectly, you wuold still have crosstalk. That's why you can't really
reproduce ALL the colors using a tricolor system - you're limited to
the inside area of the triangle defined by the primarie's places in
the chromacity diagram. Most notably, pure spectral colors are always
outside. But most (virtually all) colors in nature DO fall within
this triangle, and with some (a lot of--) tweaking, one can get very
good results.
Colorimetrists represent the 'outside' colors by assigning negative
values to the channels. But that's only a mathematical trik, useful
for computation, because in practice, you can't squeeze 'minus red'
out of your monitor...
When you try color metrics (measuring differences between colors),
things get even more scary. Colorimetrisis found out that the human
chromacity space IS 3D, but its metrics are non-euclidean....
Do you still dare to scan? :-)
Marko Cebokli
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This archive was generated by hypermail 2b29 : Wed Dec 13 2000 - 09:22:24 PST