Very Computer

There's a number of color spaces that will give you something like that. Try
the color space FAQ :

http://www.faqs.org/faqs/graphics/colorspace-faq/

Usually there's just a few equations that relate the space to RGB so writing
the code is pretty easy.

Are you really sure that you want to do that?

I've got the algorithms if you really want them, but you should be aware
that there will be some rather substantial errors in the conversion of
double (L*a*b*) to integer (RGB) variables.

I've actually written some software to demonstrate these losses.

Kelvin Craig
Datacolor International

--
Charles Poynton

<http://www.inforamp.net/~poynton/>

I'm sure I've put this up once already.  But maybe I didn't.....

Assuming a D65 illuminant, and NTSC primaries (taken from the Adobe Tiff
6.0 documentation)

RGB -> Lab

First convert to CIE XYZ tristimulus

X = (0.6070*R) + (0.1740*G)+(0.2000*B)
Y = (0.2990*R) + (0.5870*G) + (0.1140 * B)
Z = (0.0000*R) + (0.0.660*G) + (1.1110 * B)

Now the fun starts......

L = 116(Y/Yn)1/3 - 16
a = 500[(X/Xn)1/3 - (Y/Yn1/3)]
b = 200[(Y/Yn)1/3 - (Z/Zn)1/3]

1/3 means to the power of 1/3 by the way!

Here the values Xn, Yn, Zn are the values for white.  For a D65 illuminant,
and a standard 2 degree observer, use

Xn = 95.02
Yn = 100.00
Zn = 108.81

For values of X/Xn < 0.01, Y/Yn < 0.01, Z/Zn < 0.01, use this formula
instead

L = 116[f(Y/Yn) - 16/116]
a = 500[f(X/Xn) - f(Y/Yn)]
b = 200[f(Y/Yn) - f(Z/Zn)]

where
f(Y/Yn) = (Y/Yn)1/3 when (Y/Yn) is greater than 0.008856, and f(Y/Yn) =
7.787(Y/Yn) + 16/116 when (Y/Yn) is less than or equal to 0.008856.

Use the same definitions for f(X/Xn) and f(Z/Zn)

You should be able to work out the reverse transformation!

Good luck