903.2962962962962962962962962963
Its been a million years since I looked at those equations. But it seems likely to be almost trivial. They chose a point where the linear segment would mesh with the cube-rootic (good name, huh?) segment, so they would be continuous and differentiable. That is, at the point
ybreak = (6/29)^3;
the linear segment has value
f_l0 = 903.3*ybreak
And the slope of that segment so dL/dy is trivially 903.3.
Now compute the same parameters at the lower end point of the nonlinear segment.
f_hi = 116*(6/29)- 16
As you see, the L* relation is continuous across the break. What is the slope of the nonlinenar segment at ybreak?
syms Y
vpa(subs(diff(116*Y^(1/3) - 16,Y),Y,ybreak))
And as you should also see, they just chose a point where the two curve segments will be both continuous and differentiable. The choice was actually pretty abitrary, made necessary by the shape of the nonlinear segment at the bottom end. There is nothing really magic about that number in terms of color. Just a choice made so the L* curve would be well behaved.
Edit: Suppose I want to design the transformation into a space I'll call L#a#b#, from XYZ.
We can look at this a different way. We want a curve that has the characteristic shape of
L(Y) = (100+a)*(Y/Y0)^(1/3) - a
over much of the range of Y. It has a cube-root shape in Y/Y0, such that when Y==Y0, so Y/Y0==1, then we get 100 out. The problem is, then Y==0, we would get -a. And we want the curve to pass through 0 at that point. This means we can choose a linear segment that goes from 0 to some break point, I'll call it b.
So at Y/Y0==b, we need the curve to be continuous and differentiable. Assume the linear segment will be of the form
L(Y) = s*(Y/Y0)
It will pass through zero perforce at y==0, and it has a slope of s.
syms a b y Y Y0 s
Llo(y) = s*y % below b
Lhi(y) = (100+a)*(y)^(1/3) - a; % above b
Now we need to choose b and s, such that these two segments are both continuous and differentiable at the point y==b.
sbsol = solve(Llo(b) == Lhi(b), subs(diff(Llo,y) == diff(Lhi,y),y,b),[s,b])
When a == 16, what do we get?
subs(sbsol.s,a,16)
format long g
24389/27
subs(sbsol.b,a,16)
216/24389
Do you see the curve shape choice made in the beginning, of (100+a)*(y/y0)^(1/3)-a, must result in that break point, when a==16? a==16 implies exactly that break point at 0.008856... as well as the coefficient 903.296.
