OK, let me make some numbers.
Now it is more clear.
For now I am just talking about the camera raw values (X) and the normalize (0 to 1) values (x) and the output of the shaper (y) that translate them to the perceptual space.
Yes if gray is 18% Gray point is 2.47 EV bellow White ( x_{gray}= 0.18 x_{white} \rightarrow y_{white}= -log_{2}(0.18)= 2.47 EV over gray point).
Your arre right the middle gray is not in the middle of the output luminosity range, despite its name (silly me).
And that is more or less what canon camera sets, as you blow up the point where you are measuring when you overexpose something less than 2.5 EV.
We will suppose 12 bit camera (0-4096 numbers).
And a White point near the extreme, say 4090, and will suppose a 18% grey point.
We will suppose a black point in the camera taken data of 5 where the noise is so huge you discern nothing.
Let me calculate the corresponding values.
So at camera X_{white}= 4090, X_{black}= 4.
To normalize the values we us the white point set in the camera (4090) so input values will be normalize from 0 to 1
Lets calculate the camera value of the gray point, as it is 18% of white point it should be:
X_{grey}= 0.18 X{white}= 0.18 \times 4090= 736.2
Let us calculate EV from white to grey point and from black to grey point:
As expected, and now BlackEV:
Much bigger than white, as it uses to be in the examples of filmic.
Roughly 10 EV as one would expect.
Lets calculate the shaper output:
So the shaper output goes from 0 corresponding to the black point value in camera to 1 corresponding to white point.
You only get negative outputs if your raw camera is less than the black point (4) and greater than one if 1 if the raw value goes above the stablished white point (4090).
If you use EV instead, it goes from 0 to DR (about 10) being grey point at -Black_{EV}
Is this OK for now?
Gray point is not at the middle of the logarithmic interval.
