I'm still not following. Where do you get 4 stops from?
Compare the formulas for determining
film speed and
standard output sensitivity (SOS) in the digital world. The first formula says that if a certain brightness H
m creates a density D=0.1 (assuming development to standard contrast), you have reached film sensitivity S
f = 0.8 / H
m. The second formula says that if you reach medium gray exposure with a certain brightness level H
SOS, your SOS speed calculates as S
SOS = 10/H
SOS. If these two speed numbers are supposed to yield comparable numbers for comparable sensitivity, a medium gray exposure should be 10/0.8 times stronger than the exposure to reach D=0.1. The number 10/0.8 = 12.5 equals 3.64 stops.
But an ideal characteristic curves doesn't drop vertically to D=0.0 below this
point 'm', it should continue more or less as a sloped line. How long would such a line extend to the left? We know its slope is 0.8/1.3, and we know it drops by another 0.1, so a simple calculation tells you that it crosses the D=0.0 line 0.1 * 1.3 / 0.8 = 0.16 log
2(H) left of point 'm'. All together, and with a perfectly straight characteristic curve, this gives you 3.64 + 0.16 = 3.8 stops below 18% gray. A characteristic curve with a pronounced toe will be flatter near that point 'm', which means it will cross the D=0.0 line even further left, giving you yet more area with barely discernible shadow detail.