Why does a brush with 50% opacity and 100% flow take 8 individual strokes to reach 100% opacity?

Question

I have a simple round brush with 100% hardness, 50% opacity and 100% flow. All other brush settings are disabled (brush dynamics, transfer, etc). Since it has 50% opacity, I would expect 2 brush strokes to equal 100% (50 + 50) opacity. However, it takes me about 8 brush strokes to reach the same level of opacity that a single brush stroke on a 100% opacity brush makes. What kind of rules does Photoshop use when adding the opacity of overlapping brush strokes? I'm using Photoshop CC.

Answer 1

Basically it blocks 50% of what is left behind, as opposed to being a pure 50% opacity in a additive way. Therefore working in an inverse exponential way towards 99.999...% opacity.

So laid on top of each other:

• 1st stroke: 50%
• 2nd stroke: 75% (50% + 50% of 50%)
• 3rd stroke: 87.5% (75% + 50% of 25%)
• 4th stroke: 93.75% (87.5 + 50% of 12.5%)
• 5th stroke: 96.875% (93.75% + 50% of 6.25%)
• 6th stroke: 98.4375% (96.875% + 50% of 3.125%)
• 7th stroke: 99.21875% (98.4375% + 50% of 1.5625%)
• 8th stroke: 99.609375% (99.21875% + 50% of 0.78125%)

etc...

Answer 2

Each stroke is moving 50% from the current color towards the brush color. The formula would be 100% * (1 - (brush opacity ^ number of strokes)). So going from white to black, you will have:

1. 50% gray
2. 75% gray
3. 87,5% gray
4. 93,75% gray
5. 96,875% gray
6. 98,4375% gray

...etc, slowly moving towards black.

I.e, you'll never actually truly reach full opacity, but at some point it will round off to 100% anyway.

Answer 3

I suspect this has to do with the limits of transparency layers. You say that it took 8 x 50% transparency to get 0% transparency.

If you have 50% transparency, then 50% of the background colour should be visible through the top layer. If you apply 50% transparency again, then 50% of that NEW background layer should be visible = 50% x 50% = 25% original background.

Repeating 8 times, we get (0.5)^8 = 1/256. Oh, that's a pretty suspicious number!

So my guess is that you have an effective 8-bit limit - you get grades of transparency from 0/255 (0%) to 255/255 (100%), and 1/256 gets rounded down to 0/255 = 0% transparency.

Hence, it takes 8 applications of 50% to reach 0% because:

1. Transparency is multiplicative, not additive
2. It takes 8 applications to reach the lower limit of colour/transparency resolution (which appears to be based on some sort of 8-bit limit)