You can drive this with IF=20mA, VF~2.1V at 100% duty to get RLI~2.4, thus power per RLI unit is ~17.5mW/RLI.

You can also drive this with IF=25mA, VF~2.3V at 50% duty to get RLI~3.1*0.85~2.6, thus power is ~11.0mW/RLI.

So because dropping the duty to 50% does not give 50% reduction in perceived brightness (only 15% drop in my experience), you can use this technique to decrease power without decreasing perceived brightness.

What you totally ignore is that "perceived brightness" is not absolute. "perceived brightness" is relative to ambient light.

Thus, if the ambient light doubles, your sign need to souble its brightness as well not in percieved, but in absolute brightness

Let the sunshine in.

Your statements about "perceived brightness" only apply when the ambient light is constant.

Erik