> >
>
> OK ... so here's what I've come up with and it seems to work.
>
> #declare Intensity = function (LD,FD) { 1/(2/((1+(LD/FD))*(1+(LD/FD)))) };
>
>  From your example I used LD=2400 and FD=4.5 and get 142756 and some
> change ;-) ... If I missed something feel free to set me straight as I'd
> like to include this in [3.4.7.9] along with a small narrative.
>
> Jim

That's exactly it.  However I would maybe modify it slightly for presentation.
You can simplify by removing the double reciprocal (just invert the original
equation).  Also, this is true for fade_power of 2, I would leave fade_power as
part of it for completeness and to avoid confusion:

#declare Intensity = function (LD,FD) {pow(1+(LD/FD),FP)/2};

Also I would perhaps call it 'intensity multiplier' or similar.  If you want a
base intensity of 2, you would multiply this factor by 2, etc.

-tgq

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