|
![](/i/fill.gif) |
Mike <pov### [at] aol com> wrote in message news:37F03EC1.A53D7734@aol.com...
> I think I know what you are saying, but the way I'm doing it fits in with
the
> new sample_media function, so each time a supersample is taken it's added
into
> the total emission. Then later it's divided by the total number of
samples for
> that interval.
>
That will cause a problem, though. Uneven sampling (which is the result of
supersampling) needs uneven weights for the averaging process. A simple
sum-and-divide will not give desired results. I know, since I tried it.
;-)
-Nathan
Post a reply to this message
|
![](/i/fill.gif) |