I'm again ;)

>   5.1) If there are only reflecting objects and no refraction then the
> amount of nodes doesn't explode, although we still have many of them.
>   Now here kicks in the max trace level. Let's just take the default of 5
> (althought this may not be enough if the objects are very reflective and
> they reflect each other).
>   The number of nodes in the second recursion level will then be:
>   nodes2 = nodes1*0.2
>
>   Let's suppose that about 10% of the reflected rays hit another reflecting
> object. This means that the number of nodes in the third recursion will be:
>   nodes3 = nodes2*0.1
>
>   The general formula for all nodes in recursion level 4 and subsequent will
> then be:
>   nodes4 = nodes3*(1.1^(reclevel-3)) = nodes3*1.21
>
>   The total number of nodes for 5 recursion levels will then be:
>   total = nodes1+nodes2+nodes3+nodes4
>
>   Now, nodes1 depends on the resolution of the image. What resolution could
> we use? I think we can't restrict ourselves to a too low resolution since
> that would limit too much. I think that a resolution of 1024x768 could be
> good to effectively test the memory consumption.
>   So: nodes1 = 1024*768+1024*768*0.25*8 = 2359296
>       nodes2 = 2359296*0.2 = 471859
>       nodes3 = 471859*0.1 = 47185
>       nodes4 = 47185*1.21 = 57093
>       total = 2935433 nodes

Why you need for the 5th recursion level five nodes? In the rendering you only
display the 5th (last) level...in't it possible to store the data in node1,
calculate node2 and add theese values to node1? I think it is possible with only
two nodes..one to store values and one (a temp node) to calculate new values.

Paul

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