[Radiance-general] Re: rtrace

Jia Hu hujia06 at gmail.com
Mon Jul 19 11:14:20 PDT 2010


Hello:

If I set -x to 0, and -y to 1, does that mean the program will not exit
until it finishes all calculations for all points?
If I set -x to 5 and -y to 1, does that mean the program will exit after the
calculation for five points (only one line) is finished? In this case, I
have to run this program again to continue the calculation for the left
points?

Generally, if I set -x to 0 and want the program finishes all calculation at
one time, do I need to set -y?

Thanks,

On Mon, Jul 19, 2010 at 2:14 AM, Greg Ward <gregoryjward at gmail.com> wrote:

> Just back from a week in the woods, and playing catch-up....
>
> From: Thomas Bleicher <tbleicher at googlemail.com>
>> Date: July 14, 2010 2:31:46 AM PDT
>>
>>
>> I have to speculate about the connection between "-x" and "-n" here
>> but i think that if you allow more processes (-n) than points in a
>> scan line (-x) then rtrace simply can't start enough processes. This
>> would mean that each point creates it's own process. Greg?
>>
>
> This is more or less correct.  What actually happens is that rays are sent
> to multiple processes, but we need to preserve the first-in/first-out (FIFO)
> order of the rays sent to rtrace, which means buffering up results on the
> output side.  Logically, if you have set -x to 5, then we need to take 5
> ordered input ray origins and directions and run in through N processes
> until we get all 5 results out so we can put them back in the same order.
>  In other words, throwing additional processes over the -x setting buys you
> nothing, because your queue is never deeper than this number.  The default
> of -x 0 (infinite) is the best, because it means rtrace can continue to
> slurp in new rays and buffer them indefinitely, outputting ordered results
> as they become available.
>
> In any case, you aren't likely to see much benefit to the rtrace -n option
> if you are only computing a handful of rays.
>
> Cheers,
> -Greg
>
>
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