It was, indeed, just a problem of printing. Numbers are big so it didn't
fit in %d. I used %lld now and I got acceptable results.
After looking a bit over viewweight again, there are two concepts that I
would like to know more about.
How does hypersampling work, and what does the variable hypersample exactly
contain?
The region struct seems to have a line of sight value that is expressed in
percentage and is then converted to a factor. Its then multiplied together
with volume and density to obtain the final weight value. What exactly is
line of sight?
Mario.
2017-07-26 18:14 GMT+02:00 Christopher Sean Morrison <brl...@mac.com>:
>
>
> On Jul 26, 2017, at 10:59 AM, Mario Meissner <mr.rash....@gmail.com>
> wrote:
>
> Sean,
>
> Today I started modifying rtexample.c so that it simulates a ray crossing
> several regions (tried with the goblet example) and returning the mass it
> saw as it crossed each region. I assumed most stuff with predefined
> constant values for now.
>
>
> Sounds good!
>
> I wonder why the last result is negative? Segment distance seems to be the
> cause, although OutHIT dist = 10134 and InHIT dist = 10114, so my
> subtraction should be returning a positive result. Will keep working on it.
>
>
> At first glance, notice how the length and volume values are the same.
> This implies maybe a data type printing problem. What are the types of
> those variables? I suspect you need %zd or %lld instead of %d. I would
> definitely figure out what is going on there before moving on. You
> obviously need non-negative lengths. :)
>
> Cheers!
> Sean
>
>
>
> Mario.
>
> 2017-07-26 15:39 GMT+02:00 Christopher Sean Morrison <brl...@mac.com>:
>
>>
>> I think I can follow you, but I'm not sure how to put contributions
>> together. In the blade example you seem to be taking the mean value between
>> them. I'll assume that's the case.
>> Attached goes a simple box with two vectors. Should the density at point
>> a be 5?
>>
>>
>> Let’s see. The gist of the original formula (which had a segment error
>> in the denominator, looking back) is to calculate the density contribution
>> from VA, the density contribution of VB, and then take their average (i.e.,
>> assume simple linear contributions from each). Simple looking at your
>> gridding without any equations, we can see that:
>>
>> contrib(a, VA) is 3.5
>> contrib(b, VB) is 6.5
>>
>> Adding them up and taking the average is thus density 5. So yes. ;)
>>
>> Likewise, would point b have a density of 6 then?
>>
>>
>> Yes. 8 + 4 / 2.
>>
>> I'm not sure if I calculated the contributions correctly either, but this
>> is the way that would make the most sense to me.
>>
>>
>> I think it’s as good a starting point as any and is relatively easy to
>> implement.
>>
>> That said, note that this method does have a potential flaw in that we’re
>> treating the density contributions uniformly when in reality, they should
>> probably be weighted by distance to the vector.
>>
>> In your example, they’re equidistant, so it works out. But consider the
>> implication of box that is 100 times taller, for example. Instead of 4 x
>> 4, it’s 4 x 400. Point b should be a value far closer to 8 than 4,
>> certainly not 6. That’s a problem that can be dealt with later, but
>> something to keep in mind.
>>
>> Cheers!
>> Sean
>>
>>
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>>
> <rtexample.c>
>
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