Daniel,
> According to your description of the algorithm only edges with incident
> faces containing the vertex will be considered (which sounded reasonable so
> far). Do you need to update your algorithm description there?
>
>
Those edges will NOT be considered. They will be deemed ineligible. I
Rakshika,
According to your description of the algorithm only edges with incident
faces containing the vertex will be considered (which sounded reasonable so
far). Do you need to update your algorithm description there?
Additionally, the line intersects the edge only in a 2D projection but
proba
Daniel,
Although the triangle 3 does not contain V, triangle 1 does. And we get to
triangle 1 through triangle 3. That is, for any edge we first check if it
contains the vertex V, or if the line from V to the edge intersects any of
the other two edges of the incident face. When the line from V to
E1 incident faces don't contain V. So, why do you consider it?
Am 20.06.2016 20:05 schrieb "Rakshika Bagavathy" <
rakshika.bagava...@gmail.com>:
> Daniel,
>
> You had asked how one line would intersect another. So i made a simple
> drawing, which i've attached here.
> Here, we need to get the clo
Daniel,
You had asked how one line would intersect another. So i made a simple
drawing, which i've attached here.
Here, we need to get the closest non-crossing edge for vertex V. And when
we are considering edge E1, the perpendicular line from V to E intersects
the edges of triangles 1, 2, and 3.
On Jun 20, 2016 12:27 PM, "Inderpreet Singh" wrote:
>
> Your PR is now merged. So you have two successful merges.
Ah! Thank you :)
I am finished with my schema, and will be sending another PR soon.
If current code of branch seems good enough, I would like you to merge
GSOC15-merged branch with m
On Thu, Jun 16, 2016 at 12:31 PM, Param Hanji
wrote:
> Hi,
>
> I tried to call boolweave and boolfinal methods using segments generated
> on the GPU. I'm pretty sure the weaving process works fine, but boolfinal()
> not so much. It returns an empty partition for every pixel, which I
> confirmed u
