CGAL users discussion list

[Oleg <>]

Hi Daisy! That's definitely a possible way to eliminate the points
that fall inside the other figure. Using locate() seems to be an
easier way, but I guess I'm doing something wrong and haven't managed
to get it working yet.

Oleg

Wednesday, April 11, 2012, 20:14, you wrote:

> Hi,
>     In my oppinion, you can have a try with the deviation.
>     Let *n* be a unit normal vector from the origin to the plane PI, and M
> be a point in 3D space. Make a line segment MQ from the M to the plane PI,
> and let MQ be perpendicular to PI at point Q. Then the projection from the
> vector QM to the normal vector n is called deviation between M and PI.
> http://cgal-discuss.949826.n4.nabble.com/file/n4548847/4.bmp 
> Obviously, we can judge the M's position, on one side of the PI, or the
> other. Especially, if we use the normalized plane equation, we can compute
> the deviation easier.
>    I am not sure if I can help you. Good luck.

> _______________________________________________
> If you reply to this email, your message will be added to the discussion below:
> http://cgal-discuss.949826.n4.nabble.com/Determining-if-a-point-is-inside-a-polyhedron-tp4523460p4548847.html

> To unsubscribe from Determining if a point is inside a polyhedron,
> visit
>

---

View this message in context:

Re[2]: Determining if a point is inside a polyhedron
Sent from the cgal-discuss mailing list archive at Nabble.com.