Check if a point is inside an arbitrary hexahedron - geometry

Question

I am working on a 3D finite element code, where i face the following problem:
If I take an arbitrary point (say x), how do I figure out, which element it is in?
This can be simplified to: How do I check if an arbitrary point (x) lies inside or outside of an (hexahedral) element?
What I already found:
Limited to cubes: How to determine a point is inside or outside a cube?
Limited to rectangular shapes: https://math.stackexchange.com/questions/1472049/check-if-a-point-is-inside-a-rectangular-shaped-area-3d
Contrary to the two approaches above, my problem does not assume right angles nor parallel faces.
Problem sketch:
Notation: (again: though the sketch shows a regular shape, our hexahedron is assumed to be of general shape)
8-node hexahedron topology, nodes: 0,..,7
axis: r,s,t
t
|
4--------|-------------7
/| | /|
/ | | / |
/ | | / |
/ | | / |
/ | | / |
/ | | / |
5----------------------6 |
| | | | |
| | o------|---------s
| | / | |
| 0------/--------|------3
| / / | /
| / / | /
| / / | /
| / / | /
| / r | /
|/ |/
1----------------------2
Data that we have available:
coordinates of the nodes (vectors P0 to P7)
coordinates of the point we want to check (lets say Px)
Additionaly we assume the nodes are ordered as sketched above.
My approach/solution so far:
calculate the surface (outward) normal vectors
Use cross products, eg. for the r_pos_normal_vec (pointing out of the plane)
r_pos_normvec = (P2-P1) x (P5-P1)
and for the r_neg_normal_vec
r_neg_normvec = (P4-P0) x (P3-P0)
similarly for the s and t directions
check two opposite corner nodes (I chose node0 and node 6)
For node0
calculate vector from P0 to Px:
P0x = Px - P0
calculate inner prodcut of P0x and surfaces adjacent to node 0
<P0x, r_neg_normal_vec>
<P0x, s_neg_normal_vec>
<P0x, t_neg_normal_vec>
For node1
same scheme as for node 0, whereas P1 instead of P0 and the positive counterparts of the normal vectors are used
Iff all 6 (3 from node0 and 3 from node1) inner products result in negative value -> the point is inside the hexahedron.
Question:
I implemented the functionality described above in my code and ran some tests.
It seems to work, from the math side I am quite confident.
Please discuss my approach, I am happy for any hints/clues/recommendations/bug fixes ...
Is there some way to make this faster?
Alternative solutions?
Note:
To speed up the algorithm a box check can be done first:
Construct a rectangular box around the hexahedron:
Get the min and max values of the node coordinates in each direction.
If the point to check (x) is outside this (larger) box, it cannot be inside the hexahedron.

Answer 1

For any convex polyhedron, establish the implicit equations of the faces (f.i. plane by three points), of the form ax+by+cz+d=0.
When you plug the coordinates of a known point inside the volume (such as the center) in the expression ax+by+cz+d, you will get a set of signs. An arbitrary point is inside if it yields the same signs.
Update:
For maximum performance, you can consider also using an axis-aligned bounding box for quick rejection. This only makes sense if many of the points are outside. Make sur to use a shortcut evaluation so that early rejection can happen.
Note that a rejection test such as X<Xmin is nothing but the above sign test against the plane of equation X-Xmin=0.

Answer 2

I personnally prefer your method, however there also is a way to approach the problem if the hexahedral restricted to parallelepiped. So you can transfer the coordinate of P in the frame $(0; e_1; e_2; e_3)$ to $(P_0, P_0P_1,P_0P_3,P_0P_4)$. We call it $(a,b,c)$, then the point is in that parallelepiped if $a,b,c > 0 a \in [0,1], a+b+c \in [0,1]$.

Answer 3

Because you mentioned that you want to be able to handle arbitrary hexahedrons, I think that your process might be improved if you adjust your s, r, and t normals to account for having slightly warped faces. I would do this by making the following change to r normals (and similar for s and t):
r_pos_normvec = (P6-P1) x (P5-P2)
r_neg_normvec = (P7-P0) x (P4-P3)
This would be important for a case where you shifted node 6 towards node 7 (say 0.9xP6) and had a point at 0.95xP6. Without the warping correction, I believe you would erroneously determine the point as inside the hexahedron.

Answer 4

Here is a python example :
def point_is_in_hexa(point,centers,normals):
vect=[]
prod=[]
for i in range(6):
vect.append(point-centers[i])
vect= np.array(vect)
for i in range(6):
prod.append(np.dot(vect[i],normals[i]))
prod=np.array(prod)
if all(prod <= 0):
is_in_hexa = 1
else:
is_in_hexa = -1
return is_in_hexa
https://github.com/fgomez03/hexacheck