Triangulation patterns in .ifc file format using coordinates and indexes - python

I've posted this in another forum as well due to the mathematical nature of the issue:
forum post
I have an .ifc file in which the raw data exported describes a wall in the xy plane by a set of coordinates and their corresponding indexes according to the link explanation:
Explanation
I have a txt where the data is divided into the coordinates in xyz space, then indexes and some other data.
I was hoping that someone can help me understand how to link the indexes to their corresponding coordinates. There are 164 coordinate pairs and 324 index pairs so it doesn't make sense to me that each index relates to only 1 coordinate pair.
The goal is to establish a relationship between indexes and coordinates such that this type of data can output the wall thickness, which is in this case '10'. I was thinking that (according to the link above) by taking the first triangle described, it should describe the edge of the wall in 3D and therefore give us one of its sides as the shortest segment in the wall which is the thickness.
I received an answer in the mentioned forum post, that I should
"...expanding out each coordinate in terms of X's, Y's, and Z's [instead of (X,Y,Z) triples) and then use every index triple to get the actual coordinate for the individual coordinate instead of one triple.
So for example you have X[], Y[] and Z[] and you have an index (a,b,c) then you find X[a], Y[b], and Z[c] not Point(a,b,c)... "
I didn't quite understand this explanation, and would appreciate any help or further explanation in order to achieve my goal.
Thank you

Let's start with the cordinates (IfcCartesianPointList3D): each one is a triplet, resulting in a Point with (x,y,z) coordinates.
Then the IfcTriangulatedFaceSet uses indices to construct triangles. It has 2 indexing modes: direct and indirect via PnIndex. The indexing mode is determined by the existence of an array for PnIndex (attribute number 5). Take note that I call these variants direct and indirect - they aren't mentioned that way in the IFC documentation.
Direct indexing
PnIndex is not set. Lets look at an (simple and constructed) example:
#100=IFCCARTESIANPOINTLIST(((0,0,0),(1,0,0),(1,1,0),(0,1,0)));
#101=IFCTRIANGULATEDFACESET(
/*reference to the points*/ #100,
/*no normals*/ $,
/*no indication if closed or open*/ $,
/*coordinate indices*/ ((1,2,3),(1,3,4)),
/*no PnIndex*/ ());
This describes a square lying in the x-y-plane. Each entry in attribute CoordIndex is a triplet giving a one-based index into a point in the IfcCartesianPointList. This means there are two triangles constructed from the following points:
(0,0,0) (1,0,0) (1,1,0)
(0,0,0) (1,1,0) (0,1,0)
Indirect indexing
Lets build further on the previous example:
#100=IFCCARTESIANPOINTLIST(((0,0,0),(1,0,0),(1,1,0),(0,1,0)));
#101=IFCTRIANGULATEDFACESET(
/*reference to the points*/ #100,
/*no normals*/ $,
/*no indication if closed or open*/ $,
/*coordinate indices*/ ((1,2,3),(1,3,4)),
/*PnIndex*/ (2,3,4,1));
This time there is PnIndex set. It adds a level of indirection to access the points. Triplets from CoordIndex point into PnIndex (1-based). The value found in PnIndex is then used to access the IfcCartesianPointList.
So for the first triangle we have: (1,2,3) in CoordIndex. These point to 2, 3 and 4 in PnIndex. These result in the following points from the point list: (1,0,0) (1,1,0) (0,1,0)
Repeating the procudure for the second triangle (1,3,4) we get values 2, 4, 1 from PnIndex and the following points: (1,0,0) (0,1,0) (0,0,0)
It is again a square, but this time with a different triangulation.
Now if you want to know your wall thickness you will need to calculate the extents from the resulting geometry. If your wall is aligned with the coordinate system axes this is easy (get the difference between the smallest and largest X, Y and Z). If it is not, you might need to transform the points or look further into 3D-extent calculations (my knowledge ends there).

In a triangulation it's roughly num of triangles = 2 * num of vertices.
A wall (e.g. a rectangle) may be described by two triangles that share an edge and the two vertices of this edge.
Instead of describing the whole model triangle by triangle, each with its three vertices, or edge by edge, it's cheaper, avoids repeating vertex data, to set an index for each vertex and set a triangle by the three indices of its vertices. This is usually called "indexed rendering".

Related

locating cell containing point in irregular 3D grid

I have an irregular 3D grid which looks something like this:
Typical dimensions of the grid are 100/100/100 cells. Each cell is spatially defined by the coords of the 8 corner nodes. The 4 vertices of the each face of a cell are not necessarily co-planar, so I represent each face as a pair of triangles and thus a cell as a polyhedron consisting of 12 triangles (2 per face). I am trying to locate the IJK index of the cell that contains an XYZ point using Python. I bisect sequentially the cell range in the I, J and K directions and test which half of the grid the point lies using the method described here Testing whether a 3D point is inside a 3D polyhedron to locate the point. Unfortunately, this does not work in some cases. In the above figure, point A is physically outside the grid but inside the current bisection range (defined by the brown dotted lines) while point B is inside the grid but outside the current range. I think the reason for this is that triangles representing the faces of the cells within the current range (eg the large brown triangles in the figure) are not co-planar with the triangles that comprise the individual cell faces within that range (eg those shaded yellow, blue etc). I have tried to show this in 2D below:
The current bisection range is shown by the brown dotted line and brown vertices. Initially, the red point is within the current range. We bisect in the X direction (bisection 1) and the red point is within the current range (dotted brown line) so we discard the right half. We now bisect in the Y direction (bisection 2) and the red point is outisde this range so we discard the top half. We eventually arrive at the final step when we have a single index in each of the I & J directions. As shown here, this places the red point in the wrong cell.
Would appreciate any suggestions for an alternative algorithm to the one I am currently trying to implement. Stepping back, I am actually interested in calculating the faces within the grid intersected by a series of line segments, so am using the "point in a polyhedron" method as an intermediate step. I looked at geomdl which could represent each face as a NURBS object but does not seem to implement intersection between a ray and a NURBS object. I also had a quick look at the Python bindings to CGAL but that looked like a massive learning curve to climb, so put that aside. Thanks in advance!

Coordinates Python

I am facing with the sorting airfoil coordinates. In particular given a set of coordinates, which are not sorted, I have to sorted them starting from the trailing edge upper surface. Here I report the code that I have developed but as you can see, the starting point do not match with what I suppose, moreover exist several oscillations as you can see in the reported figure (and a detail, in blue the starting point after the sort).
Can someone suggest me what I miss? How can I do?
Thanks you in advance.
def sort_airfoil(points):
x0 = np.mean(-points[:,1])
y0 = np.mean(points[:,2])
r = np.sqrt((-points[:,1]-x0)**2 + (points[:,2]-y0)**2)
tempx=-points[:,1]
xmax=np.max(tempx)
ind_max=np.where(tempx==xmax)
ymax=np.max(points[ind_max,2])
ind_max_t=np.where((tempx>0.95*xmax) & (tempx<xmax))
ymax_t=points[ind_max_t,2]
ymin=np.min(ymax_t)
indx_temp=np.where(points[:,2]==ymin)
xmin=np.max(tempx[indx_temp])
xmed=(xmin+xmax)/2
ymed=(ymin+ymax)/2
print(x0,y0)
print(xmin,ymin)
print((xmin+xmax)/2, (ymin+ymax)/2)
angle0=np.arctan2((ymed-y0),(xmed-x0))
print("angle", angle0)
angles = np.where((points[:,2]-y0) > 0, np.arccos((-points[:,1]-x0)/r), 2*np.pi-np.arccos((-points[:,1]-x0)/r))
angles=angles-angle0
for i in range(len(angles)):
if angles[i]<0:
angles[i]=angles[i]+2*np.pi
elif angles[i]>2*np.pi:
angles[i]=angles[i]-2*np.pi
mask = np.argsort(angles)
x_sorted = points[mask,1]
y_sorted = points[mask,2]
points_new=np.zeros([len(points), 3])
points_new[:,0]=points[:,0]
points_new[:,1]=x_sorted
points_new[:,2]=y_sorted
return points_new
The issue comes from the algorithm itself: it only work when the points form a convex polygon. However, the shape is concave.
More specifically, the first sorted points (and the last ones) form a zigzag-shaped lines because there is two sets of points (green arrows) interleaving with growing angles (red arrow) from the median point (red line).
Note the points are horizontally flipped on the gathered point from the question. Thus the points are sorted clockwise.
One simple solution is to split horizontally the shape in many set of point (eg. 10 set) so that each set form a convex shape. Then, the parts can be merged to form the final shape. The merge consists in finding the points at the "edge" of each locally-sorted set of points (parts) and reorder the partially sorted array of points consequently.
More specifically, the points of each part are split in 2 sub-sets: the upper ones and the lower ones. You can find them easily by selecting the 2 left-most points of a right part with the right-most points of a left part. The 2 top-most points needs to be connected each other and the same for the 2 bottom-most points. Thus, the sequence of the two upper sets of points needs to be reordered so they are contiguous and the same for the lower part.
Here is an example:
Note that if you are unsure about how to split the points in many parts so that each one form a convex-shaped sets of points, then you can: split the shape in n parts, check if the set of points form a convex shape by computing a convex hull (eg. using a Graham scan) and split evenly the parts that are concave (recursively). This is quite expensive, but more robust.

Implementing collisions using the Separating axis theorem and Pygame [duplicate]

This is what I am currently doing:
Creating 4 axis that are perpendicular to 4 edges of 2 rectangles. Since they are rectangles I do not need to generate an axis (normal) per edge.
I then loop over my 4 axes.
So for each axis:
I get the projection of every corner of a rectangle on to the axis.
There are 2 lists (arrays) containing those projections. One for each rectangle.
I then get the dot product of each projection and the axis. This returns a scalar value
that can be used to to determine the min and max.
Now the 2 lists contain scalars and not vectors. I sort the lists so I can easily select the min and max values. If the min of box B >= the max of box A OR the max of box B <= the min of box A then there is no collision on that axis and no collision between the objects.
At this point the function finishes and the loop breaks.
If those conditions are never met for all the axis then we have a collision
I hope this was the correct way of doing it.
The python code itself can be found here http://pastebin.com/vNFP3mAb
Also:
http://www.gamedev.net/page/reference/index.html/_/reference/programming/game-programming/collision-detection/2d-rotated-rectangle-collision-r2604
The problem i was having is that the code above does not work. It always detects a a collision even where there is not a collision. What i typed out is exactly what the code is doing. If I am missing any steps or just not understanding how SAT works please let me know.
In general it is necessary to carry out the steps outlined in the Question to determine if the rectangles "collide" (intersect), noting as the OP does that we can break (with a conclusion of non-intersection) as soon as a separating axis is found.
There are a couple of simple ways to "optimize" in the sense of providing chances for earlier exits. The practical value of these depends on the distribution of rectangles being checked, but both are easily incorporated in the existing framework.
(1) Bounding Circle Check
One quick way to prove non-intersection is by showing the bounding circles of the two rectangles do not intersect. The bounding circle of a rectangle shares its center, the midpoint of either diagonal, and has diameter equal to the length of either diagonal. If the distance between the two centers exceeds the sum of the two circles' radii, then the circles do not intersect. Thus the rectangles also cannot intersect. If the purpose was to find an axis of separation, we haven't accomplished that yet. However if we only want to know if the rectangles "collide", this allows an early exit.
(2) Vertex of one rectangle inside the other
The projection of a vertex of one rectangle on axes parallel to the other rectangle's edges provides enough information to detect when that vertex is inside the other rectangle. This check is especially easy when the latter rectangle has been translated and unrotated to the origin (with edges parallel to the ordinary axes). If it happens that a vertex of one rectangle is inside the other, the rectangles obviously intersect. Of course this is a sufficient condition for intersection, not a necessary one. But it allows for an early exit with a conclusion of intersection (and of course without finding an axis of separation because none will exist).
I see two things wrong. First, the projection should simply be the dot product of a vertex with the axis. What you're doing is way too complicated. Second, the way you get your axis is incorrect. You write:
Axis1 = [ -(A_TR[0] - A_TL[0]),
A_TR[1] - A_TL[1] ]
Where it should read:
Axis1 = [ -(A_TR[1] - A_TL[1]),
A_TR[0] - A_TL[0] ]
The difference is coordinates does give you a vector, but to get the perpendicular you need to exchange the x and y values and negate one of them.
Hope that helps.
EDIT Found another bug
In this code:
if not ( B_Scalars[0] <= A_Scalars[3] or B_Scalars[3] >= A_Scalars[0] ):
#no overlap so no collision
return 0
That should read:
if not ( B_Scalars[3] <= A_Scalars[0] or A_Scalars[3] <= B_Scalars[0] ):
Sort gives you a list increasing in value. So [1,2,3,4] and [10,11,12,13] do not overlap because the minimum of the later is greater than the maximum of the former. The second comparison is for when the input sets are swapped.

Area of polygon with list of (x,y) coordinates

It might seem a bit odd that I am asking for python code to calculate the area of a polygon with a list of (x,y) coordinates given that there have been solutions offered in stackoverflow in the past. However, I have found that all the solutions provided are sensitive to the order of the list of (x,y) coordinates given. For example, with the code below to find an area of a polygon:
def area(p):
return 0.5 * abs(sum(x0*y1 - x1*y0
for ((x0, y0), (x1, y1)) in segments(p)))
def segments(p):
return zip(p, p[1:] + [p[0]])
coordinates1 = [(0.5,0.5), (1.5,0.5), (0.5,1.5), (1.5,1.5)]
coordinates2 = [(0.5,0.5), (1.5,0.5), (1.5,1.5), (0.5,1.5)]
print "coordinates1", area(coordinates1)
print "coordinates2", area(coordinates2)
This returns
coordinates1 0.0
coordinates2 1.0 #This is the correct area
For the same set of coordinates but with a different order. How would I correct this in order to get the area of the non-intersecting full polygon with a list of random (x,y) coordinates that I want to make into a non-intersecting polygon?
EDIT: I realise now that there can be multiple non-intersecting polygons from a set of coodinates. Basically I am using scipy.spatial.Voronoi to create Voronoi cells and I wish to calculate the area of the cells once I've fed the coordinates to the scipy Voronoi function - unfortunately the function doesn't always output the coordinates in the order that will allow me to calculate the correct area.
Several non-intersecting polygons can be created from a random list of coordinates (depending on its order), and each polygon will have a different area, so it is essential that you specify the order of the coordinates to build the polygon (see attached picture for an example).
The Voronoi cells are convex, so that the polygon is unambiguously defined.
You can compute the convex hull of the points, but as there are no reflex vertices to be removed, the procedure is simpler.
1) sort the points by increasing abscissa; in case of ties, sort on ordinates (this is a lexicographical ordering);
2) consider the straight line from the first point to the last and split the point sequence in a left and a right subsequence (with respect to the line);
3) the requested polygon is the concatenation of the left subsequence and the right one, reversed.

Separating Axis Theorem and Python

This is what I am currently doing:
Creating 4 axis that are perpendicular to 4 edges of 2 rectangles. Since they are rectangles I do not need to generate an axis (normal) per edge.
I then loop over my 4 axes.
So for each axis:
I get the projection of every corner of a rectangle on to the axis.
There are 2 lists (arrays) containing those projections. One for each rectangle.
I then get the dot product of each projection and the axis. This returns a scalar value
that can be used to to determine the min and max.
Now the 2 lists contain scalars and not vectors. I sort the lists so I can easily select the min and max values. If the min of box B >= the max of box A OR the max of box B <= the min of box A then there is no collision on that axis and no collision between the objects.
At this point the function finishes and the loop breaks.
If those conditions are never met for all the axis then we have a collision
I hope this was the correct way of doing it.
The python code itself can be found here http://pastebin.com/vNFP3mAb
Also:
http://www.gamedev.net/page/reference/index.html/_/reference/programming/game-programming/collision-detection/2d-rotated-rectangle-collision-r2604
The problem i was having is that the code above does not work. It always detects a a collision even where there is not a collision. What i typed out is exactly what the code is doing. If I am missing any steps or just not understanding how SAT works please let me know.
In general it is necessary to carry out the steps outlined in the Question to determine if the rectangles "collide" (intersect), noting as the OP does that we can break (with a conclusion of non-intersection) as soon as a separating axis is found.
There are a couple of simple ways to "optimize" in the sense of providing chances for earlier exits. The practical value of these depends on the distribution of rectangles being checked, but both are easily incorporated in the existing framework.
(1) Bounding Circle Check
One quick way to prove non-intersection is by showing the bounding circles of the two rectangles do not intersect. The bounding circle of a rectangle shares its center, the midpoint of either diagonal, and has diameter equal to the length of either diagonal. If the distance between the two centers exceeds the sum of the two circles' radii, then the circles do not intersect. Thus the rectangles also cannot intersect. If the purpose was to find an axis of separation, we haven't accomplished that yet. However if we only want to know if the rectangles "collide", this allows an early exit.
(2) Vertex of one rectangle inside the other
The projection of a vertex of one rectangle on axes parallel to the other rectangle's edges provides enough information to detect when that vertex is inside the other rectangle. This check is especially easy when the latter rectangle has been translated and unrotated to the origin (with edges parallel to the ordinary axes). If it happens that a vertex of one rectangle is inside the other, the rectangles obviously intersect. Of course this is a sufficient condition for intersection, not a necessary one. But it allows for an early exit with a conclusion of intersection (and of course without finding an axis of separation because none will exist).
I see two things wrong. First, the projection should simply be the dot product of a vertex with the axis. What you're doing is way too complicated. Second, the way you get your axis is incorrect. You write:
Axis1 = [ -(A_TR[0] - A_TL[0]),
A_TR[1] - A_TL[1] ]
Where it should read:
Axis1 = [ -(A_TR[1] - A_TL[1]),
A_TR[0] - A_TL[0] ]
The difference is coordinates does give you a vector, but to get the perpendicular you need to exchange the x and y values and negate one of them.
Hope that helps.
EDIT Found another bug
In this code:
if not ( B_Scalars[0] <= A_Scalars[3] or B_Scalars[3] >= A_Scalars[0] ):
#no overlap so no collision
return 0
That should read:
if not ( B_Scalars[3] <= A_Scalars[0] or A_Scalars[3] <= B_Scalars[0] ):
Sort gives you a list increasing in value. So [1,2,3,4] and [10,11,12,13] do not overlap because the minimum of the later is greater than the maximum of the former. The second comparison is for when the input sets are swapped.

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