I am creating a script to generate cylinders in a 3D space, however, I would like for them to not occupy the same region in space (avoid overlapping).
The cylinders are defined by a start and end point, and all have a fixed radius.
I am storing the existing cylinder in an array called listOfCylinders which is an nDim array of shape (nCylinders, 2Points [start, end], {x,y,z} coordinates of each point)
I was able to cook up:
def detect_overlap(new_start, new_end, listOfCylinders):
starts = listOfCylinders[:, 0]
ends = listOfCylinders[:, 1]
radius = 0.1
# Calculate the distance between the new cylinder and all the existing cylinders
dists = np.linalg.norm(np.cross(new_end - new_start, starts - new_start), axis=1) / np.linalg.norm(new_end - new_start)
# Check if any of the distances are less than the sum of the radii
if np.any(dists < (2*radius)):
return True
# If no overlap or intersection is found, return False
return False
But this is not accountting for situations where there is lateral overlaping.
Does anyone have a good algorithm for this?
Best Regards
WLOG one of the cylinders is vertical (otherwise rotate space). If you look at the projections of the apparent outline onto XY, you see a circle and a rectangle ended with ellipses. (For simplicity of the equations, you can also make the second cylindre parallel to XZ.)
If these 2D shapes do not overlap, your are done. Anyway, the intersection of a circle and an ellipse leads to a quartic equation.
You can repeat this process, exchanging the roles of the two cylinders. This gives a sufficient condition of non-overlap. Unfortunately, I am not sure it is necessary, though there is a direct connection to the plane separation theorem.
For a numerical approach, you can proceed as follows:
move the cylindre in the canonical position;
generate rectangles on the oblique cylindre, by rotation around the axis and using an angular parameter;
for all sides of the rectangles, detect interference with the cylindre (this involves a system of a quadratic inequation and two linear ones, which is quite tractable);
sample the angular parameter densely enough to check for no valid intersection.
I guess that a complete analytical solution is possible, but complex, and might anyway lead to equations that need to be solved numerically.
Related
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.
I'm a newbie to computer graphics so I apologize if some of my language is inexact or the question misses something basic.
Is it possible to calculate face normals correctly, given a list of vertices, and a list of faces like this:
v1: x_1, y_1, z_1
v2: x_2, y_2, z_2
...
v_n: x_n, y_n, z_n
f1: v1,v2,v3
f2: v4,v2,v5
...
f_m: v_j, v_k, v_l
Each x_i, y_i , z_i specifies the vertices position in 3d space (but isn't neccesarily a vector)
Each f_i contains the indices of the three vertices specifying it.
I understand that you can use the cross product of two sides of a face to get a normal, but the direction of that normal depends on the order and choice of sides (from what I understand).
Given this is the only data I have is it possible to correctly determine the direction of the normals? or is it possible to determine them consistently atleast? (all normals may be pointing in the wrong direction?)
In general there is no way to assign normal "consistently" all over a set of 3d faces... consider as an example the famous Möbius strip...
You will notice that if you start walking on it after one loop you get to the same point but on the opposite side. In other words this strip doesn't have two faces, but only one. If you build such a shape with a strip of triangles of course there's no way to assign normals in a consistent way and you'll necessarily end up having two adjacent triangles with normals pointing in opposite directions.
That said, if your collection of triangles is indeed orientable (i.e. there actually exist a consistent normal assignment) a solution is to start from one triangle and then propagate to neighbors like in a flood-fill algorithm. For example in Python it would look something like:
active = [triangles[0]]
oriented = set([triangles[0]])
while active:
next_active = []
for tri in active:
for other in neighbors(tri):
if other not in oriented:
if not agree(tri, other):
flip(other)
oriented.add(other)
next_active.append(other)
active = next_active
In CG its done by polygon winding rule. That means all the faces are defined so the points are in CW (or CCW) order when looked on the face directly. Then using cross product will lead to consistent normals.
However many meshes out there does not comply the winding rule (some faces are CW others CCW not all the same) and for those its a problem. There are two approaches I know of:
for simple shapes (not too much concave)
the sign of dot product of your face_normal and face_center-cube_center will tell you if the normal points inside or outside of the object.
if ( dot( face_normal , face_center-cube_center ) >= 0.0 ) normal_points_out
You can even use any point of face instead of the face center too. Anyway for more complex concave shapes this will not work correctly.
test if point above face is inside or not
simply displace center of face by some small distance (not too big) in normal direction and then test if the point is inside polygonal mesh or not:
if ( !inside( face_center+0.001*face_normal ) ) normal_points_out
to check if point is inside or not you can use hit test.
However if the normal is used just for lighting computations then its usage is usually inside a dot product. So we can use its abs value instead and that will solve all lighting problems regardless of the normal side. For example:
output_color = face_color * abs(dot(face_normal,light_direction))
some gfx apis have implemented this already (look for double sided materials or normals, turning them on usually use the abs value ...) For example in OpenGL:
glLightModeli(GL_LIGHT_MODEL_TWO_SIDE, GL_TRUE);
I want to calculate the intersection over union IoU between two rectangles with axes not aligned, but with an angle of the axes smaller than 30 degrees. An approximate value is also seeked.
One possible solution is to check if the angle between the two rectangles is less than 30 degree and than rotate them parallel to aligne the axis. From here it is easy to calculate the IoU.
Another possibility is to use monte carlo methods for the intersection ( generate a point, find if the point is under some line of one rectangle and above some line of the other), but this seems expensive because I need to use this calculation a large number of times.
I was hopping that there is something better out there; maybe a geometry library, or maybe an algorithm from the computer vision folks.
I am trying to learn grasping positions using deep neural networks. My algorithem should predict a bounding box (rectangle) for an object in an rgb image. For any image I have also the ground truth (another rectangle) bounding box. From this two rectangles I need the IoU.
Any idea?
Since you're working in Python, I think the Shapely package would serve your needs.
There is quite effective algorithm for calculation of intersection between two convex polygons, described in O'Rourke book "Computational Geometry in C".
C code is available at the book page (convconv).
Algorithm traverse edges of the first polygon, checking orientations of the second polygon vertices in order to detect intersections. When two consequent vertices lie on the different sides of the edge, intersection occurs (there is a lot of trick cases). Algorithm outline is here
You can consider a number of numerical approaches, practically "rendering" the rectangles into some "canvas"/canvases, and traverse the pixels for making your statistics. The size of the canvas should be the size of the bounding box for the entire scene, practically you can find that via picking the minimum and maximum coordinates occurring for each axis.
1) "most CG" approach: really get a rendering library, render one rectangle with red, other rectangle with transparent blue. Then visit each pixel and if it has a non-0 red component, it belongs to the first rectangle, if it has a non-0 blue component, it belongs to the second rectangle. And if it has both, it belongs to the intersection too. This approach is cheap for coding, but requires both writing and reading the canvas even in the rendering phase, which is slower than just writing. This might be even done on GPU too, though I am not sure if setup costs and getting back the result do not weight out the benefit for such a simple scene.
2) another CG-approach would be rendering into 2 arrays, preferably some 1-byte-per-pixel variant, for the sake of speed (you may have to go back in time a bit in order to find such dedicated rendering libraries). This way the renderer only writes, into one array per rectangle, and you read from two when creating the statistics
3) as writing and reading pixels take time, you can do some shortcut, but it needs more coding: convex shapes can be rendered via collecting the minimum and maximum coordinates per scanline, and just filling between the two. If you do it yourself, you can spare the filling part and also the read-and-check-every-pixel step at the end. Build such min-max list for both rectangles, and then you "just" have to check their relation/order for each scanline, to recognize overlaps
And then there is the mathematical way: this is not really useful, see EDIT below while it is unlikely that you would find some sane algorithm for calculating intersection area, specifically for the case of rectangles, if you find such algorithm for triangles, which is more probable, that would be enough. Both rectangles can be split into two triangles, 1A+1B and 2A+2B respectively, and then you just have to run such algorithm 4 times: 1A-2A, 1A-2B, 1B-2A, 1B-2B, sum the results and that is the area of your intersection.
EDIT: for the maths approach (though this also comes from graphics), I think https://en.wikipedia.org/wiki/Sutherland%E2%80%93Hodgman_algorithm can be applied here (as both rectangles are convex polygons, A-B and B-A should produce the same result) for finding the intersection polygon, and then the remaining task is to calculate the area of that polygon (here and now I think it is going to be convex, and then it is really easy).
I ended up using Sutherland-Hodgman algorithm implemented as this functions:
def clip(subjectPolygon, clipPolygon):
def inside(p):
return(cp2[0]-cp1[0])*(p[1]-cp1[1]) > (cp2[1]-cp1[1])*(p[0]-cp1[0])
def computeIntersection():
dc = [ cp1[0] - cp2[0], cp1[1] - cp2[1] ]
dp = [ s[0] - e[0], s[1] - e[1] ]
n1 = cp1[0] * cp2[1] - cp1[1] * cp2[0]
n2 = s[0] * e[1] - s[1] * e[0]
n3 = 1.0 / (dc[0] * dp[1] - dc[1] * dp[0])
return [(n1*dp[0] - n2*dc[0]) * n3, (n1*dp[1] - n2*dc[1]) * n3]
outputList = subjectPolygon
cp1 = clipPolygon[-1]
for clipVertex in clipPolygon:
cp2 = clipVertex
inputList = outputList
outputList = []
s = inputList[-1]
for subjectVertex in inputList:
e = subjectVertex
if inside(e):
if not inside(s):
outputList.append(computeIntersection())
outputList.append(e)
elif inside(s):
outputList.append(computeIntersection())
s = e
cp1 = cp2
return(outputList)
def PolygonArea(corners):
n = len(corners) # of corners
area = 0.0
for i in range(n):
j = (i + 1) % n
area += corners[i][0] * corners[j][1]
area -= corners[j][0] * corners[i][1]
area = abs(area) / 2.0
return area
intersection = clip(rec1, rec2)
intersection_area = PolygonArea(intersection)
iou = intersection_area/(PolygonArea(rec1)+PolygonArea(rec2)-intersection_area)
Another slower method (don't know what algorithm) could be:
from shapely.geometry import Polygon
p1 = Polygon(rec1)
p2 = Polygon(rec2)
inter_sec_area = p1.intersection(rec2).area
iou = inter_sec_area/(p1.area + p2.area - inter_sec_area)
It is worth mentioning that in just one case with bigger polygons (not my case) the shapely module had an area twice greater than the first method. I didn't test both methods intensively.
This might help
What about using Pythagorean theorem ? Since you have two rectangles, when they intersect, you will have one or more triangles, each with one angle of 90°.
Since you also know the angle between the two rectangles (20° in my example), and the coordinates of each rectangle, you can use the the appropriate function (cos/sin/tan) to know the length of all the edges of the triangles.
I hope this can help
I've been tasked with writing a python based plugin for a graph drawing program that generates an STL model of a graph. A graph being an object made up of vertices and edges, where a vertex is represented by a 3D ball (a tessellated icosahedron), and an edge is represented with a cylinder that connects with two balls at either end. The end result of the 3D model is that it will get dumped out to an STL file for 3D printing. I'm able to generate the 3D models for the balls and cylinders without any issues, but I'm having some issues generating the overall model, and getting the balls and cylinders to connect properly.
My original idea was to create tessellated icosahedrons at the origin, then translate them out to the positions of the vertices. This works fine. I then, for each edge, I would create a cylinder at the origin, rotate it to the correct angle so that it points in the correct direction, then translate it to the midpoint between the two vertices so that the ends of the cylinders are embedded in the icosahedrons. This is where things are going wrong. I'm having some difficulties getting the rotations correct. To calculate the rotations, I'm doing the following:
First, I find the angle between the two points as follows (where source and target are both vertices in the graph, belonging to the edge that I'm currently processing):
deltaX = source.x - target.x
deltaY = source.y - target.y
deltaZ = source.z - target.z
xyAngle = math.atan2(deltaX, deltaY)
xzAngle = math.atan2(deltaX, deltaZ)
yzAngle = math.atan2(deltaY, deltaZ)
The angles being calculated seem reasonable, and as far as I can tell, do actually represent the angle between the vertices. For example, if I have a vertex at (1, 1, 0) and another vertex at (3, 3, 0), the angle edge connecting them does show up as a 45 degree angle between the two vertices. (That, or -135 degrees, depending which vertex is the source and which is the target).
Once I have the angles calculated, I create a cylinder and rotate it by the angles that have been calculated, like so, using some other classes that I've created:
c = cylinder()
c.createCylinder(edgeThickness, edgeLength)
c.rotateX(-yzAngle)
c.rotateY(xzAngle)
c.rotateZ(-xyAngle)
c.translate(edgePosition.x, edgePosition.y, edgePosition.z)
(Where edgePosition is the midpoint between the two vertices in the graph, edgeThickness is the radius of the cylinder being created, and edgeLength is the distance between the two vertices).
As mentioned, its the rotating of the cylinders that doesn't work as expected. It seems to do the correct rotation on the x/y plane, but as soon as an edge has vertices that differ in all three components (x, y, and z), the rotation fails. Here's an example of a graph that differs in the x, and y components, but not in the z component:
And here's the resulting STL file, as seen in Makerware (which is used to send the 3D models to the 3D printer):
(The extra cylinder looking bit in the bottom left is something I've currently left in for testing purposes - a cylinder that points in the direction of the z axis, located at the origin).
If I take that same graph and move the middle vertex out in the z axis, so now all the edges involve angles in all three axis, I get a result something like the following:
As show in the app:
The resulting STL file, as show in Makerware:
...and that same model as viewed from the side:
As you can see, the cylinders definitely aren't meeting up with the balls like I thought they would. My question is this: Is my approach to doing this flawed, or is it some small but critical mistake that I'm making somewhere in my rotations? I'm pretty sure it isn't a problem with the rotation functions themselves, as I've been able to independently verify that they work as expected. I also tried creating a rotate function that takes in a yaw, pitch, and roll and does all three at once, and it seemed to generate the same result, like so:
c.rotateYawPitchRoll(xzAngle, -yzAngle, -xyAngle)
So... anyone have any ideas on what I might be doing wrong?
UPDATE: As joojaa pointed out, it was a combination of calculating the correct angles as well as the order that they were applied. In order to get things working, I first calculate the rotation on the x axis, as follows:
zyAngle = math.atan2(deltaVector.z, deltaVector.y)
where deltaVector is the difference between the target and source vectors. This rotation is not yet applied though! The next step is to calculate the rotation on the y axis, as follows:
angle = vector.angleBetweenVectors(vector(target.x - source.x, target.y - source.y, target.z - source.z), vector(target.x - source.x, target.y - source.y, 0.0))
Once both rotations are calculated, they are then applied... in the reverse order! First, the x, then the y:
c.rotateY(angle)
c.rotateX(-zyAngle) #... where c is a cylinder object
There still seems to be a few bugs, but this seems to at least work for a simple test case.
Rotation happens in successive order, so the angles affect each other. It is not possible to use a Euler model to rotate them at once. This is why you can not just calculate the rotations based on the first static situation. Just imagine turning a cube so that it is standing on its corner upright. Yes the first rotation is 45 but the second is not since the cube is already turned by that time (draw a each step of the sequence and see what happens). Space rotations aren't trivial.
So you need to rotate one angle then re calculate the second angle and so forth. This is also why your first rotation works right. You only need 2 rotations unless your interested in making sure the rotation around the shaft has a certain direction.
I would suggest you use axis angles or matrices instead to do this. Mainly because in axis angles this is trivial the angle is the dot between the along tube start and end vectors and the axis is the cross between those 2. You can then convert those to Euler angles if you need. But probably you can just use the matrix directly. For ideas on how conversions and how the rotation could directly be calculated see: transformations.py by Christoph Gohlke. Also see the accompanying c source.
I think i need to expand this answer a bit
There is a really easy way out for this question that sidesteps all your and many other persons problems. The answer is do not use Euler angle rotation. Ive used a lot of brainpower to try to explain Euler rotations to problems that are ultimately solved more easily without Euler rotations. To justify i will leave just one reason for this if you want more think up of some more answers.
The reason most to use Euler rotation sequences is that you probably don't understand Euler angles. There are in fact only a handful of situations where they are good. No self respecting programmer uses Euler rotations to solve this issue. What you do is you use vector math instead.
So you have the direction vector from the source to target which is usually calculated:
along = normalize(target-source)
this is simply one of your matrix rows (or column notation is up to model maker), the one that corresponds to your cylinders original direction (the rows are just x y z w), then you need another vector perpendicular to this one. Choose a arbitrary vector like up (or left if your along is pointing close to up). cross product this up vector by your along for the second row direction. and finally put your source as the last row with 1 in the last column. Done fully formed affine matrix describing the cylinders prition. Much easier to understand since you can draw the vectors.
There are shorter ways but this one is easy to understand.
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.